Nu
cle
ar
Ph
ys
ics
Laboratory Manual for Nuclear Physics
La
b
M
an
ua
l
Shobhit Mahajan
shobhit.mahajan@gmail.com
Version 3.0
Last modified on January 16, 2021
Shobhit Mahajan
Lab Manual for Nuclear Physics
Contents
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81
2 RADIOACTIVITY
2.1 Radioactivity . . . . . . . . . . .
2.1.1 Measure of radioactivity .
2.1.2 Activity Law & Half Life
2.2 Nuclear Decay . . . . . . . . . .
2.2.1 Alpha Decay . . . . . . .
2.2.2 Beta Decay . . . . . . . .
2.2.3 Gamma Decay . . . . . .
2.3 References . . . . . . . . . . . . .
2.4 Questions . . . . . . . . . . . . .
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98
98
99
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117
La
b
M
an
ua
l
Nu
cle
ar
Ph
ys
ics
1 STATISTICS & ERROR ANALYSIS
1.1 Probability . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Random Variables . . . . . . . . . . . . . .
1.2 Uncertainty in Measurement . . . . . . . . . . . . .
1.2.1 Uncertainity, Accuracy & Precision . . . . .
1.2.2 Systematic & Random Errors . . . . . . . .
1.2.3 Significant Digits . . . . . . . . . . . . . . .
1.2.4 Reporting of Uncertainties & Rounding Off
1.2.5 Permutations & Combinations . . . . . . .
1.3 Statistical Analysis of Data . . . . . . . . . . . . .
1.3.1 Histograms & Distribution . . . . . . . . .
1.3.2 Parent & Sample Distribution . . . . . . . .
1.3.3 Mean & Deviation . . . . . . . . . . . . . .
1.4 Distributions . . . . . . . . . . . . . . . . . . . . .
1.4.1 Binomial Distribution . . . . . . . . . . . .
1.4.2 Poisson Distribution . . . . . . . . . . . . .
1.4.3 Normal Distribution . . . . . . . . . . . . .
1.5 Error Estimation . . . . . . . . . . . . . . . . . . .
1.5.1 Propagation of Errors . . . . . . . . . . . .
1.6 Estimation and Error of the Mean . . . . . . . . .
1.6.1 Method of Maximum Likelihood . . . . . .
1.6.2 Estimated Error in the Mean . . . . . . . .
1.7 Method of Least Squares . . . . . . . . . . . . . . .
1.8 Goodness of Fit . . . . . . . . . . . . . . . . . . . .
1.9 References . . . . . . . . . . . . . . . . . . . . . . .
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3 INTERACTION WITH MATTER
3.1 Introduction . . . . . . . . . . . . . . . . . . . . .
3.1.1 Cross Section . . . . . . . . . . . . . . . .
3.2 Interaction of Charged Particles with Matter . .
3.2.1 Interaction of Heavy charged particle with
3.2.2 Interaction with matter of electrons . . .
3.2.3 Interaction of gamma rays with matter . .
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1
Shobhit Mahajan
3.3
3.4
Lab Manual for Nuclear Physics
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
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153
153
5 Experiment: GM Characteristics
5.1 Introduction . . . . . . . . . . . . .
5.2 Precautions . . . . . . . . . . . . .
5.2.1 Health Effects of Radiation
5.3 Experiment . . . . . . . . . . . . .
5.3.1 Purpose . . . . . . . . . . .
5.3.2 Method . . . . . . . . . . .
5.3.3 Sample Data . . . . . . . .
5.4 Questions . . . . . . . . . . . . . .
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Nu
cle
ar
Ph
ys
ics
4 G-M COUNTER
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Detector Models . . . . . . . . . . . . . . . . . . . . .
4.3 Ionisation of Gases . . . . . . . . . . . . . . . . . . . .
4.3.1 Townsend Avalanche . . . . . . . . . . . . . . .
4.3.2 Kinds of Detectors & Detector Regions . . . .
4.4 GM Counter . . . . . . . . . . . . . . . . . . . . . . .
4.4.1 Geiger Discharge . . . . . . . . . . . . . . . . .
4.4.2 Quenching . . . . . . . . . . . . . . . . . . . . .
4.4.3 Dead Time & Recovery Time . . . . . . . . . .
4.4.4 Geiger Counting Plateau & Operating Voltage
4.4.5 Counting Efficiency . . . . . . . . . . . . . . .
4.5 References . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Questions . . . . . . . . . . . . . . . . . . . . . . . . .
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175
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191
7 Experiment: Absorption
7.1 Introduction . . . . . .
7.2 Experiment . . . . . .
7.2.1 Purpose . . . .
7.2.2 Method . . . .
7.2.3 Sample Data .
7.3 Questions . . . . . . .
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the Inverse Square Law for
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198
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La
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M
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6 Experiment: GM Counter: Counting
6.1 Introduction . . . . . . . . . . . . . .
6.2 Experiment . . . . . . . . . . . . . .
6.2.1 Purpose . . . . . . . . . . . .
6.2.2 Method . . . . . . . . . . . .
6.2.3 Sample Data . . . . . . . . .
6.2.4 Error Estimation . . . . . . .
6.3 Questions . . . . . . . . . . . . . . .
of γ
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8 Experiment: Verification of
8.1 Introduction . . . . . . . .
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8.2.1 Purpose . . . . . .
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8.2.3 Sample Data . . .
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9 Experiment: To Determine the Range of β rays in Aluminum and to determine the End Point
Energy
207
9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
2
Shobhit Mahajan
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208
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215
10 Experiment: Gamma Ray spectrum using a Scintillation Counter
10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2.1 Inorganic Scintillators . . . . . . . . . . . . . . . . . . . . . . .
10.2.2 Organic Scintillators . . . . . . . . . . . . . . . . . . . . . . . .
10.2.3 Photomultiplier Tube . . . . . . . . . . . . . . . . . . . . . . .
10.2.4 Gamma Ray Spectrum . . . . . . . . . . . . . . . . . . . . . . .
10.3 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.4 Sample Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.5 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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217
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257
9.3
Experiment . . . . .
9.2.1 Purpose . . .
9.2.2 Method . . .
9.2.3 Sample Data
Questions . . . . . .
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A p-Value Tables
B Using MS Excel
B.1 Simple calculations . . . .
B.2 Plotting Data . . . . . . .
B.2.1 Error Bars . . . .
B.2.2 Formatting Graphs
B.3 Fitting Data . . . . . . . .
B.4 Statistical Analysis . . . .
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262
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275
Using Gnuplot
C.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . .
C.2 Plotting with inbuilt functions of GNUPLOT . . . . . . .
C.2.1 Interactive plotting . . . . . . . . . . . . . . . . .
C.3 Saving Plots . . . . . . . . . . . . . . . . . . . . . . . . . .
C.3.1 Customization . . . . . . . . . . . . . . . . . . . .
C.4 Plotting using data from a file . . . . . . . . . . . . . . . .
C.5 Plotting using data from file and fitting to a smooth curve
C.5.1 Curve Fitting & Interpolation . . . . . . . . . . . .
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283
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296
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C
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Nu
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ar
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9.2
Lab Manual for Nuclear Physics
La
b
D Radioactive Decay Equilibrium
D.1 Introduction . . . . . . . . . . . . . . . . . . .
D.2 Bateman Equation . . . . . . . . . . . . . . .
D.3 Different Kinds of Equilibrium . . . . . . . .
D.4 Numerical Integration of Bateman Equations
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E Theory of Alpha Decay
306
E.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306
E.2 1-d Tunnel Effect For Rectangular Barrier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
E.3 Tunnel Effect with Nuclear Potential Barrier: Geiger-Nuttall Law . . . . . . . . . . . . . . . . . . . . . . . 310
F
Fermi’s Theory of Beta Decay
F.1 Fermi’s Golden Rule . . . . . .
F.2 Fermi’s Theory of Beta Decay .
F.2.1 Matrix Element . . . . .
F.2.2 Density of Final States .
F.2.3 Decay Rate . . . . . . .
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314
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G Semi-Classical Theory of Gamma Decay
326
G.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326
3
Shobhit Mahajan
Lab Manual for Nuclear Physics
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La
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Nu
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G.2 Density of Final States . . . .
G.3 Interaction Hamiltonian . . .
G.3.1 Dipole Approximation
G.4 Transition Rate & Lifetime .
4
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326
327
327
329
Shobhit Mahajan
Lab Manual for Nuclear Physics
A Note to the Reader
This Manual is intended for use in the Nuclear Physics Laboratory of the M.Sc (Previous) class.
Nu
cle
ar
Ph
ys
ics
The Manual is organised into 10 Chapters. The first 4 Chapters provide the theoretical background
for the experiments which are performed in the laboratory. Chapter 1 is a fairly detailed introduction
to the Statistical Tools which are required to analyse the experiments. It includes topics like distributions, Error Analysis and Goodness of Fit etc. Most of these topics are familiar to you from your
undergraduate days. However, they are presented in enough detail here and they do not assume any
prior knowledge of statistics.
M
an
ua
l
Chapter 2 and 3 are a review of nuclear physics concepts which are required for the experiments.
These include radioactive radiation (alpha, beta and gamma rays) as well as their interaction with matter. This material, once again, should be familiar to you and it is presented here for review. However,
again, the material is complete and does not assume any prior knowledge. Most of the experiments
carried out in the laboratory use a Geiger Muller counter. This is discussed in detail in Chapter 4.
Although the first 4 chapters provide you with enough information to be able to do the experiments,
they are Not meant to be a substitute for the books which discuss each of these topics in detail. There
are many excellent books available on these topics and you are encouraged to go through them.
b
Chapters 5 − 10 are detailed discussions of the experiments which are available in the laboratory.
For each experiment, the procedure is discussed and sample data is given. This sample data is then
analysed and the errors and results are obtained. It is important for you to understand well the calculations given here so that you can do the same with the data that you obtain in the laboratory yourself.
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Each Chapter has some questions in the end which you are encouraged to attempt to answer to test
your understanding of the theory and the experiments.
There are 7 Appendices in the Manual. There is an appendix which discusses the use of Microsoft
Excel program to do data analysis and plot graphs etc. There is also an appendix which discusses the
use of GNUPLOT to plot graphs in case you want to use a Linux platform. In addition, the detailed
theories for radioactive equilibrium and alpha, beta and gamma decay are also given in the Appendices.
The computers in the Laboratory and possibly the one you have at home are using the Windows
operating system which do not have a native C compiler or GNUPLOT. However, it is easy to install
a Linux emulator on your machine.
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Go to https://www.cygwin.com/ and download and install the program Cygwin. This will come
in two versions- Cygwin32 and Cygwin64. If you have a 64-bit computer then install Cygwin64, else
install Cygwin32. Once installed, run the program as you run any windows program, that is by double
clicking on the icon. This will open a small window. In this type startx. This will open a terminal
on your screen, exactly like the one you see in the laboratory on the Linux machines. It is as if, your
Windows machine has turned into a Linux machine. You can run all the Linux OS programs like Gnuplot, gcc, emacs etc. in this terminal. Once you are finished, just type exit and you will return to
your Windows environment.
We would very much like to get your suggestions regarding how to improve this Manual. In addition,
if there are any errors or misprints that are spotted in the Manual, we would like to hear from you.
Please send a mail with the suggestions/errors etc. to shobhit.mahajan@gmail.com making sure
you quote the version number of the Manual as well as the Modification date of the Manual you are
using. The version number and date are on the title page of the Manual.
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January 16, 2021
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Finally, it is important to realise that this manual can only help you with your experiments. Ultimately, it is essential that you perform the experiments yourself and then do the data analysis with the
help of the tools discussed in the Manual. Unless you do the actual experiment and the calculations
yourself, you will never learn the subject.
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Radiation Safety Instructions.
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General Precautions
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1. Handle the radioactive sources with utmost care and respect. Don’t bend or try to break them.
2. As far as possible, always handle the sources with gloves or a forceps, especially for sources
with more than 15µCi strength.
3. Sources should be kept as far from the body as possible.
4. Although the sources in the laboratory are always in their holders, it is in general important
to never touch the source using bare hands. Always use forceps to handle sources.
5. Do not eat or drink during the lab. Please do not keep any edible material or even drinking
water on your work table. Keep your bags with the food and water on the table on the side.
6. When bringing or returning the source to the source room, please be extremely careful to not
let it fall.
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7. Do not leave the source lying around the work table. Always keep it carefully while using it
and return it promptly after you are finished.
8. Do not keep the source in your pocket or in close contact with your body.
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9. Wash your hands after the experiment and before eating or drinking anything.
Units used for measuring radiation
Quantity
Activity
Definition
becquerel: 1 decay per second
curie=3.7 × 101 0 Bq
Exposure
roentgen= 1 esu of charge produced in 1 cc of air at STP
Absorbed Dose
Gray=1 J/kg
Equivalent dose Siervert=Absorbed dose × weight factor (weight factor = 1 for β and 20 for α)
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Symbol
Bq
Ci
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Gy
Sv
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Natural Sources of Radiation
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We are continuously exposed to radioactive radiation from natural sources. The natural sources of
radiation are mostly natural Uranium, 238U, 232Th and their decay products. Apart from these, there
is 40K, 14C etc. A person weighing around 70 kg, contains about 4500 Bq of 40K which is a gamma
and beta emitter. Thus we are getting a self dose of about 0.16 mSv y−1 . We normally take in about
100 Bq per day of 40K from milk, banana and green leafy vegetables. This maintains the amount of
40
K in the body. the normal ambient background dose rate is 0.1µ Gy h−1 . In addition, when we get
X-rays or other scans done, we get radiation. For instance, a chest X-ray exposes us to about 0.06 mSv
while a abdomen CT scan exposes us to 10 mSv.
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Dose Limits
The Atomic Energy Regulatory Board sets some dose limits for exposure.
Whole Body Exposure (Effective Dose)
Occupational Workers
Apprentices
100 mSv in 5 years with an average of 6 mSv in a year
20 mSv & a maximum of 30 mSv
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Type of Limit
Annual Dose Limit
General Public
1 mSv in a year
For parts of the body (Equivalent Dose)
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Occupational Workers
Lens of the Eye
150 mSv
Skin
500 mSv
Hands & Feet
500 mSv
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Apprentices General Public
50 mSv
15 mSv
150 mSv
50 mSv
150 mSv
50 mSv
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Lab Manual for Nuclear Physics
Chapter 1
1.1
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STATISTICS & ERROR ANALYSIS
Probability
Since we are going to be dealing with probability in the subsequent sections, let us give a brief background. The probability of an event refers to the likelihood that the event will occur. Mathematically,
the probability that an event will occur is expressed as a number between 0 and 1. The sum of probabilities in any statistical experiment is always 1, a statement of the fact that something will certainly
happen. Let us illustrate how one can calculate probabilities.
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Consider first the case of an experiment with n possible outcomes which are each equally likely. Now
if we take a subset r of these and call them successes then clearly the probability of success in the
experiment is nr . Thus, if there are 10 balls, 7 white and 3 black in a bag, then the probability of getting
3
.
a black ball if one is picked out at random from the bag is 10
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There is another approach to probability where one talks about relative frequencies. Suppose I count
the number of cars passing a particular point on a road at a particular interval of time and notice how
many of them are white. Suppose on the first day, I see 5 white cars out of a total of 20 cars, while
on the second day I count 9 white cars out of 30 while on the third day I find 3 white cars out of 5
and so on. Clearly, the relative frequency of white cars is different on different days. However, one
could find for instance that if I repeat this experiment many many times, then the relative frequency is
0.26. Then the Law of Large Numbers says that the relative frequency of an event will converge on
the probability of that event as the number of trials increases. This is what we anyway feel using our
common sense. If we have, for instance, a fair coin, then we know that the probability of getting a head
or a tail is 12 . However, when we start tossing the coin, it might happen that we get a series of heads
or tails which might lead one to believe that the probabilities of these events (that is getting a head or
a tail) are not equal. But we do know intuitively that if we perform a large number of coin tosses, then
the number of heads (or tails) is going to be 12 of the total number of tosses. This is precisely what the
Law of Large Numbers says- the average of the results of a large number of trials will be close
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to the expected value and the two will converge as the number of trials increases.
Some definitions in probability theory are useful:
1. Two events are mutually exclusive or disjoint if they cannot occur at the same time.
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2. The probability that Event A occurs, given that Event B has occurred, is called a conditional
probability. The conditional probability of Event A, given Event B, is denoted by the symbol
P (A|B).
3. The complement of an event is the event not occurring. The probability that Event A will not
occur is denoted by P (A0 ).
4. The probability that Events A and B both occur is the probability of the intersection of A and B.
T
The probability of the intersection of Events A and B is denoted by P (A B). If Events A and B
T
are mutually exclusive, P (A B) = 0.
5. The probability that Events A or B occur is the probability of the union of A and B. The probability
S
of the union of Events A and B is denoted by P (A B) .
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6. If the occurrence of Event A changes the probability of Event B, then Events A and B are dependent. On the other hand, if the occurrence of Event A does not change the probability of Event
B, then Events A and B are independent.
These definitions allow us to write down the rules for probability.
P (A) = 1 − P (A0 )
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Subtraction: The probability that event A will occur is equal to 1 minus the probability that event
A will not occur.
Multiplication: The probability that Events A and B both occur is equal to the probability that
Event A occurs times the probability that Event B occurs, given that A has occurred.
P (A
\
B) = P (A)P (B|A)
Addition: The probability that Event A or Event B occurs is equal to the probability that Event
A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur.
P (A
[
B) = P (A) + P (B) − P (A
\
B)) = P (A) + P (B) − P (A)P (B|A)
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These rules are fairly obvious intuitively and the easiest way to prove them is to use Venn diagrams
where the results quoted above are immediately clear.
1.1.1
Random Variables
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Discrete Random Variable
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When the value of a variable is determined by a chance event, that variable is called a random variable. Random variables can be discrete or continuous.
Within a range of numbers, discrete variables can take on only certain values. Suppose, for example,
that we flip a coin and count the number of heads. The number of heads will be a value between 0
and +∞. Within that range, though, the number of heads can be only certain values. For example,
the number of heads can only be a whole number, not a fraction. Therefore, the number of heads is a
discrete variable. And because the number of heads results from a random process - flipping a coin - it
is a discrete random variable.
Continuous Random Variable
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1.2
1.2.1
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Continuous variables, in contrast, can take on any value within a range of values. For example,
suppose we randomly select an individual from a population. Then, we measure the age of that person.
In theory, his/her age can take on any value between 0 and +∞, so age is a continuous variable. In
this example, the age of the person selected is determined by a chance event; so, in this example, age
is a continuous random variable.
Note that discrete variables can be finite or infinite. Thus, for instance the number of heads in coin
flips can be infinite while the number of aces that I can choose from a deck of cards is finite (0, 1, 2, 3, 4).
Continuous variables can always take an infinite number of values while some discrete variables can take
infinite number of values.
Uncertainty in Measurement
Uncertainity, Accuracy & Precision
All measurements that we do have some degree of uncertainty. This uncertainty might come from a
variety of sources about which we will talk later. But the fact that needs to be emphasised is that all
measurements have some uncertainty and an analysis of this uncertainty is what we call error analysis.
Any measured value that we quote must be accompanied by our estimate of the level of certainty or confidence associated with that measurement. This fact is absolutely essential since without this the basic
question of science, namely “does the result of our experiment agree with the theory” can not be an-
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swered. To decide whether the proposed theory is valid or not, this question would need to be answered.
When we carry out an experiment to measure a quantity, we of course assume that some ‘true’ or
exact value exists. We of course may or may not know this value but we always attempt to find the
best value possible given the limitations of our own experimental setup. Typically, every time we carry
out the experiment, we will find a different value and so the question is how do we report our best
estimate of this ‘true’ value? Usually, this is done as
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measurement = best estimate ± uncertainty
For example, let us assume you want to find the weight of your mobile phone. By simply putting
in your hand, you can estimate it to be between 100 and 200 grams. But that is not good enough. So
you go to a balance in the laboratory and it gives you a reading of 145.55 grams. This value is much
more precise than the original estimate you obtained, but how does one know that it is accurate?
One way is to repeat your measurement several times and suppose you get the values 145.59, 145.53
and 145.51 grams. Then one could say that the weight of the phone is 145.55 ± .04 grams. But
now suppose you go to another balance and find a value of 144.15 grams? Now one is faced with a
problem since your original best estimate is very different from this measurement. So what does one do?
To understand this, we need to understand first the difference between precision and accuracy .
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Precision & Accuracy
• Accuracy is how close the measured value is to the true or the accepted value of
a quantity.
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• Precision is a measure of how well the result can be measured, irrespective of the
true value. It reflects the degree of consistency and agreement between repeated,
independent measurements of the same quantity as well as the reproducibility of
the results.
• Any statement of uncertainty associated with a measurement must include factors
which affect both accuracy and precision. After all, it is a waste of time if we
determine a result which is very precise but highly inaccurate or its converse, that
is, a result which is very accurate but highly imprecise.
In our example above, we have no way of knowing whether our result is accurate or not unless we
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compare it with a known standard. For instance, we could use a standard weight to determine if the
balances used in our measurement are accurate or not. Suppose one performs two experiments and get
two values of g, the acceleration due to gravity. One experiment gives g = 9.7 m s−2 while the other
other gives g = 9.6345 m s−2 . Clearly the first result is more accurate while the second one is more
precise.
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Precision is often reported experimentally as relative uncertainty defined as
Precision: Relative Uncertainty =
uncertainty
measured value
(1.1)
Thus in our example of the weight of the mobile phone, the relative uncertainty is
Relative Uncertainty =
0.04
= 0.00027 = .027%
145.55
Accuracy on the other hand is reported usually as relative error which is defined as
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Accuracy:
Relative Error =
measured value - expected value
expected value
(1.2)
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In our example above, if we think that the expected value of the weight of the mobile phone is 145.50
grams, then the relative error is
145.55 − 145.50
0.05
=
= 0.034%
145.50
145.50
Thus we see that any measurement needs to be both precise and accurate for it to be good. The
idea of making good measurements is directly related to errors in measurement. Errors in measurement
can be broadly classified into two categories- random and systematic.
1.2.2
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Relative Error =
Systematic & Random Errors
Systematic Errors are errors which will make the results obtained by us differ from the “true” value
of the quantity under consideration. They are reproducible inaccuracies which are difficult to detect and
also cannot be analysed statistically. An important thing to realise is that systematic errors cannot
be reduced by repeated measurements.
Random Errors are errors which are fluctuations in the observations which are statistical in nature.
They can be analysed statistically and furthermore, they can be reduced by repeated measurements
and taking averages as we shall see later.
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To illustrate this distinction think of the following experiment. Suppose I wish to find the time
period of a pendulum by timing some number of oscillations with the help of a stop watch. There could
be several sources of error- One source of error will be my reaction time, that is the time between my
seeing the pendulum bob reaching the extreme position and my starting the watch and again at a later
point my observing the bob and my stopping the watch. Obviously, if my reaction time was always
exactly the same, then the time delay wouldn’t matter since they would cancel. However, we know
that in practice, my reaction time will be different. I may delay more in starting, and so underestimate
the time of a revolution; or I may delay more in stopping, and so overestimate the time. Since either
possibility is equally likely, the sign of the effect is random. If I repeat the measurement several times,
I will sometimes overestimate and sometimes underestimate. Thus, my variable reaction time will show
up as a variation of the answers found. By analyzing the spread in results statistically, I can get a very
reliable estimate of this kind of error.
Now suppose that the watch that I use is slow or fast. Then, no matter how many times I repeat the experiment (of course with the same watch), I can never know the amount of such an error.
Further, the error’s sign will always be the same- either the watch will be fast or slow leading to either an overestimate or underestimate of the rate of revolution. This is an example of a systematic error.
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Systematic & Random Errors
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In general, there is no set prescription for eliminating systematic errors and mostly one has to use
common sense to know if there are any systematic errors and to eliminate them. Random errors on the
other hand are usually easier to study and eliminate or reduce. But one should remember that in many
situations, the accuracy of a measurement is dominated by possible systematic errors in the instrument
rather than the precision with which we can make the measurement.
To summarise
Systematic Errors are reproducible inaccuracies that are difficult to detect and cannot
be analyzed statistically. If a systematic error is identified when calibrating against a
standard, applying a correction or correction factor to compensate for the effect can
reduce the bias.
Random Errors are statistical fluctuations in the measured data due to the precision
limitations of the measurement device. Random errors can be evaluated through
statistical analysis and can be reduced by averaging over a large number of observations
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The fundamental aim of an experimentalist is to reduce as many sources of error as s/he can, and
then to keep track of those errors that can’t be eliminated.
1.2.3
Significant Digits
Whenever one writes the results of an experiment, the precision in the experiment is normally indicated
by the number of digits which one reports the result with. The number of significant figures depends
on how precise the given data is. The following rules are helpful:
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Significant Figures
1. The leftmost NONZERO digit is ALWAYS the MOST significant digit.
2. When there is NO decimal point, the rightmost NONZERO digit is the least
significant digit.
3. In case of a decimal point, the rightmost digit is the least significant digit EVEN
IF IT IS A ZERO.
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4. The number of digits between the most and least significant digits are the number
of significant digits.
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Thus for instance, 22.00, 2234, 22340000, 2200. all have four significant digits while 2200 has only
two significant digits because of the absence of the decimal point. When one is adding, subtracting,
multiplying or dividing numbers, then the result should be quoted with the least number of significant
figures in any one of the quantities being used in the operation of adding, multiplying etc. In your intermediate calculations, always keep ONE MORE significant digit than is needed in the final answer.
Also when quoting an experimental result, the number of significant figures should be one more than is
suggested by the experimental precision.
In reporting measurements of experiments, one has to exercise special care. Thus for instance, if
one sees a result as 8120, we are not sure whether the result actually has 3 or 4 significant digits. By
the rules above, the 0 in the end is just a place value. However, it might actually transpire that the
experiment actually did measure the last digit to be 0. To get around this problem, one always uses
the scientific notation. Thus if the expreimental result is only precise to three significant figures, one
would report it as
8.12 × 103
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or if it was precise to 4 significant digits, we would report it as
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8.120 × 103
Two things that need to be always avoided are
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• Writing more digits in an answer (intermediate or final) than justified by the
number of digits in the data.
• Rounding-off, say, to two digits in an intermediate answer, and then writing three
digits in the final answer.
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While dropping off some figures from a number, the last digit that one keeps should be rounded
off for better accuracy. This is usually done by truncating the number as desired and then treating
the extra digits (which are to be dropped) as decimal fractions. Then, if the fraction is greater than
1
, increment the truncated least significant figure by one. If the fraction is smaller than 21 , then do
2
nothing. If the fraction is exactly 12 , then increment the least significant digit only if it is odd.
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What about reporting results which arise from addition or multiplication of two measurements with
different significant figures. For instance, consider a measurement of two masses of 9.9 gm and 0.3163
gm. How does one report the total mass? To illustrate this, let us write the numbers as 9.9 ∗ ∗ ∗ ∗ gm
and 0.3163∗ gm. Then if we add them up we see that result we get is 10.2 ∗ ∗ ∗ ∗ gm that is to say that
we are only sure of the first decimal place number and so we should report this as 10.2 gm. Similarly,
if we had two numbers which we need to multiply, say 3.413∗ and 2.3∗, the product will be 7.8 ∗ ∗ ∗ ∗∗
which will be reported as 7.8.
When multiplying two quantities with different significant
digits, the product has the same number of significant digits as that of the number with
the least number of significant digits.
1.2.4
Reporting of Uncertainties & Rounding Off
We are now in a position to give some basic rules and conventions of how experimental results should
be reported. We have already seen that every measurement will have an uncertainty or error associated
with it. This error could be random or systematic. This uncertainty indicates to us the precision or
reliability of the measurement. However, it is obvious that the magnitude of uncertainty by itself does
not give us the complete picture. An uncertainty of .1 seconds for instance, would be a very precise
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measurement if the quantity being measured is 100 seconds say but not very precise if we were measuring 1 second. Thus a more useful concept is that of fractional uncertainty or relative uncertainty that
we saw in Eq(1.1). The percentage uncertainty then is a better indicator of the precision involved in
the particular experiment.
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A more serious question is that of reporting uncertainties. Although there is no definite theoretical
reason to expect that uncertainties should be reported to a certain number of significant figures, the
general consensus is that uncertainties in measurements should be reported to 1 significant
figure. Note that this is NOT something that one can prove or disprove. It is more of a convention
which is followed. Of course, if the leading significant figure in the uncertainty that we are reporting
has the value 1 or 2 then by keeping only one significant digit, we might be making a mistake. In these
cases, it is best to report it to two significant digits.
Once we know how the uncertainty in the measured quantity is to be reported, then this also tells us
how we need to report our answers. For instance, if the uncertainty is 0.4 meters in the measurement
of some length L , then it would be meaningless to report the result as
L = 100.4135 ± 0.4 m
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Clearly, the number of digits we retain in the result is governed by the uncertainty in the measurement. In this example,we should report it as
L = 100.4 ± 0.4 m
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Next suppose the quantity we are measuring has a value x = 12.0349 cm with an uncertainty of
δx = 0.153 cm. How does one report this result? The most significant digit in the uncertainty is 1 and
hence we should keep two significant digits. Thus δx = 0.15 cm and the result should be
x = (12.03 ± 0.15) cm
Next suppose we have a measurement as z = 1.43 × 106 seconds with an uncertainty of δz = 2 × 104
seconds. The correct way to report this is
z = (1.43 ± 0.02) × 106 s
Similarly, the mass of the electron is measured as
m = 9.11 × 10−31 kg
with
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δm = 2.2345 × 10−31 kg
By our rule above, we see that we should report the uncertainty as ±2 × 10−31 . This means that the
measured quantity should also retain only one significant digit, that is 9. Thus we should report this as
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m = 9 ± 2 × 10−31 kg
1.2.5
Permutations & Combinations
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Please keep this in mind when you report your results. Typically, since your calculators give you many more digits for any calculation, please use these rules to determine
how you should report the results.
To complete the review, let us define permutations and combinations.
Suppose we have a set A of elements (a1 , a2 , a3 ). Then a permutation of A is a particular ordering of
its elements. Thus in our example there are 6 permutations.
(a1 , a2 , a3 ), (a1 , a3 , a2 ), (a2 , a1 , a3 ), (a2 , a3 , a1 ), (a3 , a1 , a2 ), (a3 , a2 , a1 )
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.
Note that in permutations ordering is very important. It should be obvious that for a set of
N objects, there are N ! permutations since for the first place we have N choices, second place (N − 1)
choices after filling the first place, (N − 2) choices for the third place after filling the first two places
etc.
For combinations, ordering is not important. One way to think about it is that permutations
are lists in which ordering is important while combinations are set of objects. Thus in our previous
example, we can ask the question as to how many combinations exist of two elements of the set A?
This is simply (a1 , a2 ), (a1 , a3 ), (a2 , a3 ). Note that each of these is a subset of A. In fact, the number of
combinations can be obtained by listing all the subsets of the set with exactly the number of elements
required for the combination.
From a set of n objects the number of permutations of r distinct elements is written as
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n
Pr =
n!
(n − r)!
(1.3)
From a set of n objects, the number of combinations of r elements from a set of n elements or the
number of subsets of r elements from a set of n elements is given by
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R
n
Cr =
n
Pr
n!
=
(n − r)!r!
r!
(1.4)
It should be obvious that the number of permutations is always larger than the number of combinations.
1.3.1
Statistical Analysis of Data
Histograms & Distribution
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1.3
The fundamental problem in reporting the results of an experiment is to estimate the uncertainty in a
measurement. It is reasonable to think that the reliability of an estimate of uncertainty in a measurement can be improved if the measurement is repeated many times. The first problem in reporting the
results of many repeated measurements is to find a concise way to record and display the values obtained.
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Suppose we measured the weights of all the new one rupee coins minted since they were introduced.
It is clear that all the coins will not have the same weight. The actual weight would depend on several
things including when it was minted and how much it has been in use etc. One way to display the
results of our measurement would be a histogram as shown in the Figure 1.1. This is for a sample of
100 coins and we have divided the weights into bins of width ∆ = 0.01 gm.
Figure 1.1: Binned histogram
We have plotted the data in such a way that the fraction of measurements that fall in each bin is
given by the area of the rectangle above the bin. That is to say that the height P (k) of the k th bin is
such that
Area = P (k) × ∆ = fraction of measurements in the k th bin
Thus, for instance, the area of the rectangle between 2.50 − 2.51 is 20 × 0.01 = .2. This means that
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20% of the coins fall in this weight range.
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We can see that such a plot gives us a good way to represent the data namely how the weights of the
coins in our sample are distributed. In most experiments, as the number of measurements increases, the
histogram begins to take on a definite simple shape, and as the number of measurements approaches
infinity, their distribution approaches some definite, continuous curve, the so-called limiting distribution as in Fig(1.2).
Figure 1.2: Limiting Distribution
An important concept that we will need to understand is that of a probability distribution.
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A probability distribution is a table or an equation that links each outcome of a
statistical experiment with its probability of occurrence.
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Recall that when the value of the variable is an outcome of a statistical experiment, then it is a
random variable. Thus, for instance we can think of a statistical experiment of tossing a coin twice.
We can get 4 possible outcomes- HH,HT,TH and TT. Now let the variable X represent the number
of heads in this experiment. Then X is a random variable and it can take 3 values, 0, 1, 2. We can
construct a table for the values x of the random variable X and the probability associated with that
value.
x
p
0
0.25
1
0.50
2
0.25
Table 1.1: Discrete Probability Distribution
This is a probability distribution. Clearly we can see that there will be discrete and continuous probability distributions depending on whether the variable is discrete or continuous. In the example above
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of the coin toss, the variable X is a discrete variable and hence this is a discrete probability distribution.
Examples of discrete distributions are the Binomial distribution and the Poisson distribution which we
shall examine shortly.
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On the other hand, if the random variable is continuous then the the probability distribution associated with it is called a continuous probability distribution. Note that a continuous probability
distribution differs from a discrete probability distribution in that for a continuous distribution, the
probability that the variable assumes a particular value is zero and hence it can’t be represented by
a table. Instead, we represent it with a function. Such a function is called a probability distribution
function.
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Properties of a Probability Distribution Function
1. Since the continuous random variable is defined over a continuous range of values
(called the domain of the variable), the graph of the density function will also be
continuous over that range.
2. Since the continuous random variable can take an infinite number of values, the
probability that it takes a specific value, say a is zero.
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3. Furthermore, the area bounded by the curve of the density function and the x-axis
is equal to 1, when computed over the domain of the variable.
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4. Finally, the probability that a random variable assumes a value between a and b is
equal to the area under the density function bounded by a and b. Note that the
area below the line x = a say, is equal to the probability that the variable X can
take any value value less than or equal to a.
An example of a continuous probability distribution that we will study will be the Normal or Gaussian distribution.
1.3.2
Parent & Sample Distribution
Any measurement of a quantity is usually expected to approximate the quantity though not be exactly
equal to it. We have already seen that every time we make a measurement, we expect some discrepancy between the measurements because of random errors and so we expect every measurement to be
different. However, as we increase the number of measurements, we see that the data is more and more
distributed around the correct value of the quantity being measured. (Of course all this is true if we
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can neglect or correct for systematic errors).
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Suppose we make an infinite number of measurements. Then in principle, we would know the exact nature of the distribution of the data points. If we had such a hypothetical distribution, then we
could determine the probability of getting any particular value of the measurement by doing a single
measurement. This hypothetical distribution is called the parent distribution . Thus, in the above
example, the parent distribution for the one rupee coins minted in a particular period would be the
weights of ALL such coins. However, in practice, we always have a finite set of measured values, as
in the example above where we have a sample of 100 coins on which we carried out the measurements.
The distributions that are obtained from the samples of the parent distribution are called sample distribution. Of course, in the limit of infinite observations, the sample distribution becomes the parent
distribution.
We can define the probability distribution function P (x) which is normalised to a unit area. This
function is defined in such a way that for the limiting distribution (that is in the limit that the number
of observations N is very large), the number of observations of the variable x between x and x + ∆x is
given by
∆N = N P (x)∆x
Mean & Deviation
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1.3.3
Mean of Sample Distribution : x̄ ≡
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b
The parent and the sample distributions discussed above can be characterised by several quantities.
We can define a mean of the sample distribution in exactly the same way as we understand it- as the
average value of the quantity. Thus the mean of the sample distribution, x̄ is
N
1 X
xi
N i=1
(1.5)
where xi are the different observed values of the variable x. Clearly, the mean of the parent distribution, µ is
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Mean of Parent Distribution :
µ ≡ lim
N →∞
N
1 X
xi
N i=1
!
(1.6)
If the measurement of interest can be made with high precision, the majority of the values obtained
will be very close to the true value of x, and the limiting distribution will be narrowly peaked about
the value µ. In contrast, if the measurement of interest is of low precision, the values found will be
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widely spread and the distribution will be broad, but still centered on the value µ. Thus, we see that
the breadth of the distribution not only provides us with a very visual representation of the uncertainty
in our measurement, but also, defines another important measure of the distribution. How can we
characterise this measure?
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The most often used parameter for characterising the breadth or dispersion of the distribution is
called the standard deviation, σ. We can define the variance σ 2 of the parent distribution as
Variance of Parent Distribution :
which is easily seen to be
2
σ = lim
N →∞
from the definition of µ in Eq(1.6).
2
σ = lim
N →∞
1 X
2
(xi − µ)
N
1 X 2
xi − µ 2
N
(1.7)
(1.8)
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This is the measure of dispersion for the parent distribution. What about the sample distribution?
The variance here is defined in an analogous way, except that the factor in the denominator is N − 1
instead of N .
"
Variance of Sample Distribution :
N
1 X
s2 =
(xi − x̄)2
N − 1 i=1
#
(1.9)
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Note that as N approaches ∞, N − 1 and N are the same. But for any finite N the difference comes
in because though the initial set of measurements were N independent measurements (since all the N
xi were independent), however, in calculating x̄, we have used one independent piece of information.
The rigorous proof of this statement is a bit tricky though not required for our purposes.
The importance of these two parameters, the mean µ and the standard deviation σ (or variance σ 2 ) is
that this is precisely the information we are trying to extract from the experiment that we perform. For
the sample distribution, s2 characterises the uncertainty associated with our experiment to determine
the true and actual values. As we shall see, the uncertainty in determining the mean of the parent
distribution is proportional to the standard deviation. Thus we conclude that for distributions which
are a result of statistical or random errors, these two parameters describe the distribution well.
How do we determine the mean and standard deviation of distributions? For this, we define a
quantity called the expectation value .
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R
The expectation value of any function f (x) of x is defined as the weighted average
of f (x) over all the values of x weighted by the probability density function p(x).
It is easy to see that the mean is the expectation value of x and the variance is the expectation value
of square of the deviations from µ. Thus for a discrete distribution from Eq(1.6), we need to replace
the observed values xi by a sum over the values of possible observations multiplied by the number of
times we expect the observation to occur. Thus
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N
N
X
1 X
1 X
µ ≡ E(X) = lim
[xj P (xj )]
xi = lim
[xj N P (xj )] = lim
N →∞ N
N →∞ N
N →∞
i=1
j=1
(1.10)
In a similar way, the variance can be written as
N
N
X
X
2
2
σ = lim
(xj − µ) P (xj ) = lim
xj P (xj ) − µ2 = E(X 2 ) − E(X)2
2
N →∞
N →∞
j=1
(1.11)
j=1
For a continuous distribution, the analogous quantities are
∞
µ ≡ E(X) =
xp(x)dx
(1.12)
−∞
and
∞
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∞
σ2 =
(x − µ)2 p(x)dx =
−∞
x2 p(x)dx − µ2 = E(X 2 ) − E(X)2
(1.13)
−∞
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Example 1.3.3.1
I throw a dice and get 1 rupee if it is showing 1, get 2 rupees if it is showing 2, get 3 if it is showing
3, etc. What is the amount of money I can expect if I throw the dice 150 times?
For one throw, the expected value is
E(X) =
X
xi P (xi )
The probability of getting any of the digits in one roll is 16 . Thus
E(X) = 1 ×
1
1
1
1
1
1
7
+2× +3× +4× +5× +6× =
6
6
6
6
6
6
2
Thus if I roll the dice 150 times, my expected payoff is
24
7
2
× 150 = 525 rupees.
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Lab Manual for Nuclear Physics
Example 1.3.3.2
I am given a probability distribution as below
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X P(X)
1
8
4
1
12
6
3
16
8
1
20
5
1
24
8
Find the variance of this distribution?
We know that the variance is given by Eq(1.11). But to use this, we first need to find the
expectation value of x or the mean. This is
E(X) =
X
xi P (xi ) = 8 ×
Now
X
[xi − E(X)]2 P (xi ) = 32.71
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σ2 =
1
3
1
1
1
+ 12 × + 16 × + 20 × + 24 ×
= 17
4
6
8
5
18
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Example 1.3.3.3
At a pediatrician’s clinic, the age of the children who come to the clinic, x years, is given by the
following probability distribution function:
3
x(2 − x) 0 < x < 2
4
= 0 otherwise
f (x) =
(1.14)
If on a particular day 100 children are brought to the clinic, how many are expected to be under
16 months old?
16 months is 43 years. So the probability of finding a child under 16 months is given by the area
under the curve of the probability distribution function between 0 and 43 . This is
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4
3
4
P (x < ) =
3
f (x)dx
0
4
3
=
3
x(2 − x)dx
4
0
4
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3
4
3
=
x(2 − x)dx
0
3 80
=
4 81
20
=
27
(1.15)
Thus for 100 children, the number we expect to be under 16 months is
100 ×
20
≈ 74.07
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What about the mean age of the children brought to the clinic? For this,we need to use Eq(1.12).
Thus
2
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E(X) =
xf (x)dx
0
2
=
x
3
x(2 − x) dx
4
0
= 1
(1.16)
This result is not surprising if we try to see how the distribution looks graphically as in Figure
1.3.
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1
f(x)
0.8
0.6
0.4
0.2
0
0.5
1
1.5
2
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0
Figure 1.3: Graph of function in Eq(1.14)
We can see that the distribution is symmetrical about x = 1 and therefore the mean must be in
the middle of the range that is x = 1. This is always true and can be used when we know that
the distribution is symmetrical.
Finally, what about the variance of the distribution of the age of the children?
We know that the variance for a continuous distribution is given by Eq(1.13). Thus
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σ 2 = E(X 2 ) − E(X)2
But
2
2
x2 f (x)dx
E(X ) =
b
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0
2
=
x
2
3
x(2 − x) dx
4
0
6
=
5
(1.17)
Thus
σ2 =
6
1
− 12 =
5
5
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Lab Manual for Nuclear Physics
Distributions
We have already seen that the results of a statistical experiment result in a distribution (either discrete
or continuous). We would be interested in three kinds of distributions, viz. Binomial, Poisson and
Gaussian or normal distributions. It is important to note where these distributions are used.
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The Gaussian distribution is the one which we encounter most frequently since this describes the
distribution of random observations in many experiments.
The Poisson distribution is generally used for counting experiments of the kind used in nuclear
physics where the data is the number of events per unit interval. In the study of random processes like
radioactivity, Poisson distribution is important as it is whenever we sort data in bins to get a histogram.
Finally, the binomial distribution is a discrete distribution which is used whenever the possible
number of final states is small. This is true for instance in coin tossing experiments or even in scattering
experiments in particle or nuclear physics.
1.4.1
Binomial Distribution
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A Binomial or Bernoulli trial is basically a statistical experiment which has the following properties:
1. The experiment consists of n repeated trials.
2. Each trial can result in just two possible outcomes. We call one of these outcomes a success and
the other a failure.
b
3. The probability of success, denoted by P , is the same on every trial.
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4. The trials are independent, that is, the outcome on one trial does not affect the outcome on other
trials.
A typical example would be repeated tosses of a coin and counting the number of heads that are
turn up. Clearly the properties mentioned above are satisfied.
We can define a Binomial random number as the number of successes, say x in a binomial
experiment with n trials. The probability distribution of such a binomial random number
is called the binomial distribution. An example of such a distribution we have already seen in the
discussion of the discrete distribution where the probability distribution is given as a table in Table 1.1.
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We can find an expression for the probability P (x; n) for x successes in a binomial experiment with n
trials by analysing an experiment of coin toss. Suppose we want to know the probability of x coins with
heads and n − x coins with tails. For this purpose, we know that there are n Cx different combinations
in which we can get the set of observations. In each of these combinations the probability of x heads
x
coming is px which in this case is 21 and the probability for n − x tails is (1 − p)n−x = q n−x which here
n−x
is 12
. With this, we can write down the probability P (x; n) of getting x successes in an experiment
with n trials, each with probability p as
n x n−x n
n!
px (1 − p)n−x
P (x; n, p) =
p q
= Cx px q n−x =
x!(n − x)!
x
Mean of Binomial Distribution
(1.18)
To find the mean of the binomial distribution, recall that the mean is just the expectation value of
the random variable. In this case, the random variable is x and the probability distribution function
Eq(1.18) is the probability of x successes in n independent trials when the probability of success in each
trial is p. Thus by the definition of expectation value, we have
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n
X
E(x) =
=
x=0
n
X
x=0
=
n
X
x
n!
px (1 − p)n−x
x!(n − x)!
n!
px (1 − p)n−x
(x − 1)!(n − x)!
(1.19)
b
x=1
xP (x; n, p)
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since the x = 0 term does not contribute. Now substitute m = n − 1 and y = x − 1. Then
m
X
(m + 1)!
E(x) =
(p)y+1 (1 − p)m−y
y!(m
−
y)!
y=0
= (m + 1)p
m
X
y=0
= np
m
X
y=0
m!
(p)y (1 − p)m−y
y!(m − y)!
m!
(p)y (1 − p)m−y
y!(m − y)!
But we know that the binomial theorem states that
29
(1.20)
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(p + q)m =
m
X
y=0
m!
py q m−y
y!(m − y)!
Thus
m
X
y=0
m!
(p)y (1 − p)m−y = (p + 1 − p)m = 1
y!(m − y)!
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and so the mean of the Binomial Distribution is given by
E(x) = np
Variance of Binomial Distribution
(1.21)
Recall that the variance of a distribution is defined as the difference of the Expectation value of the
square of the random variable and the square of the expectation value, that is
σ 2 = E(x2 ) − E(x)2
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Consider
E(x(x − 1)) =
n
X
[x(x − 1)]n Cx px (1 − p)n−x
x=0
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b
=
=
n
X
x=0
n
X
x=2
x(x − 1)
n!
px (1 − p)n−x
x!(n − x)!
n!
px (1 − p)n−x
(x − 2)!(n − x)!
= n(n − 1)
n
X
x=2
= n(n − 1)p
2
= n(n − 1)p2
(n − 2)!
px (1 − p)n−x
(x − 2)!(n − x)!
n
X
x=2
m
X
y=0
(n − 2)!
p(x−2) (1 − p)n−x
(x − 2)!(n − x)!
(m)!
p(y) (1 − p)m−y
(y)!(m − y)!
2
= n(n − 1)p (p + 1 − p)m
= n(n − 1)p2
(1.22)
where we have used the substitution, y = x − 2 and m = n − 2. Now variance is
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σ 2 = E(x2 ) − E(x)2
= E(x(x − 1)) + E(x) − E(x)2
= n(n − 1)p2 + np − n2 p2
= np(1 − p)
(1.23)
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Thus we have the important result that the variance of a Binomial Distribution is given by
σ 2 = np(1 − p)
(1.24)
Note that here we have used the fact that the Expectation value of the sum of two random variables
is the sum of the Expectation values of the variables. That is if X and Y are two random variables,
then
E(X + Y ) = E(X) + E(Y )
In our case we have used the fact that
E(x2 ) = E(x(x − 1)) − E(x)
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To show this in general is fairly straightforward. Consider two random variables X and Y which can
take values, x1 , x2 , · · · and y1 , y2 , · · · . Now
E(X + Y ) =
XX
j
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b
=
=
(xj + yk )P (X = xj , Y = yk )
k
XX
j
X
xj P (X = xj , Y = yk ) +
XX
j
k
xj P (X = xj ) +
j
X
yk P (X = xj , Y = yk )
k
yk P (Y = yk )
k
= E(X) + E(Y )
where we have used the fact that
X
P (X = xj , Y = yk ) = P (X = xj )
k
and
X
P (X = xj , Y = yk ) = P (Y = yk )
j
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because the sum over all values of y implies that the joint probability distribution P (X, Y ), that is
k P (X = xj , Y = yk ) will reduce to P (X = xj ).
P
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Example 1.4.1.1
An unbiased coin is tossed ten times. What is the probability of getting less than 3 heads?
The probability of finding less than 3 heads in 10 tosses, is the probability of finding less than or
equal to 2 heads, P (H ≤ 2). This will be the sum of probabilities of finding no heads, 1 head and
two heads. Thus
P (H ≤ 2) = P (H = 0) + P (H = 1) + P (H = 2)
Now the probability of finding x heads in n tosses is given by Eq(1.18). In our case, n = 10,
p = q = 12 . Thus
10
n x n−x
10
1
1
P (H = 0) =
p q
=
=
x
0
2
1024
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Similarly
9
10
n x n−x
10
1
1
=
P (H = 1) =
p q
=
x
1
2
2
1024
b
2 8
n x n−x
10
1
1
45
P (H = 2) =
p q
=
=
x
2
2
2
1024
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Thus the total probability of getting less than 3 heads in 10 tosses is
P (H = 0) + P (H = 1) + P (H = 2) =
1
7
[1 + 10 + 45] =
1024
128
Example 1.4.1.2
Here is a game to test sixth sense. Take 4 cards numbered 1 to 4. One person picks a card at
random and another person tries to identify the card. What is the probability distribution that
the second person would identify the card correctly if the test is repeated 4 times?
Let P (X = x) be the probability of correctly identifying x cards after 4 attempts. Then, by the
binomial probability distribution function, this is given by
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n x n−x
4
P (X = x) =
p q
==
(0.25)x (0.75)4−x
x
x
since the probability of identifying the correct card in 4 attempts, out of 4 cards is
x = 0, 1, 2, · · · , 4. Thus the probability of getting one card right in 4 attempts is
1
4
= 0.25. Here
4
P (1) =
(0.25)(0.75)3 = 0.4219
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The probability distribution is given by
x
0
1
2
3
4
P(x)
0.3164
0.4219
0.2109
0.0468
0.0039
If this is done with say 100 people, we can see that the number of people getting 1 card correct is
100 × 0.4219 ∼ 42.
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Example 1.4.1.3
A biased that is an unfair dice is thrown fifty times and the number of sixes seen is ten. If the
dice is thrown a further fifteen times find:
(a)the probability that a six will occur exactly thrice;
b
(b)the expected number of sixes;
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(c)the variance of the expected number of sixes.
The experiment is clearly a Bernoulli or Binomial trial. If the success is taken to be getting a six,
then the probability p is given by
p=
10
= 0.2
50
Now if X is defined as the number of sixes in 15 trials, then
X = B(15, p)
We want the probability for getting exactly 3 sixes in 15 trials. Thus, x = 3 and
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10
B(15, ) =
50
3 12
15
1
4
≈ 0.2475
3
5
5
The expected number of sixes will be the expectation value E(X). This is, from Eq(1.21),
E(X) = np = 15 ×
1
15
=
=3
5
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Finally, the variance is given by Eq(1.23)
σ 2 = np(1 − p) = 15 ×
1.4.2
Poisson Distribution
1 4
× = 2.4
5 5
A Poisson distribution is the probability distribution that results from a Poisson experiment. A Poisson
experiment has the following properties:
1. The experiment results in outcomes that we can call successes or failures.
2. The average number of successes µ that occur in a specified region is known.
3. The probability that a success will occur is proportional to the size of the region.
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4. The probability that a success will occur in an extremely small region is virtually zero.
5. What happens in a specified region is independent of what happens in any other
region.
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b
A Poisson random variable is the number of successes that result from a Poisson experiment. The probability distribution of a Poisson random variable is called a Poisson
distribution. A Poisson distribution is an approximation to the binomial distribution when the average number of successes, that is µ is much smaller than the possible number, that is when µ n. In
such cases, evaluation of the binomial probability is extremely complicated and tedious.
We have already discussed the Binomial Distribution which is the result of a Bernoulli trial. Recall
that in a Bernoulli trial, we take a sample of definite size and count the number of times a certain event
occurs. We thus know the number of times the event did occur and the number of times the event did
not occur. However, there are some instances when we do know the number of times the event occurs
but do not know the number of times that the event did not occur. As an example, think of watching
a thunderstorm for an hour. I can count the number of times that I see a lightning flash, say 15 times.
However I cannot say how many times the lightning flash did not occur. This is the case where isolated
events are occurring in a continuum of time. Or for instance, suppose under a microscope I look for
the number of malarial parasites in a blood sample. Here what we have is isolated events (the sighting
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of a malarial parasite) in a continuum of volume or area of the slide. In such cases, we cannot use the
Binomial distribution because we do not know the value of n in (p + q)n . It is in these kinds of cases
that we can use the Poisson Distribution.
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As an example of a Poisson distribution, consider the flux of cosmic rays reaching the earth. This
is known to be around 1 per cm2 per minute. Now consider a detector with a surface area of 40 cm2 .
We expect to detect 40 cosmic rays per minute in this detector. Now suppose we record the number
of cosmic rays detected in a 20 second interval. On an average we then expect about 13.3 cosmic rays.
However, when we do the experiment over many 20 second intervals, we will detect something like
13, 15, 14, 12 etc and occasionally even 9 or 18 cosmic rays. We can plot a histogram of this, that is
plot the number of times nx that we observe x rays in this fixed interval of time. Or, if we divide
the number of times nx by the total number of intervals N , then we can get the probability Px of
observing x cosmic rays in this experiment. If our number of intervals N is large, then this probability distribution will be a Poisson distribution. Whenever we observe independent random events that
occur at a constant rate such that the expected number of events is µ, we get a Poisson distribution.
In the case of cosmic rays, the events are obviously random and clearly independent since the arrival
of one cosmic ray does not depend on the arrival of others. Further, the rate of arrival is almost constant.
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Another example can be a scattering experiment with a beam of B particles incident on a thin foil
and the probability p of any one interaction taking place is very small. Then we know that the number
of observed interactions r will be binomially distributed where we will take the number of trials as B
and the probability of success (interaction) as p. It turns out that when p is very small, then the values
of Px will be like a Poisson distribution with a mean given by Bp (which we have seen is the mean of
the binomial distribution Eq(1.21)).
There are many real life instances where Poisson distribution is used. Thus, for instance, for a telephone exchange, the number of calls coming in some unit of time can be modeled as a Poisson variable
provided that the number of subscribers to that exchange is large and the subscribers act independently.
Or an insurance company might model the number of wierd accidents (say falling off your bed to hurt
yourself) if the population is large, the probability of such an event is small and each event (of a fall)
is independent of any other event.
Thus we see that a binomial distribution goes to a Poisson distribution when the number of trials N
increases while the probability of success p decreases in such a way that N p is a constant, say µ. With
this, we can derive the probability distribution function for the Poisson distribution.
Consider the case of a binomial distribution where p 1 and we consider the situation where n → ∞
but np remains finite. Recall that np is the mean of the binomial distribution (Eq1.21). The probability
function for the binomial distribution is
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P (x; n, p) =
1
n!
px (1 − p)−x (1 − p)n
x! (n − x)!
But
n!
= n(n − 1)(n − 2) · · · (n − x + 2)(n − x + 1)
(n − x)!
Now since x n, each of these x factors is very nearly n and hence this becomes
Now
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n!
≈ nx
(n − x)!
(1 − p)−x ≈ (1 + px) ≈ 1
since p → 0. Thus we now have the probability function as
P (x; n, p) =
Consider now
1 x x
n p (1 − p)n
x!
µ
1
(1 − p) = lim[(1 − p) ] =
= e−µ
p→0
e
n
1
p
µ
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since the mean, µ = np for a Binomial distribution.
Thus we get the Probability Distribution function for the Poisson distribution
PP (x; µ) = lim PB (x; n, p) =
p→0
µx −µ
e
x!
b
R
(1.25)
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This is the probability of obtaining x events in the given interval. Remember that x is positive
integer or zero.
Another way to see this is as follows:
We know the Binomial probability distribution function as
P (x; n, p) =
1
n!
px (1 − p)n−x
x! (n − x)!
We set np = µ. Then
P (x; n, p) =
µ x 1
n!
µ n−x
1−
x! (n − x)! n
n
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which is simply
P (x; n, p) =
µx
n!
1 µ n µ −x
1
−
1
−
x! (n − x)! nx
n
n
Now as n → ∞, we have
Thus,
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µ
→0
n
n!
1
→1
(n − x)! nx
µ n
1−
→ e−µ
n
µ −x
→1
1−
n
P (x; n, p) =
Mean of Poisson Distribution
µx e−µ
x!
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The mean of the distribution is the expectation value of the random variable. Thus
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b
∞ X
µx −µ
E(x) =
x e
x!
x=0
∞ X
µx−1
−µ
= µe
(x − 1)!
x=1
= µe−µ
∞
X
µy
y=0
y!
= µ
(1.26)
Thus we see that the mean of the Poisson distribution is
R
E(x) = µ
Variance of the Poisson Distribution
37
(1.27)
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We know that the variance is defined in terms of difference of the expectation value of the square of
the variable and square of the expectation value. That is
σ 2 = E(x2 ) − E(x)2
Consider
E(x ) =
=
=
=
=
x
x
2 µ −µ
e
x!
x=0
∞ x
X
2 µ −µ
0+
x
e
x!
x=1
∞ y+1
X
2 µ
−µ
(y + 1)
e
(y + 1)!
y=0
∞ X
µy µ
2
−µ
(y + 1)
e
(y + 1)(y)!
y=0
∞ X
µy −µ
e
µ
(y + 1)
(y)!
y=0
"∞ X
#
∞ y
X
µy −µ
µ −µ
µ
(y)
e
e
+
(y)!
(y)!
y=0
y=0
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=
∞ X
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2
= µE(x) + µ
= µ(µ + 1)
(1.28)
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R
b
Since the sum of the probability distribution function over all x is unity and therefore the second
term in the square bracket is unity. Thus we see that the variance is simply
σ 2 = E(x2 ) − E(x)2 = µ(µ + 1) − µ2 = µ
(1.29)
This is a remarkable result. The mean and the variance of the Poisson distribution is the
same. This gives rise to the famous square root rule. In an experiment where the distribution
satisfies Poisson distribution conditions, for instance in the counting of N independent events in a fixed
interval, we can estimate the mean of the distribution to be N . Then, as we saw above, the variance is
√
√
also N and therefore σ = N . The statistical errors in them would be N and we would quote our
√
results as N ± N . We will return to this later in the Chapter when we discuss error estimation.
Example 1.4.2.1
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A factory produces resistors and packs them in boxes of 500. If the probability that a resistor is
defective is 0.005, find the probability that a box selected at random contains at most two resistors
which are defective.
If we take X as the ‘number of defective resistors in a box of 500’, then
X = B(500, 0.005)
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since the trials are obviously Binomial. Now in this case, we can see that the number of trials,
n = 500 is large and the probability of success p = 0.005 in each trial is low, so the Binomial
distribution can be approximated by a Poisson distribution with a mean
µ = np = 500 × 0.005 = 2.5
Then the probability of finding two or less defective resistors in a box of 500 is
P (X ≤ 2) = P (0) + P (1) + P (2)
2.50 e−2.5
0!
P (1) =
2.51 e−2.5
1!
P (2) =
2.52 e−2.5
2!
b
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or
P (0) =
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where
P (X ≤ 2) = 6.625 × e−2.5 ≈ 0.543
Example 1.4.2.2
A bank manager opens on an average 3 new accounts per week. Use the Poisson distribution to
calculate the probability that in a given week she will open 2 or more accounts but less than 6
accounts.
To use the Poisson distribution function, we need to know the mean. In this case, the mean is
given as 3 accounts per week. Thus we can use Eq(1.25) with µ = 3. Then the probability of
opening 2 or more accounts but less than 6 in a week is simply
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P (2 < x < 6, 3) = P (2, 3) + P (3, 3) + P (4, 3) + P (5, 3) =
e−3 32 e−3 33 e−3 34 e−3 35
+
+
+
= 0.7167
2!
3!
4!
5!
Example 1.4.2.3
Thirty sheets of plain glass are examined for defective pieces. The frequency of the number of
sheets with a given number of defects per sheet was as follows:
Frequency
0
8
1
5
2
4
3
7
4
4
5
2
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No. of defects
Table 1.2: Distribution of Defects in Glass sheets
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What is the probability of finding a sheet chosen at random which contains 4 or more defects?
µ=
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From Table 1.2 we see that the total number of sheets is 30 while the total number of defects in
these 30 sheets is 60. To use the Poisson distribution function, we need to find the mean number
of defects. We know that there are 60 defects in 30 sheets of glass. Thus the mean number of
defects per sheet is
60
=2
30
The probability then of finding a sheet with 4 or more defects is
P (x ≥ 4) = 1 − P (x < 4)
= 1 − P (0) − P (1) − P (2) − P (3)
−2 0
e (2)
e−2 (2)1 e−2 (2)2 e−2 (2)3
= 1−
+
+
+
0!
1!
2!
3!
= 0.1431
Example 1.4.2.4
At the ITO intersection, vehicles pass through at an average rate of 600 per hour.
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(a)Find the probability that none passes in a given minute.
(b)What is the expected number passing in five minutes?
The average number of vehicles per minute is simply
600
= 10
60
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µ=
Thus the probability that no vehicle passes in a given minute is
P (0, 10) =
e−10 100
= 4.53 × 10−5
0!
The expected number of vehicles passing in five minutes is
E(X = 5) = 10 × 5 = 50
1.4.3
Normal Distribution
b
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The Normal Distribution is extremely important in probability theory and statistics and forms the
cornerstone of most of statistical analysis. For our purposes, its importance lies in the fact that many
real-world phenomena involve random quantities that are distributed in an approximately normal fashion. For instance, the errors in a scientific measurement are approximately normal. It is often called
Gaussian distribution and also referred to as “bell-shaped distribution”, because the graph of its probability density function resembles the shape of a bell.
La
The Normal or Gaussian Distribution is an approximation to the Binomial distribution for the limiting case when the number of possible observations, that is n goes
to infinity AND the probability of success in each measurement is finite and remains constant, that is when np 1. It is also the limiting case of the Poisson distribution when
the mean µ becomes large.
The Gaussian probability density is defined as
R
"
2 #
1
1 x−µ
PG = √ exp −
2
σ
σ 2π
41
(1.30)
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As one can see this is a continuous distribution function and thus describes the probability of getting
a value x from a random observation from a parent distribution of mean µ and standard deviation σ.
Properly defined, we should talk about a Gaussian Probability Distribution Function, such that
the probability dPG of the random observation having a value between x and x + dx i.e.
dPG (x; µ, σ) = pG (x; µ, σ)dx
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With this probability distribution function, we can now see how a Poisson distribution goes to a
Gaussian distribution for large mean. Consider first a Poisson distribution and a normal distribution
both with mean 1. Thus µ = 1 and in the case of the Gaussian distribution, σ = 1. Then the probability
that n ≤ 2 that is n ≤ µ + 1σ for the Poisson distribution is
P (1 ± 1) = P (0) + P (1) + P (2) = .736
where P (n) is the Poisson distribution function of Eq(1.25). With the Gaussian distribution function
(Eq(1.30)), we get
P (0 < x < 2) = 0.68
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Thus we see that for small means, the two distributions are fairly different. What about for large
√
mean? Let us take µ = 15
= 7.5. Then σ = 7.5 = 2.7. Then the corresponding probabilities for
2
µ + 1σ are
P (7.5 ± 2.7) = P (0) + P (1) + · + P (10) = 0.64
and for the Gaussian distribution
P (0 < x < 10.2) = 0.68
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b
which is very similar. In general, for µ > 5, the Gaussian distribution is a good approximation to the
Poisson distribution. This fact will be important to us since we will see that in counting experiments,
it will become easier to use the Gaussian distribution in cases where the mean number of counts is large.
Properties of a Normal Distribution
1. The total area under the normal curve is equal to 1.
2. The probability that a normal random variable x equals any particular value is 0.
3. The probability that x is greater than some value a equals the area under the normal curve
bounded by a and ∞.
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4. The probability that x is less than a equals the area under the normal curve bounded by a and
−∞.
5. About 68% of the area under the curve falls within 1 standard deviation of the mean.
6. About 95% of the area under the curve falls within 2 standard deviations of the mean.
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7. About 99.7% of the area under the curve falls within 3 standard deviations of the mean.
A convenient form of the normal distribution is the Standard Normal Distribution. To obtain
this, we simply use the substitution
x−µ
(1.31)
σ
in Eq(1.30). Then the probability distribution function for the standard normal distribution becomes
z=
R
2
z
1
exp −
pG (z)dz = √
dz
2
2π
(1.32)
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It is important to see that since for all the values of X, the normal variable falling between x1 and
x2 have corresponding Z (the standard normal variable) values between z1 and z2 , it means that the
area under the X curve between X = x1 and X = x2 equals the area under the Z curve between Z = z1
and Z = z2 . Therefore, we have, for the probabilities,
b
P (x1 < X < x2 ) = P (z1 < Z < z2 )
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Mean of the Standard Normal Distribution
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∞
E(x) =
xpG (x)dx
−∞
∞
x2
x exp −
2
dx
−∞

1
= √ 
2π
0
−∞
= 0

2
2
∞
x
x
x exp −
dx + x exp −
dx
2
2
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1
= √
2π
0
(1.33)
Thus we see that the mean of standard Normal distribution is 0.
Variance of the Standard Normal Distribution
We know that
σ 2 = E(x2 ) − E(x)2
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Now
∞
E(x2 ) =
x2 pG (x)dx
−∞
∞
2
x
x exp −
dx
2
b
2
−∞
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1
= √
2π

0
2
∞ 
2 x
x x exp −
dx
2
1
x
= √  x x exp −
dx +
2
2π
−∞
0


0
2 0
2
2 ∞ ∞
2
1
x
x
x
x
= √  −x exp −
+
exp −
dx + −x exp −
+
exp −
dx
2
2
2
2
2π
−∞
0
−∞
1
= √
2π
∞
exp −
0
2
x
2
dx
−∞
= 1
(1.34)
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using Integration by parts ( udv = uv −
function is normalised to 1.
vdu) and also the fact that the probability distribution
Therefore
R
σ 2 = E(x2 ) − E(x)2 = 1 − 0 = 1
(1.35)
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Thus we see that the standard normal distribution has a mean equal to 0 and a variance
equal to 1.
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The difference between the standard normal distribution and the normal distribution can be seen
from the curves for probability distribution. Consider a normal distribution with µ = 2, σ = 13 . The
probability distribution function will look like Figure 1.4
Figure 1.4: Normal Distribution with mean = 2 and σ =
1
3
The corresponding standard normal distribution with µ = 0, σ = 1 will resemble Figure 1.5
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Lab Manual for Nuclear Physics
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Shobhit Mahajan
Figure 1.5: Standard Normal Distribution with mean =0 and σ = 1
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The two graphs obviously have very different µ and σ but have identical shapes and a shifting of
the axes will give one from the other. It is also easy to see that the area under the two curves between two equivalent points is the same. Thus, for instance, the area of the Normal distribution (with
µ = 2, σ = 13 between 0.5σ to 2σ to the right of the mean will be the area from x1 = µ + σ2 = 2 + 16
to x2 = µ + 2σ = 2.66. The area under the Standard normal distribution would be the area from
z1 = µ + σ2 = 0 + 0.5 = 0.5 to z2 = µ + 2σ = 0 + 2 = 2.0.
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The Standard Normal distribution with the probability distribution function given by Eq(1.32), gives
us the probability of finding the value. This is also the area under the curve from 0 to the value. This is
usually tabulated in a z-table which can be looked up as a standard reference as given below in Figure
1.6.
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Shobhit Mahajan
Figure 1.6: z-tables for Standard Normal Distribution§
§(Source: http:// www.katyanovablog.com/ picsgevs/ normal-distribution-table)
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Example 1.4.3.1
The mean weight of 1000 parts produced by a machine was 30.05 gm with a standard deviation
of 0.05 gm. Find the probability that a part selected at random would weigh between 30.00 gm
and 30.15 gm?
30.00 is 1σ that is 0.05 below the mean. Similarly, 30.15 is 2σ = 0.1 above the mean. Thus
P (30.00 < X < 30.15) = P (−1 < Z < 2)
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b
since recall that the area under the normal gaussian curve between two points x1 and x2 is the
same as under a standard normal curve between two points z1 and z2 which are related to x1 and
x2 by the transformation Eq(1.31). And 1σ for the standard normal distribution is 1 from the
mean that is 0 and 2σ is 2. These values can be looked up from the standard tables.
P (−1 < Z < 2) = 0.3413 + 0.4772 = 0.8185
So the probability is 0.8185.
What about the Gaussian distribution? The Mean of the Gaussian distribution is obtained easily
now either by a substitution as given above into the standard normal distribution or by direct calculation.
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1
√
=
σ 2π
∞
−∞
∞
(x − µ)2
x exp −
2σ 2
dx
(x − µ)2
(x − µ) exp −
2σ 2
−∞
1
= 0+µ √
σ 2π
= µ
∞
−∞
1
dx + √
σ 2π
∞
(x − µ)2
µ exp −
dx
2σ 2
−∞
(x − µ)2
dx
exp −
2σ 2
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1
√
E(x) =
σ 2π
(1.36)
since the distribution function is normalised to 1.
We can also easily see this by using the Standard Normal distribution. Recall that the standard
Normal distribution and a Gaussian distribution are related by Eq 1.31,
x−µ
σ
Thus for a normal random variable X, we can write it as a linear combination of the standard normal
variable Z as
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z=
x = σz + µ
Now using the linearity of expectation values, we have
La
b
E(x) = E(σz + µ)
= σE(z) + µ
= µ
since E(z) = 0 as we saw above.
Thus we see that for a Gaussian distribution,the mean is given by
R
E(x) = µ
We can also find the Variance of Gaussian Distribution easily.
48
(1.37)
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σ 2 = E((x − µ)2 )
1
√
=
σ 2π
= σ
(x − µ)2
dx
(x − µ) exp −
2σ 2
2
−∞
∞
(y)2
y exp −
2
2
dy
−∞
2
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σ2
= √
2π
∞
(1.38)
since the integral is simply the variance of the Standard Normal Distribution given in Eq(1.34) which
is 1. Thus we see that the Variance of the Gaussian distribution is σ 2 .
The probability distribution function for the Normal or Gaussian distribution (Eq(1.30) is normalised
to 1. That is
∞
PG (x)dx = 1
−∞
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From the definition of the probability distribution function, we know that the probability that any
one x value lies between the limits x = µ − ∆ and x = µ + ∆ is simply the the area under the Gaussian
curve between these limits. If one computes (by integration) such areas for various choices of ∆, one
can show that the probability of finding any one measurement of x between various limits, measured
as multiples of the standard deviation, σ, is given by the data given in Table 1.3.
Probability
0.50
0.68
0.80
0.90
0.95
0.99
0.999
Interval
µ − 0.674σ < x < µ + 0.674σ
µ−σ <x<µ+σ
µ − 1.282σ < x < µ + 1.282σ
µ − 1.645σ < x < µ + 1.645σ
µ − 1.960σ < x < µ + 1.960σ
µ − 2.576σ < x < µ + 2.576σ
µ − 3.291σ < x < µ + 3.291σ
Table 1.3: Normal Distribution: Probabilities with intervals
How can one interpret this table? The table indicates that we can be 95% confident that any one
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measurement that we make in the experiment (assuming all measurements are distributed normally)
will lie within approximately 2σ of the mean. Thus, the probability column can be taken as the
confidence level and the interval column as the confidence interval.
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This interpretation and analysis looks very straightforward. However, there is a problem- the problem is that the µ and the σ that we are using in the Gaussian distribution is the parent mean and
standard deviation. This means, as we have discussed above, that this will only be valid if we make an
infinite number of measurements!
We will address this issue in Section 1.6.
1.5
Error Estimation
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The basic aim in any experiment is to measure a quantity and also estimate the uncertainties in the
measurements. We also need to understand the sources of the uncertainties. Lastly we need to know
how to combine uncertainties in measurements of more than one quantity into the error in the quantity which is calculated from these measurements. This is what we now discuss. Throughout this
section, we will only be dealing with statistical or random errors. Systematic errors will
be assumed to have been taken care of.
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First of all, it is important to understate a crucial fact which allows one to use the statistical methods
discussed above to analyse the errors in any experiment. The crucial result is that any measurement
subject to many small random errors will be distributed normally. This follows from the
Central Limit Theorem which states that the distribution of the sum of a large number of random
variables will tend towards a normal distribution. We may think of a measurement as being the result
of a process namely our carrying out many small steps in the experiment. Each step in the process
may lead to a small error with a probability distribution. When we sum the error over all steps to get
final error, the Central Limit Theorem guarantees that this will lead to a normal distribution no matter
what the error on the individual steps works out to be. So as a result of this, we generally expect
normal distributions to describe errors. Note that this simple, yet powerful fact allows us to use the
whole machinery of normal distributions and statistics to analyse errors.
In the case of observations that we are taking are collections of finite number of counts over finite
intervals, then the underlying distribution we know is Poisson. In this case, we know that the observed
values would be distributed around the mean in a Poisson distribution. (Recall that the random variable
in a Poisson distribution can only take positive values, including zero since it is defined as the number
of successes in a Poisson experiment.) In fact, in any experiment where data is grouped in bins to form
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Shobhit Mahajan
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a histogram, the number of events in each bin will be distributed according to a Poisson distribution.
This allows us a tremendous simplification. We know that for a Poisson distribution, the standard
deviation is, Eq(1.29), simply the square root of the mean of the Poisson distribution.
√
µ
σ=
Recall that we have defined relative uncertainty, Eq(1.1), as
uncertainty
measured value
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Relative Uncertainty =
The measured value for us is the mean, that is µ. The uncertainty in this is obviously the standard
deviation, that is σ. Thus, we can say that
relative uncertainty =
σ
1
=√
µ
µ
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In our counting experiments, for instance, this means, that as we increase the number of counts per
interval (that is increase the mean µ), the relative uncertainty goes down as the square root of the mean.
This is actually referred to in general as the Square root rule for Counting Experiments which
states that the uncertainty in the any counted number of random events, which is used as an estimate
of the true average number, is equal to the square root of the counted number. For our purposes, this is
extremely important in our counting experiments since the process which we are measuring, namely the
counts are random and distributed in a Poisson distribution in any time interval. However, care should
be taken to note that this uncertainty is only in the counted variable and not in any derived
quantity. Thus, for instance if we were to measure N counts in an interval of T seconds, to get the
rate R per second,
N
T
Now to find the uncertainty in R, we know that the uncertainty is in the measured random variable
√
N and it is N . Thus the number of counts in time T is really
La
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R=
N±
√
N
From this the rate can be seen to be simply
√
N± N
R=
T
and NOT
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r
N
N
R=
±
T
T
since only the quantity N is counted and hence the uncertainty in N is the square root of N .
1.5.1
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This example above then leads us to the issue of how to estimate the error in any derived quantity
from the errors in the measured quantities?
Propagation of Errors
Consider an experiment where we measure some quantities, for instance the number of counts and the
distance and use these measured quantities to determine a quantity which is a function of the two
measured quantities. Let the desired quantity by x and the measured quantities, u and v be such that
x = f (u, v)
(1.39)
Suppose further that the most probable value of x, x̄ is such that
x̄ = f (ū, v̄)
(1.40)
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To determine the most probable value of x, we take the measurements of u and v, that is ui and vi and
determine the different xi . That is
xi = f (ui , vi )
(1.41)
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b
We have already seen that in the limit of an infinite number of measurements, the sample distribution
will go to the limiting distribution or parent distribution and the average of x, that is x̄ will be the
mean of the distribution. In that limit, we can use the calculated sample mean, x̄ to find the variance
σx2 as
σx2
= lim
N →∞
1 X
2
(xi − x̄)
N
(1.42)
Also, expanding the function in Eq(1.39) in a Taylor series around the averages of u and v,
xi − x̄ ' (ui − ū)
∂x
∂u
+ (vi − v̄)
∂x
∂v
(1.43)
Of course the partial derivatives have to be evaluated when the other variable is kept fixed at the
mean value.
Now combining Eq(1.42) and Eq(1.43), we have
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Shobhit Mahajan
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Lab Manual for Nuclear Physics
2
∂x
∂x
1 X
(ui − ū)
+ (vi − v̄)
' lim
N →∞ N
∂u
∂v
"
2
#
2
X
∂x
∂x
∂x
∂x
1
(ui − ū)2
+ (vi − v̄)2
+ 2(ui − ū)(vi − v̄)
(1.44)
' lim
N →∞ N
∂u
∂v
∂u
∂v
Clearly, the first two terms are related to the variance of u and v, that is
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σu2
1 X
2
(ui − ū)
N
(1.45)
1 X
2
(vi − v̄)
N
(1.46)
σv2
(1.48)
= lim
N →∞
and
σv2
= lim
N →∞
We also define a new quantity covariance between the two variables u and v as
2
σuv
= lim
N →∞
Thus the variance for x is given by
R
'
σu2
∂x
∂u
2
+
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σx2
1 X
(ui − ū)(vi − v̄)
N
∂x
∂v
2
+
2
2σuv
∂x
∂u
∂x
∂v
(1.47)
This is known as the error propagation equation.
R
La
b
The covariance term which is a measure of the correlation of the variations in u and v. In most
experiments the fluctuations in the measured variables are uncorrelated and so on the average the
covariance term vanishes for a large number of observations. We shall neglect this in our further
discussion. Thus we have in general, the error propagation equation as
σx2
'
σu2
∂x
∂u
2
+
σv2
∂x
∂v
2
+ ···
(1.49)
where the quantity x could be a function of any number of uncorrelated, independent
variables.
Let us see what the error propagation equation looks like in some common cases.
1. Sums & Differences
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Let
x=u+a
where a is a constant. Then since
∂x
∂u
(1.50)
= 1, we get
σx = σu
(1.51)
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We can also find the relative uncertainty (Eq(1.1))
σx
σu
σu
=
=
x
x
u+a
2. Weighted Sums & Differences
(1.52)
Suppose x is the weighted sum of two variables u and v.
x = au + bv
where a and b are constants. Then, since
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∂x
∂u
∂x
∂v
=a
=b
we get using Eq(1.48)
b
σx2 = a2 σu2 + b2 σv2
La
assuming no correlation.
3. Multiplication & Division
Suppose the quantity of interest x is defined by
x = uv
where u and v are measured quantities. In this case, we can see that
∂x
∂u
and
54
=v
(1.53)
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Lab Manual for Nuclear Physics
∂x
∂v
=u
and therefore the error propagation equation tells us that
σx2 = v 2 σu2 + u2 σv2
(1.54)
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or
σx2
σu2 σv2
=
+ 2
x2
u2
v
For the case of division, we have
x=
then
(1.55)
u
v
and therefore
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∂x
1
=
∂u
v
∂x
u
=− 2
∂v
v
σx2 =
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b
or
σu2 u2 σv2
+ 4
v2
v
σx2
σu2 σv2
=
+ 2
x2
u2
v
(1.56)
Note the very important difference between Eqs(1.56, 1.55) and Eq(1.53). In the case of weighted
sums and differences the absolute errors are relevant while in this case, it is only the fractional
errors in u and v which are related to the fractional error in x.
4. Powers
Suppose
x = aub
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where a and b are constants. Then
∂x
∂u
= abub−1 =
bx
u
and, using Eq(1.49)
The relative uncertainty in x is
bx
u
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σx = σu ×
σx
b
= σu
x
u
(1.57)
To illustrate the use of the Error Propagation Equation, let us consider a few examples.
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Example 1.5.1.1
Consider an experiment where we count N1 = 945 counts in a 10 second interval in an experiment
and then N2 = 19 counts in a 10 second interval. We have already found a background reading of
NB = 14.2 counts for the same 10 second interval by carrying out a separate experiment carefully.
Thus we assume that there is no uncertainty in the background reading. Now in the
first time interval, the corrected counts are
x1 = N1 − NB = 930.8 counts
La
b
with an uncertainty of
σx1 = σN1 =
√
945 ' 30.7 counts
and a relative uncertainty of
σx1
30.7
=
= 0.032 ' 3.1%
x
930.8
On the other hand, in the second interval, the figures are
x2 = 19 − 14.2 = 4.8 counts
with an uncertainty of
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Lab Manual for Nuclear Physics
σx2 = σN2 =
√
19 ' 4.4
and a relative uncertainty of
σx2
4.4
=
= 0.91
x
4.8
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Example 1.5.1.2
Now suppose in the previous example, the background reading also was subject to some uncertainty. Then we have the formula
x=u+v
and thus the error propagation becomes
σx2 = σu2 + σv2
since the partial derivatives are unity. The uncertainty in x is
σx =
p
σu2 + σv2
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In the above experiment for the first reading, if the background was not error free, then we will
have the error in the net count to be
σ x1 =
q
p
2
2
σN
+
σ
=
(30.7)2 + (3.7)2 = 30.97
N
1
B
b
and so we should report our net counts as
La
930.8 ± 30.9
Example 1.5.1.3
Suppose we want to find the ratio of the activity of two sources by measuring the counts from
them. Then suppose the counts from the first source is N1 = 586 and that from the second is
N2 = 265. Then the ratio of activity is given by
R=
N1
586
=
= 2.211
N2
265
The error in R by Eq(1.56) is given by
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σx2
N2
1
1
σu2 σv2
N1
+
= 2.3 × 10−2
=
+ 2 = 2+ 2 =
2
2
x
u
v
N1
N2
586 265
and therefore
σR = R ×
√
2.3 × 10−2 = 2.211 × 0.151 = 0.333
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and therefore the ratio of activities should be quoted as
R = 2.21 ± 0.33
Example 1.5.1.4
We now consider a simple example where we can repeatedly use the error propagation formulas
given above to get the error in a compound quantity. Consider the simple experiment of finding
g, the acceleration due to gravity. For this, I throw a mass from top of the telescope tower and
measure h the height of the tower and also t, the time it takes for the mass to reach the ground.
I get
t = 1.9 ± 0.1 s
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and
h = 18.5 ± 0.2 m
Now I can use the formula h = 21 gt2 to determine g as
2h
= 10.24 m s−2
t2
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b
g=
I need to know the uncertainty in h and t which are given and then use the error propagation
rules to find the uncertainty in g.
We can write
g=
x
y
where
x = 2h
and
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y = t2
Now we can use Eq (1.56) to get
σg2
σx2 σy2
=
+ 2
g2
x2
y
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To get the error in x, we have to use Eq(1.53) to get
σx2 = 4σh2
Similarly for the error in y, we can use Eq(1.57) to get
2σt
σy
=
y
t
Thus we have
σg2
4σh2 4σt2 σh 2
=
+ 2 =
+
g2
4h2
t
h
2σt
t
2
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Now
0.2
σh
=
= 0.011 = 1.1%
h
18.5
and
La
Therefore,
b
σt
0.1
=
= 0.052 = 5.2%
t
1.9
σg
g
2
=
σ 2
h
h
+
2σt
t
2
= .0112 + 4 × .0522 = .0109
and therefore
σg = .11 × 10.24 m s−2 = 1.07 m s−2
Therefore we should quote our result as
g = 10.24 ± 1.07 m s−2
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We can see that the accepted value of g, that is 9.8 ms−2 is within the margin of error. Also, since
the contribution to the error in g from t is much more, one should improve the measurement of t
rather than try to make the measurement of h better since that contributes very little to the error.
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What we see from this example is that we can repeatedly use the various error propagation formulas
discussed above to get the error in any compound expression. However, this is something that
we need to be careful when using. The error propagation formulae that we have used
work ONLY when the quantities are independent. Here for instance, we could write
the quantity g as xy because we know that the errors in x (which is basically the
error in h) are independent of the error in the denominator y (which is basically the
error in t). In another situation where we have the SAME variable appearing in the
numerator and the denominator, this assumption is NOT true. In that case, we need
to use the general error propagation equation Eq(1.48). We shall encounter such a
situation in Chapter 6.
Example 1.5.1.5
A radioactive sample is counted for one minute and we observe 900 counts. We then remove the
sample and count the background rate for 10 minutes to observe 100 counts. What is the net
count rate and the uncertainty in it?
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There are two measured quantities in this experiment- the gross count rate and the background
count rate. (We assume that the time has no uncertainty). The uncertainty in each of these
measured quantities, we have seen is simply the square root of the measurement. Thus
√
900 = 900 ± 30 counts per minute
b
Gross count rate = RG = 900 ±
La
√
Background count rate = RB = 100± 100 = 100±10 in 10 minutes = 10±1 counts per minute
Net count rate = RN = RG − RB = 890 counts per minute
What about the uncertainty in RN ? For this we need to use the error propagation equation since
remember, RN is a derived quantity and NOT a measured quantity. The defining relation for
RN is given above and so we get
σRN =
q
σR2 G + σR2 B =
60
p
(30)2 + (1)2 ≈ 30
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Lab Manual for Nuclear Physics
Thus we should quote our result as 890 ± 30.
Example 1.5.1.6
A Geiger counter is placed near a suspected source of radioactivity and it records 58 counts in 30
seconds. The source is removed and the background count is found to be 48 counts in 30 seconds.
Can we be sure that the source is truly radioactive?
RT =
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The total count rate is
NT
58
=
= 116 counts per minute
T
0.5
The uncertainty in this measurement is
σRT
σN
=
=
T
The background count rate is
RB =
√
NT
58
=
≈ 15 counts per minute
T
0.5
√
NB
48
=
= 96 counts per minute
T
0.5
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and the uncertainty is
σRB
σN
=
=
T
√
NB
48
=
≈ 14 counts per minute
T
0.5
√
Our net computed source count rate is thus
La
b
RN = RT − RB = 20 counts per minute
Once again, this net count rate is NOT a measured but a derived quantity. So we need to use
the error propagation equation to find the uncertainty in this given the uncertainties in the total
count rate and the background count rate. Thus
σRN
q
√
= σR2 T + σR2 B = 152 + 142 ≈ 21 counts per minute
Thus we have to report our result as 20 ± 21 counts per minute and conclude that we can not say
for sure whether the source is radioactive or not.
Example 1.5.1.7
Finally, let us consider a very simple example where there are both systematic and random errors.
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Suppose I want to measure an unknown resistance R2 . I can for instance use the following circuit.
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(Adapted from ‘A Practical Guide to Data Analysis for Physical Science Students’ by L. Lyons)
Figure 1.7: Resistance Measurement
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We measure the resistance R1 , voltages V1 and V2 using meters and then deduce the value of R2
using Ohm’s law
R2 =
V2 − V1
R1
V1
(1.58)
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b
Clearly, the experiment involves measurement of 3 quantities and each of these will have random
and systematic errors. Random errors we already know happen because of statistical fluctuations.
What about the systematics? In this particular case, there could be many sources of systematic
errors. Thus, for instance, the voltmeters V1 and V2 may not be calibrated properly and/or the
resistance meter might also not be calibrated. We could also have the resistance being temperature dependent and hence not be very accurate. Or there could be some stray capacitances in
case we are using an AC source. We would need to determine or estimate the amount of these
systematic errors.
Now suppose, we performed the experiment and obtained the readings for R1 , V1 and V2 as follows:
R1 = (2.0 ± 0.1Ω) ± 1%
V1 = (1.00 ± 0.02V) ± 10%
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V2 = (1.30 ± 0.02V) ± 10%
The results of the measurements are reported with the average value plus/minus the random error
and the second error in percentage is the estimate of the systematic errors. The random errors
come as previously discussed from statistical fluctuations and are known accurately. The systematic errors quoted above are assumed to have been estimated using a variety of methods.
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Now as we have been discussing, the nature of random and systematic errors is different. However
sometimes we may want the error in our measurement as a single figure. In this case, since the
two sources of error, random and systematic are uncorrelated, we should add them in quadrature
just as we saw in the error propagation equation. Thus we get
R1 = 2.0 ± 0.1Ω
V1 = 1.00 ± 0.10V
V2 = 1.30 ± 0.13V
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But note that the quantity of interest is not V2 or V1 but V2 − V1 . What about the error in
this? This depends crucially on whether we measure the voltages using the same equipment or
separate equipment. If the same equipment is used, then the errors are obviously correlated while
if separate instruments are used, then the errors are uncorrelated. If different meters are used,
then
V2 − V1 = (1.30 ± 0.13) − (1.00 ± 0.10) = 0.30 ± 0.16 V
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where we have added the errors in quadrature. Now if the same meter is used then the errors
are correlated and we have to include this effect. Thus, if there is a 10% systematic uncertainty
in each measurement, then the overall result will also have a 10% uncertainty. Thus,
V2 − V1 = (1.30 ± 0.02) − (1.00 ± 0.02) ± 10% = 0.30 ± 0.03 ± 10% = 0.30 ± 0.04 V
where the errors have been added in quadrature.
We are finally ready to see what the error in the quantity of interest, namely R2 is.
V2 − V1
V2
R2 =
R1 =
− 1 R1
V1
V1
h
i
Think of this expression as the product of two measured quantities, VV12 − 1 and R1 . If we know
the errors in both these, then we can use Eq(1.55) to find the error in R2 . The error in R1 is
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i
h
known as above. The error in VV21 − 1 is simply the error in VV21 . Again the error in this quantity
can be found from Eq(1.56) once the errors in V2 and V1 are known.
The error in VV12 can be calculated assuming that the same meter is used to measure both these
quantities. Then
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V2
(1.30 ± 0.02)
=
= 1.30 ± 3% = 1.30 ± 0.03
V1
(1.00 ± 0.02)
since the systematic errors cancel out using the same meter. Thus the value of
Therefore
V2
− 1 = 0.30 ± 0.03
V1
R2 = [0.30 ± 0.03] [2.0 ± 0.1] = 0.60 ± 0.07Ω
1.6
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What if we had done the following: We have the error in V2 − V1 above. We have the errors in R1
and V11 . Why couldn’t we have used the error propagation equation for these three quantities and
found the error in R2 ? This would be wrong because we cannot assume, under any circumstances
that the errors in V2 − V1 and V11 are uncorrelated since V1 comes in both of these and no matter
how we measure V1 , the errors will always be correlated. Thus one has to be careful when using
related quantities in determining the errors in a calculated quantity.
Estimation and Error of the Mean
La
b
Recall that when we do any measurement, we typically end up with a sample distribution of the data
points of the quantity being measured. We saw that in the limit of an infinite number of measurements, this sample distribution goes to the parent distribution. Of course, the aim of any experiment
is ultimately to determine the parameters of the parent distribution. We also saw that a random set of
observations or measurements are distributed according to a Gaussian (or Poisson distribution). The
question we ask is what is the best estimate of the mean µ of the parent distribution?
1.6.1
Method of Maximum Likelihood
Suppose we conduct an experiment where we obtain N data points which are randomly selected from
the parent distribution. This set of N points of course is our sample distribution. The question
we need to answer is how are the parameters, x̄ and s, that is the average and standard
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deviation of the sample distribution related to µ and σ, the mean and standard deviation of
the parent population? If the parent distribution is Gaussian with mean µ and standard deviation
σ, then the probability of finding x1 (one of the data points in the sample distribution) in the range x1
and x1 + dx1 , which we will define as the probability of finding x1 , is simply
"
2 #
1
1 x1 − µ
P (x1 ) = √
exp −
2
σ
σ 2π
(1.59)
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Similarly, the probability of finding x2 in the range x2 and x2 + dx2 is
"
2 #
1
1 x2 − µ
P (x2 ) = √
exp −
2
σ
σ 2π
(1.60)
and so on. Note that each of these observations are independent and hence these probabilities are
uncorrelated. Thus, the probability of finding x1 between x1 and x1 + dx1 , x2 between x2 and x2 + dx2
etc is simply
Pµσ (x1 , x2 , x3 , · · · , xN ) = P (x1 )P (x2 )P (x3 ) · · · P (xN ) =
1
√
σ 2π
N
"
1X
exp −
2
xi − µ
σ
2 #
(1.61)
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It is important to stress that this probability is for a specific value of µ and σ. But we don’t know
what that value is! So we start off by guessing a value of µ and σ, say µ0 and σ. We compute the
probability Pµ0 ,σ given Eq(1.61). Then we guess another value of µ, say µ00 with the same σ and compute
Pµ00 ,σ . We do this for several values of µ with the same σ. Then we choose that value of µ for which
the probability is a maximum. This is known as the Method of Maximum Likelihood.
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Now it is clear that for the probability (as a function of µ) in Eq(1.61) to be a maximum, the
argument of the exponential should be a minimum since the probability is a constant times a negative
exponential. Let
"
2 #
1 X xi − µ
Y =−
2
σ
Then, we want to find the value of µ for which
"
2 #
dY
d 1 X xi − µ
=−
=0
dµ
dµ 2
σ
or
X xi − µ σ
or
65
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X
xi = N µ
which gives us
R
P
xi
= x̄
N
µ=
(1.62)
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Thus, we see that the most probable value of the population mean µ is precisely the average or sample mean x̄ of the sample distribution.
This is an extremely powerful result. It basically allows us to get an estimate of the parent population mean from a sample distribution which is all what we can possibly obtain from any experiment.
What about the error in this mean?
1.6.2
Estimated Error in the Mean
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We know that the sample standard deviation, s, is the average uncertainty associated with each of the
measurements x1 , x2 , · · · , xN . A legitimate question from our point of view is also, what is the estimated
uncertainty or standard deviation of in our determination of the mean µ for the parent distribution or
x̄ for the sample distribution? Recall that our assumption was that each of the data points xi is from
the same parent distribution and hence it is characterised by the same standard deviation σ.
σµ2 =
"
X
σi2
∂µ
∂xi
2 #
(1.63)
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b
To find the uncertainty in the mean, we can use the error propagation equation Eq(1.49) to find σµ
using Eq(1.62). Thus we have
Now, as we have supposed, all the data points have the same σ, that is σi = σ for all i. We also have
∂µ
∂
=
∂xi
∂xi
1 X
1
xi =
N
N
Thus, we get
R
"
σµ2 =
X
σi2
1
N
2 #
Thus we see that the uncertainty in the mean is
66
=
σ2
N
(1.64)
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R
σ
σµ = √
N
(1.65)
R
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The quantity σµ is called the standard deviation of the mean or the standard error . Thus we
see that if we take N measurements of some quantity x and obtain x1 , x2 , · · · xN , then we can state
our result as the best estimate of x which we have seen is the mean of the sample population and the
uncertainty as Eq(1.65).
Best estimate of a variable from N measurements, xi
x = x̄ ± σµ
where
x̄ =
and
1 X
xi
N
σx
σµ = √
N
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σx is simply the sample standard deviation s.
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This result has an enormous significance for us. The standard deviation σx or s represents the average uncertainty in the individual measurements x1 , x2 , · · · , xN . Thus, if we were to make some more
measurements (using the same technique), the standard deviation σx = s would not change appreciably. On the other hand, the standard deviation of the mean, σµ would slowly decrease as we increase
N . This decrease is just what we would expect. If we make more measurements before computing an
average, we would naturally expect the final result to be more reliable, and this improved reliability is
just what the denominator guarantees. Stated another way, by taking more and more measurements,
we smoothen the distribution (the histogram of the data points) and also can determine the peak (the
mean) in an improved fashion. But note that the increase in precision is only growing as the square
root of the number of measurements and so there is a limitation in by how much we can increase the
precision by taking more and more measurements.
We can now return to the question that we had raised in the Section on Gaussian distribution? How
does one establish confidence limits in the absence of an infinite number of observations? Now that we
have the best estimate of the standard deviation of the mean, we can see that the best way to report our
experimental observations of a parameter x whose mean x̄ has been determined experimentally from
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taking N measurements is
best value of x = x̄ ± σµ
Using the table for Gaussian probability intervals, we can say that this way of reporting the result
is at 68% confidence level. We can increase the confidence level to 95% by reporting it as
best value of x = x̄ ± 2σµ
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Note once again that the values for confidence limits used are for the Gaussian distribution and not
for the sample distribution which is obtained by a finite set of measurements whose standard deviation is
s. The relationship between confidence limits and the sample deviation s was obtained by W.S. Gosset
and is known as the Student’s t-distribution. Basically, there exists a function which can only be
evaluated numerically that allows us to compute a single number which relates s to the confidence level.
What we can then say is that the true value of the parameter x fell in the interval
t
t
x̄ − √ < true value of x < x̄ + √
N
N
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where N is the number of measurements and t can be found from the tables at various confidence
levels and values of N . We will not go into the details of the t- distribution but only note that in
the limit that N → ∞, the values of t at various confidence levels approaches that of the Gaussian
distribution as it should.
1.7
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Basically, for our purposes it is enough to note that for a small number of observations, the predictions from Student’s t-distribution are more accurate since the Gaussian probability distribution
overestimates the confidence level associated with a given range. To put it another way, the Student’s
t-distribution probability requires a larger uncertainty estimate than the Gaussian probability and the
two only coincide for a very large number of measurements.
Method of Least Squares
In an experiment suppose we find a collection of data points (xi , yi ) and we want to find a relationship
between them. The theoretical relationship between these two quantities, x and y is supposed to be
linear. On plotting the points (xi , yi ), we see that the relationship looks somewhat linear. In this case,
we can ask two questions- firstly, assuming the relationship to be linear, can we find the actual relationship, that is y = M x + C which best fits the data? This of course means finding the best estimates
of the constants M , the slope of the straight line and the C, the y intercept. The finding of the best fit
straight line to the data is called linear regression or the least square fit for a line. Once we have
found the best fit straight line, then we can also ask the question about the goodness of fit, that is to
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say, how well does our data fit the assumed straight line? Let us first see how to find the best fit linear
relation between the two quantities x and y.
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We assume that the uncertainty in measuring the variable x is negligible while there is an appreciable
uncertainty in the measuring the variable y. In most experiments, this assumption is a reasonable one
since usually, the uncertainty in one of the variables is much larger than that in the other one. For instance, when we measure speed, the distance measurement has much larger uncertainties than the time
variable. Or, in our case for instance, when we plot the characteristic of the number of counts versus
voltage, the error in the voltage is negligible. We also assume that the measurements of the variable
y is governed by a Gaussian distribution of the same standard deviation, σy for all measurements yi of y.
Now the linear relationship we have assumed between x and y is
y = Mx + C
If we knew M and C, then for any value of x, say xi , we would know the true value of the quantity
y, that is
yiT ≡ True Value of yi = M xi + C
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But we have assumed that the measurements of y are distributed according to a Gaussian distribution,
and therefore all the measurements of yi will be distributed in a Gaussian distribution centered on this
“true value” which would be the mean, µ. It also follows that the probability of finding a particular
value of yi is
PM,C (yi ) =
σy
1
√
"
1
exp −
2
2π
yi − M xi − C
σy
2 #
(1.66)
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Similarly, we can obtain the probabilities of obtaining all the values of y, namely yi , y2 , · · · , yN . Since
these values are all independent, the probability of obtaining our complete set of measurement, that is
yi , y2 , · · · , yN is simply the product of each of these probabilities.
"
#
N
1
1 X (yi − M xi − C)2
PM,C (y1 , y2 , · · · , yN ) ∝ N exp −
σy
2 i=1
σy2
(1.67)
Let us define a quantity χ2 as
χ2 =
N
X
(yi − M xi − C)2
σy2
i=1
Then the probability can be easily written as
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(1.68)
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2
1
χ
PM,C (y1 , y2 , · · · , yN ) ∝ N exp −
σy
2
(1.69)
∂χ2
∂M
C
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We now use the same technique as we did in the previous section to find the best values of M and
C, that is the method of Maximum Likelihood. The best values of these quantities would be those for
which the probability in Eq(1.69) is a maximum. Or, since the probability is proportional to a negative
exponential, those values of M and C for which χ2 in Eq(1.68) is a minimum. To get those values of
the parameters M and C, we differentiate χ2 w.r.t them and equate to zero. Thus
N
2 X
=− 2
xi (yi − M xi − C) = 0
σy i=1
and
∂χ2
∂C
=−
M
N
2 X
(yi − M xi − C) = 0
σy2 i=1
(1.70)
(1.71)
These equations can be rewritten in a more suggestive form as
C
X
xi + M
X
CN + M
X
x2i =
X
xi y i
xi =
X
yi
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Solving these simultaneous equations, we get
P
P P
N xi yi − xi yi
P
P
M=
N x2i − ( xi )2
and
P
b
C=
P
P P
y i − xi xi y i
x2i
P
P
N x2i − ( xi )2
(1.72)
(1.73)
(1.74)
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Defining a quantity ∆ as
∆=N
X
x2i −
2
X
xi
P
P
we get
M=
N
P
xi yi −
∆
xi
yi
(1.75)
and
P
C=
x2i
P
P P
y i − xi xi y i
∆
(1.76)
With these estimates of the slope M and the intercept C, we get the Least Square fit of the straight
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line or the line of regression of y on x.
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The question now remains as to what is the uncertainty in the measurements of the quantity y?
Before we do this, it is important to remember that the measurements y1 , y2 , · · · , yN are NOT the
measurements of the same quantity. So, a spread in their values is NOT a measure of the uncertainty in
their values. The measurement of each yi is, we have already assumed normally distributed around the
best fit value which we have seen to be M xi + C with a standard deviation which is assumed to be σy .
The deviations of the measured value from the true value or best fit value would thus be yi − M xi − C
and these too would be normally distributed around a mean of 0 and with a standard deviation of σy .
To obtain the best estimate for σy , we again need to use the Method of Maximum Likelihood with
the probability of finding the measured values y1 , y2 , · · · , yN which is given in Eq(1.69). But this time,
the parameter we are interested in is σy and so we need to find the most likely value of σy which will
maximise the probability in Eq(1.69). Thus we differentiate Eq(1.69) with respect to σy to get
N σy2 =
or
r
σy =
X
(yi − M xi − C)2
1 X
(yi − M xi − C)2
N
(1.77)
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It turns out that just as in Eq(1.9), we needed to divide by N − 1 instead of N because the number
of independent values had been reduced in computing the mean, in this case too, since we need to find
two parameters M and C (from the measurements) we need to divide by N − 2 instead of N . In any
statistical calculation, we can find the number of degrees of freedom by taking the number of independent
measurements and subtracting the number of parameters determined using these measurements. Thus
we finally get the expression for σy , the uncertainty in the measurements yi , y2 , · · · , yN is
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r
σy =
1 X
(yi − M xi − C)2
N −2
(1.78)
This finally leaves us with the issue of the uncertainty in the quantities that we have determined,
that is M and C. For this, we know the expressions for M and C in terms of yi , namely Eq(1.75) and
Eq(1.76). Thus, knowing the uncertainty in yi , that is σy which we have just determined, we can easily
use the error propagation equation Eq(1.48) to determine the uncertainty in M and C. These are given
by
r
σM = σy
and
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N
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(1.79)
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rP
σC = σy
1.8
x2i
∆
(1.80)
Goodness of Fit
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Let us consider a typical counting experiment where we obtain counting statistics. The experimental
data consists of a series of measured or observed quantities. Let us assume that we have N independent
measurements of the same quantity. These could be anything- in our case, these could be the number
of counts in a specified time interval that we have taken repeatedly. This set of observations allow us to
prepare a sample distribution which as we have already seen can be characterised by two quantitiesthe sample mean, x̄ and the sample variance s2 .
The underlying distribution which describes the process is, as we have seen, called the parent distribution which is characterised by the mean, µ and the variance σ 2 . This distribution is either Poisson
or Gaussian, depending on the value of the mean µ since we recall that for large values of µ typically
larger than 20, these two distributions become almost identical. But the question is that we don’t know
these parameters of the parent distribution. So how do we proceed?
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We take the sample mean x̄ and assume it to be the mean of the underlying parent distribution.
This is something which is a good approximation as we saw in the section on Method of Maximum
Likelihood. With this mean for the parent distribution, we compute the actual variance σ 2 . We also
have the sample variance s2 . If our estimate of the sample distribution being a good representative of
the parent distribution is good, then we expect that the two, that is the sample variance and the true
variance should be close to each other. This comparison is done quantitatively by a procedure known
as the Chi-squared test.
La
Basically, what we are trying to see is whether an obtained or measured set of frequencies in a random
sample and what we expect from an assumed statistical hypothesis match and how well they match. In
our present case, the Chi Squared test allows us to determine how well the observed sample distribution and the assumed parent distribution (in this case a Poisson distribution with a mean µ = x̄ ) match.
We define χ2 as
N
χ2 =
1X
(xi − x̄)2
x̄ i=1
(1.81)
If we recall the definition of the sample variance s2 from Eq(1.9), we can rewrite this in terms of the
sample variance as
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R
χ2 =
(N − 1)s2
x̄
(1.82)
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In our case of the underlying distribution being a Poisson distribution,we know that the variance
of the parent distribution σ 2 is simply the mean of the distribution µ. But we have already chosen
2
the mean µ to be the same as the sample mean x̄. Thus, the deviation of the ratio sx̄ from unity is
a measure of how much the sample variance differs from the variance. In terms of the χ2 , we can say
that if the sample distribution is truly Poisson, then χ2 = N − 1. Any departure from this would be a
measure of how much the sample distribution differs from a Poisson distribution.
Thus we see that χ2 is a statistic that tells us about the dispersion of the observed frequencies from
the expected frequencies. It is usually convenient to define a quantity called the degrees of freedom
ν as
ν = N − Nc
(1.83)
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where N is the number of sample frequencies and Nc is the number of constraints. One way to think
of the number of constraints is that it is the number of parameters which have been calculated from
the data to determine the probability distribution. Thus, in the case above of the Poisson distribution,
we have calculated one parameter, the sample mean x̄ from the data and therefore in that case, the
number of degrees of freedom ν is simply N − 1. With this, we can define a quantity called reduced
Chi sqaured or χ̃2 as
R
χ̃2 =
χ2
ν
(1.84)
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This reduced chi squared clearly has an expectation value equal to 1. If the calculated values of χ̃2
are much larger than 1 then we can say that either our measurements are not good, or the underlying
probability distribution that we have assumed is incorrect. If the value of χ̃2 is very small then again
there is some problem with the experiment.
Another way to think of the the χ2 squared test is to think of trying to fit a model to some given
data. Recall the quantity χ2 that we encountered while discussing the Method of Least Squares in
Eq(1.68). There, we had defined
χ2 =
N
X
(yi − M xi − C)2
σy2
i=1
for the case of the least square fitting of a straight line. It is obvious that in the general case, suppose
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we have some observations yobs and we assume an underlying model for the data which yields the values
yth for the data points, then we can define χ2 as
2
χ =
N
X
(yobs − yth )2
σi2
i=1
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Typically, we can have various competing models for our data or, what is the same thing, several different values of some model dependent parameter. To decide which model (or the value of a parameter)
fits the data best, we compute the χ2 from the data and from that determine the reduced chi squared,
χ̃2 . From our discussion above, it is clear that we should choose that model, or the value of the model
parameter, for which the reduced chi squared χ̃2 is closest to 1.
The χ2 probability distribution function is given by
Pχ (x2 ; ν) =
2
ν−2
1
− x2
2
2
e
x
2 Γ( ν2 )
ν
2
We can show easily that with this probability distribution function, that
χ¯2 ≡ E(χ2 ) = ν
and
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V ar(χ2 ) ≡ E((χ2 )2 ) − (E(χ2 ))2 = 2ν
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This is evident from Fig 1.8.
Figure 1.8: Probability distribution for χ2 for different degrees of freedom
We can look up tables of χ2 (for instance in Appendix A or at en.wikibooks.org/wiki/Engineering
Tables/Chi-Squared Distibution or www.pd.infn.it/∼lunardon/didattica/docsper2/TavoleChi2.pd) to
find out the probability associated with any value of χ̃2 . The tables tabulate the values of
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∞
2
Probabilityν (χ̃ ≥
χ̃2o )
P (x2 ; ν)dx2
=
χ2o
or
2
Probabilityν (χ̃ ≥
χ̃2o )
1
= ν ν
2 2 Γ( 2 )
∞
x2
ν−2
2
x2
e− 2 dx2
(1.85)
χ2o
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where ν is the number of degrees of freedom, χ̃2o is the calculated value of the reduced chi-squared
from the observed data and χ̃2 is the expected value of the reduced chi-squared from our model.
Clearly, from this we can say that if our model is correct, we expect
χ2 ∼ ν ±
√
2ν
Figure 1.9: p-value and area under the curve
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Given this, we can ask the question that we started with- do our observations correspond to our
underlying model (in our case, a Poisson distribution)? Or to put it another way, what is the probability
that our observed value of χ2 or a larger one, could arise purely by chance? This is the probability in
Eq(1.85) and is called the p-value. It is basically the area of the probability distribution function curve
from the observed value of χ2 to ∞ as can be seen in the Fig 1.9.
Basically, we know that if the observations were a good approximation to the underlying model, then
χ̃ should be close to 1 and from the graph is it clear that the value of the probability is around 0.5.
If the fit is not good, the value of χ̃2 will be larger and the probability smaller. Thus suppose in our
experiment, our data gives an observed value of χ2 = 220.1 and there are 199 degrees of freedom. Then
looking up the tables, we see that the value of Probabilityν (χ̃2 ≥ χ̃2o ) is 0.12 or 12% roughly. What this
means is that if our observations were indeed from a parent Poisson distribution, then if we repeat our
experiment many times, and get different data, in roughly 12% of the experiments, we will get these
values.
2
Another way to see this is as follows: Suppose we carry out an experiment and the observations give
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us a χ2 value of 1.80. Suppose that the number of degrees of freedom in the experiment is 1. Then we
see that the reduced chi-squared, χ̃2 is 1.80. If we believe that the underlying distribution is Gaussian
(or Poisson with a large mean, which as we have seen goes to a Gaussian distribution), then can we say
that the underlying assumption of a Gaussian distribution is ruled out?
If our assumption about the Gaussian nature of the underlying distribution is correct, then we can
see from the tables, that probability of obtaining a reduced chi-squared of 1.80 or larger is simply
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Probabilityν (χ̃2 ≥ 1.80) ≈ 18%
That is, if our results were governed by an underlying Gaussian distribution, then there is a 18%
probability that we would obtain a value of χ̃2 ≥ 1.80. This seems like a reasonable agreement. However, what we can say from this is that the data does not support rejecting the the hypothesis
that the underlying distribution is Gaussian. We cannot say whether the hypothesis is true. Only that
we cannot reject it.
In general, we need to decide on a cut-off below which we will not reject the hypothesis. Usually,
this is taken as either 5% or 1% and the result is quoted as being at 5% or 1% significance level. To
reiterate, what this means is that if
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Probabilityν (χ̃2 ≥ χ̃2o ) < 5%
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we reject our expected or hypothesised distribution at 5% significance. As an example of reading
the tables, given in Appendix A, let us assume we have an experiment with 10 degrees of freedom and
suppose we obtain a value of reduced chi-squared as 2.2. Then from the table in Appendix I, we see
that the probability of obtaining χ2 ≥ 2.2 is slightly less than 2%. With this, we can safely say that
we can reject the hypothesis at 5% significance but not at 1% significance. To put it another way, we
reject our hypothesis at 95% confidence level but not at 99% confidence level.
The way we use the test is as follows:
1. Suppose we do an experiment and get some observed values for the outcomes, wi .
2. We usually don’t know the probability mass functions, that is the probabilities pi associated with
these outcomes.
3. We choose a hypothetical distribution- typically binomial or Poisson, with some parameters.
4. Then our Null Hypothesis H0 is that the observations follow the hypothetical distribution.
5. Our alternate hypothesis HA is that the data is drawn from some other distribution.
6. We construct the test statistic, χ2 as follows. We take the observations, Oi that is the number
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of counts for each outcome, wi . We then compute the expected number of counts Ei assuming
that the underlying distribution is the hypothetical distribution, that is assuming that the Null
Hypothesis is true. Then
χ2 =
X (Oi − Ei )2
Ei
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7. We next need to know the degrees of freedom ν to determine the χ2 distribution, which recall is the
distribution of the test statistic. This is usually done by taking the total number of observations
and subtracting out that number that is required to compute Ei .
8. If H0 is true, then χ2 will follow the χ2 distribution for ν. This means that the conditional
probability of getting χ2 given the Null hypothesis H0 , f (χ2 |H0 ) will have the same pdf as
Y ∼ χ2 (ν).
9. Finally the p value is simply
p = P (Y > χ2 )
To understand the concept of p-values, let us consider a simple example.
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Suppose we have a company which sells bags of 100 toffees. The company claims that the toffees are
30% red, 60% green and 10% white in any bag. I pick a bag of toffees made by the company and find
that it has 50 red toffees, 45 green toffees and 5 white toffees. Is this consistent with the company’s
claims at 5% significance level?
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Let us see how it fits the steps needed for the test given above. The experiment in this case is the
taking of a random bag of toffees. The Null hypothesis, H0 is obviously that the data is consistent
with the hypothetical distribution, that is the distribution claimed by the company. The alternate
hypothesis is of course that it does not follow that distribution. Next we apply the χ2 test. We need
to determine the degrees of freedom first. The number of variables is obviously 3 since there are three
kinds of toffees. So the degrees of freedom are 3 − 1 = 2. We next find the expected values of the three
variables according to the hypothetical underlying distribution, that is the distribution claimed by the
company. These are obviously
Er = 100 × .3 = 30
Eg = 100 × 0.6 = 60
Ew = 100 × 0.1 = 10
The observed values obtained in our experiment are
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Or = 50
Og = 45
Ow = 5
From these we can calculate the χ2 as
X (Oi − Ei )2
Ei
= (50 − 30)2 /30 + (45 − 60)2 /60 + (5 − 10)2 /10 = 19.58
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χ2 =
Lastly we need to find the p-value. This is easily found from the tables for χ2 distribution with 2
degrees of freedom.
p = P (χ2 > 19.58) = 0.0001
Since the p-value is smaller than the significance level (0.05) we can reject the null hypothesis that
the sample/data follows the claimed/hypothetical distribution.
Notice carefully the meaning of the p-value. Assuming that the Null hypothesis is true, if we carried
out the experiment many times, we will get these observations 0.01% of the time.
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Let us consider another example of the use of p-values. Consider a counting experiment where we
take the number of counts in 100 , 1 minute intervals. We know that the counting statistics should
follow a Poisson distribution. The data is as follows:
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Counts/minute (xi )
0
1
2
3
4
5
Occurrences (fi ) Expected Number of Occurrences
7
7.5
17
19.4
29
25.2
20
21.7
16
14.1
14
12.1
P
We can see that the sample mean, or the average is simply x̄ = 2.59 that is x̄ = Pxfi fi i . If we
assume that the underlying distribution is Poisson and choose the mean of the Poisson distribution as
our sample mean, we can calculate the expected number of occurrences as shown in Column 3 of Table
1.8. To calculate the reduced chi-squared, we also need the number of degree of freedom. There are
6 bins that we have with 0, 1, 2, 3, 4, 5 counts. We have used two degrees of freedom to calculate the
mean and the variance (which in the case of a Poisson distribution is the same, that is σi2 = x̄ since
we have assumed that the best guess for the mean of the underlying Poisson distribution is the sample
mean.) and thus the number of degrees of freedom is ν = 4. We can now easily calculate the reduced
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chi-squared, χ̃2 using Eq 1.81 and Eq 1.84 as
N
χ̃2 =
χ2
1 X (yobs − yth )2
=
ν
ν i=1
σi2
We find that χ̃2 = 0.35. This value is less than one and so we believe that the agreement with our
hypothesis is good and the data seems to support that the underlying distribution is indeed Poisson.
To see this using p-values, we use Appendix A for ν = 4 to see that
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Probability(χ̃2 ≥ 0.35) ≈ 85%
The p-value in this case is 0.85. Once again, this means that if the hypothesis is true, we should get
these observations 85% of the time if we carry out the experiment many times. This confirms what we
have seen namely that the agreement between experiment and hypothesis is very good.
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The p-value is the probability (measured from 0 to 1, or 0% to 100%) that the hypothesis that
the data corresponds to the model is true. You can reject the hypothesis if the p-value found for a
calculation is less than 0.05(5%) or less than 0.01(1%). For example, a p-value of 0.03 when comparing
the observed data and the theoretically expected data (Poisson distribution in our case) means that
if we carry out the experiment many times, in roughly 3% of the trials will be get data this or more
extreme than the one that we have obtained in our experiment. To emphasise, if we get a p-value of
0.01 then this means the following: that there is a 1% chance of obtaining a set of measurements at
least this different (that is this different or more different) from the model, assuming the model is true.
It does NOT mean that the probability that the model is true is 1% or that the probability that the
model is false is 99%.
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Since the concept of p-values is very important in deciding about our true distributions from our
sample, let us reiterate what exactly the steps are:
1. We carry out an experiment and get some data.
2. We suspect that the underlying distribution which would result in this data is some known distribution. This is our NULL Hypothesis.
3. We find the χ2 statistic from the data and the hypothetical distribution. Suppose this value is χ20 .
4. We find the number of degrees of freedom. This is typically the number of independent parameters
we have. Usually, if we have N variables (data points), we need to have some relations to estimate
our hypothetical distribution. Thus, we might need to find the sample mean, the sample variance
etc. For every such estimate, the number of independent variables is reduced by 1.
5. Knowing the χ20 and the degrees of freedom, we can find the reduced χ2 or χ̃20 .
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6. In the χ2 distribution, we plot this test statistic and find the probability of finding a value of χ20
or χ̃20 greater than or equal to this value. This is obviously
∞
f (χ2 )dχ2
p=
χ20
This is what we are calling our p value.
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7. We next decide on a significance level, α. This is the probability that we wrongly reject the Null
Hypothesis when it is true. Thus for a good test, we want this quantity to be small. Typically
values of α chosen are 0.05, 0.01 and 0.1 corresponding to 5, 1 and 10% significance. We can find
the value of our χ2 corresponding to these values of α. These will simply be
∞
f (χ2 )dχ2 = α
χ2α
Thus for α = 0.05, χ20.05 will be found from the above integral. Note that the values of χ2 to the
right of this value of χ2α have a probability of 0.05 to occur.
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8. Now suppose we find that χ20 > χ2α . Alternatively, if our p ≤ α, this means that the test statistic
is to the right of the value corresponding to the significance level. We can then conclude one of
two things: First, that the model is correct, that is the hypothesis is valid but we are getting such
extreme values in our sample purely by chance. The probability of this happening is obviously
α since the region to the right of χ2α has that probability. Or we can conclude that our model
is not correct or alternatively, we can reject the hypothesis. The probability of this happening
is obviously 1 − α if there are only these two possibilities. Thus we can say that we can
be 100 × (1 − α) percent confident of rejecting our hypothesis. As mentioned above, typically
α is chosen to be 0.05, 0.01 or 0.1 corresponding to rejecting the hypothesis at 95%, 99% or 90%
confidence levels respectively.
9. Finally if our χ20 < χ2α or if our p ≥ α, all we can say is that we can NOT reject our Null hypothesis
that our model describes the sample data.
One should however keep certain things in mind while using the Chi-squared test. One, we are assuming the errors in the data are Gaussian. If the errors have been under-estimated then an improbably
high value of chi-squared can be obtained. On the other hand, if the errors have been over-estimated
then an improbably low value of chi-squared can be obtained. In normal experiments, some errors can
sometimes be non-Gaussian, a model is typically only rejected for very low values of p such as 0.001.
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References
1. “Data Reduction & Error Analysis for the Physical Sciences”, D. Keith Robinson & Phillip R.
Bevongton, Mcgraw Hill (2003).
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2. “An Introduction to Error Analysis: The Study of Uncertainties in Physical Measurements”, John
R. Taylor, University Science Books (1997).
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3. “Radiation Detection & Measurement”, Glenn F. Knoll, Wiley India (2009). Chapter 3.
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Chapter 2
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RADIOACTIVITY
Learning Objectives
1. To understand the nature of radioactive decay.
2. To study quantitatively the phenomenon of radioactivity including half life, decay
constant etc.
2.1
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3. To learn about the nature and properties of different kinds of radiation in radioactivity.
Radioactivity
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Radioactivity, discovered in 1896 by Becquerel has played an important role in our understanding
of the nature of matter at the subatomic level. Radioactivity has certain characteristic features which
were inexplicable when it was discovered but can be easily explained within the context of quantum
mechanics and relativity. Thus, for instance, the fact that a radioactive nucleus can spontaneously
decay and liberate energy without any excitation from outside can only be understood by thinking of
the equivalence of mass and energy. The completely random or probabilistic nature of the decay process
cannot be understood classically but comes out naturally within the framework of quantum mechanics.
There are basically five kinds of radioactive decay as listed in the Table (2.1).
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Decay
Transformation
Example
Reason for Instability
Alpha Decay
A
A–4
4
ZX → Z-2Y + 2He
A
A
–
ZX → Z+1Y + e + ν̄
A
A
+
ZX → Z-1Y + e + ν
A
–
A
ZX + e → Z-1Y
A
A *
ZX → ZX + γ
238
234
4
92U→ 90Th + 2He
14
14
–
6C → 7N + e
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+
29Cu → 28Ni + e
64
–
64
29Cu + e → 28Ni
87
87 *
38Sr → 38Sr + γ
Nucleus is too large
Beta Decay
Positron Emission
Electron Capture
Gamma Decay
Nucleus has too many neutrons relative to protons
Nucleus has too many protons relative to neutrons
Nucleus has too many protons relative to neutrons
Nucleus has excess energy
Table 2.1: Radioactive Decays§
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§(Adapted from “Concepts of Modern Physics” by Beiser,Mahajan & Rai Choudhury)
We shall discuss these different kinds of phenomenon in some detail later.
2.1.1
Measure of radioactivity
A quantitative measure of the radioactivity of a sample is activity which is defined as the rate at
which the atoms of the sample decay. If N is the number of atoms (or nuclei) present at time t, then
its activity R is defined as
R
dN
dt
(2.1)
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R=−
Clearly, since the derivative is negative, the negative sign in the definition makes the activity a
positive quantity. The SI unit of activity is becquerel (Bq) which is defined as
1 Bq = 1 decay s−1
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The traditional unit of activity is the curie (Ci) which was defined originally as the activity of 1
gram of radium 236
88Ra, but is now defined as
1 Ci = 3.70 × 1010 decays s−1 = 37 GBq
As we shall see, the radiations which come out when a radioactive nucleus decays are ionizing in
nature and when they pass through living tissue, they can damage the tissue. Sometimes the damage
maybe slight and the body heals itself. But sometimes the damage can be severe and have long term
disastrous consequences which include cancer and other illnesses.
Radiation dose is measured in a unit called sieverts (Sv) which is defined as the amount of radiation
(of any kind) which has the same biological effect as the absorption of 1 joule of X-rays or gamma rays.
Typically, a safe exposure to radiation is taken to be about 1 milliSv per year. This does not include the
background radiation to which we are anyway exposed. The background radiation that we experience
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is from the radionuclides in the rocks and earth and buildings as well as due to cosmic rays hitting the
atmosphere. To put this in perspective, a typical X-ray exposes us to about 0.02 mSv while a CT scan,
which is basically several X-ray exposures, typically can expose us to 5 − 8 mSv.
2.1.2
Activity Law & Half Life
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When we measure the activities or the rate of decay of radionuclides, we find that the rate falls off
exponentially with time. This can be encapsulated mathematically as
R(t) = R0 e−λt
(2.2)
where the constant λ is characteristic of the radionuclide and is called the decay constant. It is
convenient to define another quantity called the half-life. T1/2 as the time in which the activity drops
to one half of its initial value. Thus, we see that by definition,
R(T1/2 ) =
or
R0
= R0 e−λT1/2
2
λT1/2 = ln 2
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and so we get
R
T1/2 =
ln 2
λ
(2.3)
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It is clear that for a radionuclide which has a large decay constant, the half life is small
and vice versa.
The empirical activity law (Eq(2.2)) can be obtained if we assume a constant probability λ per unit
time for the decay of every nucleus in the sample. Then, in time dt, the probability for decay of any
one nucleus is simply λdt. This is the probability for any one nucleus to decay. If we have N nuclei,
the number dN that will decay in time dt will be clearly
dN = N λdt
Now since the number of nuclei (of a particular kind which we had initially) is decreasing, this
expression must have a negative sign for it to make sense. Thus we get
dN = −N λdt
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(2.4)
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We can define another quantity called activity A of a radionuclide. This is simply the rate of decay
of the sample. Thus clearly
R
A=−
dN
= λN
dt
(2.5)
Integrating this expression gives us the Radioactive decay law as
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R
N (t) = N0 e−λt
(2.6)
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It is important to note that the whole phenomenon of radioactivity is statistical in nature. As noted
above, every single nucleus has a definite probability of decay but which particular one decays in a
particular interval of time is essentially random in nature. All we can say is that if we had many nuclei
present (which is always the case in any experiment that we do), the fraction that will decay in any time
period will be approximately the same as the probability of an one nucleus to decay. As we have seen in
Chapter 1, we can model this statistical phenomenon as a Poisson distribution and the probabilities will
then be given by the distribution function for the Poisson distribution. Thus if we say that a particular
sample has a half life of 1 hour, all it means is that every single nucleus in that sample has a 50%
probability or chance of decaying in 1 hour. And recall that the decay probability is constant and thus
if a particular nucleus does not decay in 1 hour, it has a 75% probability of decaying in 2 hours and
NOT a 100% probability.
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We can also define a quantity called Mean Lifetime which is simply the reciprocal of the decay
constant. Thus
T =
T1/2
1
=
= 1.44T1/2
λ
ln 2
(2.7)
Finally, a quantity which is used sometimes to describe a radioactive source is specific activity
which is defined as the activity per unit mass. The mass of the radioactive nuclide with N nuclei and
having a molecular weight (or atomic weight) M , will be given by
mass =
NM
AAv
Thus, using Eq(2.8), Eq(2.1), Eq(2.2) and Eq(2.4), we have
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(2.8)
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λN AAv
NM
λAAv
=
M
specific activity =
(2.9)
where AAv is the Avogadro’s number or 6.02 × 1023 nuclei mole−1 .
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Example 2.1.2.1
Consider a sample of 113
49In weighing 2µg with a half life of 1.6582 hours. Calculate the number of
atoms remaining in the sample after 4 hours as well as the specific activity of the sample. Assume
that the daughter nuclides are stable.
We first need to calculate the number of atoms initially in the sample. Given the mass (M ) and
the atomic weight (Aw ) of the sample, this is easily done
N0 =
M
AAv = 1.066 × 1016 atoms
Aw
We next need the decay constant λ. Knowing the half life τ or T1/2 , we know from Eq 2.3 that
ln 2
= 1.16 × 10−4 s−1
τ
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λ=
Then using the Radioactive Decay Law (Eq 2.6), we get the number of atoms after 4 hours as
N = N0 e−λt = 2.006 × 1015 atoms
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Specific Activity can be found from the activity since we know that the specific activity is simply
the activity per unit mass. Activity is given by Eq 2.5 as
A = λN = 2.32 × 1011 decays per second
Thus
Specific Activity =
A
= 1.16 × 1011 decays per second per microgram
M
We now have the definitions of quantities used to describe radioactivity.
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It is also important to remember that typically, a radioactive nuclide decays and produces a daughter
nuclide and some radiation( alpha, beta and gamma radiation). The daughter nuclide is also usually
radioactive and decays itself producing another nuclide which also may or may not be radioactive. Thus,
typically there is a radioactive series or a decay chain in which different nuclides are being produced
and are decaying. Of course, the time scales of the production of the nuclides depends on the half lives
of the parent nuclides while their decay time scales are related to their own half lives. This decay chain
can be analysed and leads to different kinds of behaviour with time. We address this in Appendix D.
2.2
Nuclear Decay
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We next turn to a discussion of the nature of particles and radiation emitted by the radionucleus
during the process of its decay by radioactivity. These we have seen can be of several types as given in
Table 2.1.
There are several facts that we know about the nucleus. Let us recall them.
1. Nuclei consist of positively charged protons and neutral neutrons. These are collectively known as
nucleons.
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2. The size of the nucleus is of the order of 1 femtometer or 10−15 m. This unit is also called a Fermi.
3. The density of nucleons is roughly the same in the inside of the nucleus and hence the nuclear
radius is proportional to A1/3 where A is the mass number. This relationship is usually written as
R = R0 A1/3 with R0 ≈ 1.2 × 10−15 m
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4. Nuclear densities are enormous- a typical nucleus will have a mass density of ∼ 1017 kg m−3 .
5. Protons and neutrons are fermions and carry spin 21 . They also possess a magnetic moment. The
unit of nuclear magnetic moment, in analogy to the Bohr magneton is the nuclear magneton
which is defined as
e~
µN =
= 3.15 × 10−8 eV T−1
2mP
and is smaller than the Bohr magneton because of the presence of the proton mass mP in the
denominator. The proton magnetic moment is µP = 2.793µN and the neutron magnetic moment
is µn = −1.913µN .
6. The nucleons experience two kinds of forces- the positively charged protons experience the normal
electrostatic repulsive force which is a long range force. Neutrons and protons between themselves
also experience another type of force arising out of ‘strong’ interactions. These forces between
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protons and neutrons are of short range. They are practically negligible at interparticle separations
of a few Fermis. Below such separations, the forces are attractive all the way down to about half
a Fermi beyond which they become repulsive. This complicated nature of the internucelon force
keeps the nucleons together . Between protons they balance out the repulsive coulomb force and
also prevents nucleons from collapsing into a much smaller size object because of the repulsive
nature of the nuclear force at very short distances.
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7. The nucleons also experience weak interactions which is also a short range force and which has a
range almost a thousand times smaller than nuclear forces.
8. The sum total of the masses of the nucleons in a nucleus is more than the mass of the nucleus.
The balance is called binding energy which can be thought of as the energy required to keep the
nucleus together. The range of binding energies is from a few MeV (as in the case of deuterium)
to more than 1.5 GeV (in the case of an isotope of Bismuth).
9. Some combinations of neutrons and protons form stable nuclei. For light nuclei, generally the
number of protons and neutrons are equal. As we go to heavier nuclei, the number of neutrons
becomes greater since neutrons only experience the short range strong nuclear force and this is
required to balance the electric repulsion of protons. Of the stable nuclei, lightest one of course is
the H-nucleus which is just a proton. As we go up from hydrogen, the light nuclei tend to have
more or less the same number of neutrons and protons
2.2.1
Alpha Decay
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10. Nuclear force is short range and nucleons interact via this force only with their nearest neighbours.
On the other hand, the electric force is present throughout the nucleus and so there comes a point
when the neutrons cannot prevent the break up of the nucleus. This is the limit of stable nuclide
which is 209
83Bi.
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Alpha particles are basically helium nuclei which are emitted by some radionuclides during radioactivity
decay. The process is
A
ZX
→
A−4
Z−2Y
+ 42He
The fundamental reason for alpha particle emission is the fact that some nuclei are too
large to be stable since the short range nuclear force cannot sufficiently counteract the
electric repulsion. Nuclei which contain more than 210 nucleons are such nuclei and they decay by
emitting alpha particles to reduce their size and thereby increase their stability. The disintegration
energy or the Q factor is basically the mass difference between the parent nuclei and the sum of masses
of the daughter nucleus and all the other decay products. That is, the Q value is the difference in the
kinetic energies T of the initial and final states. As an example, consider a simple nuclear reaction
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where a projectile a strikes a target nuclei A and produces two products, b and B. Then the Q value
for the reaction is given by
R
Q = [ma + mA − (mb + mB )] c2 = Tfinal − Tinitial
(2.10)
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Since the nucleus, both the parent and the daughter nucleus in the case of alpha decay are so much
heavier than the alpha particle, there is very little recoil of the daughter nucleus. Energy momentum
conservation then gives us the kinetic energy of the alpha particle as
A−4
Q
A
This should be obvious from Eq 2.10 and the fact that alpha decay entails a nucleus of mass number
A decaying to a daughter nucleus of mass number A − 4 and a helium nucleus of mass number 4. The
energies are small enough that non-relativistic momentum conservation can be used though for energy
conservation we clearly need to take into account the mass difference since that is the only source of
kinetic energy for the decay products. Clearly, since almost all alpha particle emitters have A > 210,
the kinetic energy of the alpha particle is roughly the disintegration energy. Typical alpha particle
energies are in the range of 4 − 6 MeV. Since the energy of the alpha particle is directly related to the
Q value, the alpha particles are monoenergetic.
K.Eα ≈
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Although a heavy, unstable nucleus can become more stable by emitting an alpha particle, the problem still remains as to how the alpha particle can escape from the nucleus. The strong nuclear forces
dominate at very short distances inside the nucleus and this leads to a potential barrier. If we model
this barrier and the alpha particle as a particle in a box, the height of the box (or the potential barrier )
turns out to be around 25 MeV. This is much more than the kinetic energy of the alpha particle. Thus,
classically, there is no possible mechanism for the alpha particle to escape from the nucleus as shown
in Figure 2.1.
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Figure 2.1: Nuclear Potential Barrier and α particle tunnelling
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However, we know that alpha particles do come out and are observed. It was Gamow who first
proposed the mechanism whereby this could be possible. Essentially, we can think of the alpha particle
tunnelling through the potential barrier, something which is permitted by quantum mechanics. Under
some very reasonable assumptions, the Gamow Theory of Alpha decay agrees remarkably well with
experimental observations regarding the behaviour of the decay constant λ with the energy of the alpha
particle. Although a detailed derivation of the relationship of λ and the energy of the outgoing alpha
particle is fairly complicated, we can attempt to give a simplified derivation of the relationship. This is
done in Appendix E.
The higher the energy of the alpha particle emitted, the higher is the decay constant or the shorter
is the half life of the parent nuclide. This is obvious since as we know from elementary quantum mechanics (for details see Appendix E) that the tunnelling probability is proportional to the energy of the
tunnelling particle. (Actually the relationship between the tunnelling probability and energy is a bit
more complicated. It turns out that the ln T ∝ −E −1/2 . See Eq E.18. ) If the tunnelling probability
is low, the decay constant will be low (Eq E.21). Typically, when the energy is more than 6.5 MeV,
the half life is of the order of a few days. On the other hand, if the energy is less than 4 MeV, the
probability of the alpha particle tunnelling through the nuclear potential barrier is very small and the
half life is very long. This is shown in Table 2.2.
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Source
Half-Life
Alpha particle Energy (MeV)
Branching Ratio
148
93 years
3.1827
100%
9
4.196
77 %
9
4.149
23 %
3
Gd
238
U
238
U
4.5 × 10 years
4.5 × 10 years
240
6.5 × 10 years
5.168
76 %
240
Pu
6.5 × 103 years
5.124
24 %
244
Cm
18 years
5.80
76.4 %
244
18 years
5.76
23.6 %
Pu
Cm
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Table 2.2: Half Life of Alpha Particle Sources§
§(Adapted from “Radiation Detection & Measurement” by Knoll )
2.2.2
Beta Decay
Another process by which an unstable nucleus can become more stable is by the process of beta decay.
This is
A
ZX
→
A
Z+1Y
+ e− + ν̄
The basic decay in this process is the decay of a neutron in the parent nuclei via
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n → p + e− + ν̄
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Initially, it was thought that the only particles produced in this reaction were a proton and an
electron. However, it was observed that the beta particles come out with a range of energies from 0
upto a maximum energy called the end point energy instead of having a monoenergetic spectrum.
This was surprising since if the only particles produced were the proton and the electron, then energy
momentum conservation tell us that the electron will be monoenergetic and that the electron and the
recoiling nucleus would be moving back to back. This was not the case. Further, the neutron, proton
and electron are all fermions and have spin 21 and hence the reaction with a neutron going to a proton
and electron would not conserve spin. Finally, as we know, the electron carries a ‘charge’ called lepton
number. The nucleons have 0 lepton number and thus producing only an electron in the decay would
violate lepton number conservation.
All these anomalies were solved by introduction of a new particle called the neutrino by Pauli.
The neutrino is assumed to be massless, neutral and carries spin 21 and lepton number of 1. Thus the
process that we have is a neutron going to a proton, an electron and an antineutrino (the antiparticle
of the neutrino). With this, all the anomalous observations could be accounted for. Thus, the Q value
of the reaction, that is the mass difference between the parent nuclei and the daughter nuclei is the end
point energy of the electron. The anti neutrino carries some kinetic energy and the actual energy of
the electron is thus the Q value minus the kinetic energy of the anti neutrino. This being a three body
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decay now (that is there are three particles in the final state), we get a continuous energy spectrum
for the electron upto the maximum of Q value. Again, since there are now three particles in the final
state, there is no reason for the electron and the recoiling nucleus to be moving back to back since the
antineutrino carries some momentum too. Finally, lepton number conservation and spin conservation
is also taken care of because of the anti neutrino.
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n → p + e− + ν̄
Figure 2.2: Energy Distribution in Beta decay of
210
Bi
§
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§(Adapted from http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/beta2.html )
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Each beta decay is characterised by a fixed Q value as we have seen above. This is normally the
case when the transition of the nucleus takes place between the ground states of both the parent and
the daughter nuclei. However, if the transition is between excited states and/or ground states, then the
spectrum will change because the Q value will be different. In practice, in some beta emitters, because
of the presence of excited states which could be populated, one gets several components with different
end-point energies.
Source
Half-Life
Endpoint Energy (MeV)
14
5730 years
0.156
32
14.28 days
1.710
C
P
36
Cl
3.08 × 10 years
0.714
45
Ca
165 days
0.252
5
Table 2.3: Beta particle Sources§
§(Adapted from “Radiation Detection & Measurement” by Knoll )
The accurate theory of Beta Decay was given by Fermi and we will not describe it here. A short
description is given in Appendix F.
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Another process which can take place inside the nucleus is positron emission. This is the conversion
of a proton into a neutron and a positron and a neutrino.
p → n + e+ + ν
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Obviously, since a proton is lighter than a neutron, this process can only take place inside a nucleus
and not for free protons. On the other hand, a free neutron does decay by emitting a proton and an
electron and antineutrino with a half life of around 10 minutes.
Figure 2.3: Energy Distribution in Beta decay of
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§
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§(Adapted from ”Concepts in Modern Physics” by Beiser, Mahajan & Rai Choudhury)
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Positron emission typically occurs in proton rich nuclei and the daughter nuclei has an atomic number less by one than the parent nuclei. There is another interesting thing about positron emission.
When a proton converts into a neutron and a positron and a neutrino inside the nucleus, the positive
charge of the atom decreases by one. To balance this, the daughter atom must get rid of one of its
orbital electrons to maintain neutrality. Thus what we have is that positron emission is only possible energetically if the parent atom is atleast as heavy as the daughter atom plus two
electron masses. What this means is that isotopes which decrease in mass by less than 2me cannot
spontaneously decay by positron emission. This also means that the Q value for any positron emission
emission process is the difference in the masses of the parent atom and the sum of the masses of the
daughter atom and 2me .
Positron Emission is used extensively nowadays in a very sophisticated imaging process called
Positron Emission Tomography . The idea here is to introduce an isotope, Fluorine-18 into a
compound of glucose (fluorodeoxyglucose). Flourine-18 decays by positron emission . The glucose compound is administered to the patient and the glucose is taken up by the cells. Concentrations of tumour
cells take up more of the glucose than normal cells. Now when the Fluorine isotope decays by positron
emission, the positron annihilates an electron in the body and this pair annihilation gives rise to 2 photons moving back to back because of energy momentum conservation. These photons detected by using
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a scintillator detector and photomultiplier tubes. The image is then reconstructed using sophisticated
algorithms. The tumor cells, because of their high concentration of the positron emitting isotope, give
rise to higher positrons which show up as different from healthy tissue.
Finally, another related process is electron capture. This is when a nucleus absorbs an inner shell
electron and a proton and the electron go to a neutron and a neutrino.
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p + e− → n + ν
Usually, since the K shell electrons are the ones which are captured, an electron from the outer
shell falls to fill the vacancy thereby releasing X-rays which are characteristic of the daughter nuclide.
Electron capture is more likely than positron emission in heavy nuclides since the inner shell electrons
are closer to the nucleus in heavy elements.
2.2.3
Gamma Decay
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We are familiar with atoms existing in excited states as well as ground state. When an electron in an
excited state returns to the ground state, a photon is emitted. In essentially the same way, a nucleus
can also exist in ground as well as excited states. When an excited nuclei returns to its ground state, it
emits photons of energies equal to the difference in the energies of the excited and ground state. We can
easily estimate this energy by using the Uncertainty Principle. A typical nucleus size is ∆x ∼ 10−15 m,
while a typical mass for a nucleon would be 1 GeV c−2 . This gives us an estimate of the energy as
~2
−10
∼ 2(∆x)
m and mass
2 m ∼ few MeV. Incidentally, a similar calculation for an atom, with ∆x ∼ 10
−2
∼ 1 MeV c for an electron would give us the energy estimate to be a few eV. Clearly, atomic transitions occur between energies separated by a few eV and those in the nucleus by a few MeV. Thus, as
we have seen, alpha and beta particles also carry energies in the MeV range.
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Electromagnetic radiation which carries a few MeV of energy is said to be in the Gamma ray region
of the electromagnetic spectrum.
To study the process of gamma decay ( that is an excited nucleus giving out a gamma ray photon
and making a transition to a lower energy state) we need to use quantum mechanics. A semi-classical
theory of gamma decay using Fermi’s Golden Rule is given in Appendix G.
In most cases, some form of beta decay results in the creation of an excited state of the daughter
nuclei. This process happens in a time scale which is characteristic of the half life of beta emitters as
in Table 2.3. However, the excited states of the daughter nuclei thus created are short lived states and
decay to the ground state by emitting gamma rays. Thus what we see typically is that a parent nuclei
emits a beta particle and gamma ray photons with a time of the order of the half life. However, and
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this is crucial, the energy of the beta particle is characteristic of the parent nuclei while that of the
gamma rays is determined by the energy levels of the daughter nuclei.
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Two typical examples of gamma decay schemes are shown in Fig(2.4) and Fig(2.5).
Figure 2.4: Decay Scheme for
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§(Source: Wikicommons)
27
Co- Gamma decay§
Mg - Gamma decay§
b
Figure 2.5: Decay Scheme for
60
La
§(Adapted from “Concepts of Modern Physics” by Beiser, Mahajan & Rai Choudhury)
For instance, in the decay of 27
12Mg shown in Fig (2.5), the half-life of the decay is 9.5 minutes and
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it can take place to either of the two excited states of 27
13Al. The excited aluminum nucleus, 13Al can
then decay by emitting one or two gamma rays and come to the ground state.
An excited nucleus can sometimes also give its energy to one of the atomic electrons around it. In that
case, the electron then is emitted with a kinetic energy equal to the nuclear excitation energy minus
the binding energy of the electron. This process, a sort of photoelectric effect for nuclear photons is
called internal conversion.
Thus we see that certain nuclei are radioactive and emit one or more of the above mentioned particles/radiation. These radiations and particles are what we use to study the properties of the parent
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and daughter nuclei. However, to detect and measure the properties of these radioactive emissions, we
need to have them interact with matter. This is what we shall now turn to.
2.3
References
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1. “Radiation Detection & Measurement”, Glenn F. Knoll, Wiley India (2009).
2. “ Concepts of Modern Physics”, Arthur Beiser, S. Mahajan & S. Rai Choudhury, Mcgraw Hill
(2015).
2.4
Questions
1. Why are some nuclei stable and others unstable?
2. What are the various reasons for nuclei to be unstable?
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3. What is the activity law in radioactivity? What is the essential assumption in obtaining the empirical activity law?
4. What are the different forces operative inside an atomic nucleus? What are their
properties? How do they explain the stable or unstable nature of a nucleus?
5. What is Binding energy of a nucleus?
b
6. What is an alpha particle? Why do nuclei decay by emitting an alpha particle?
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7. What are the typical energies of an alpha particle emitted by a radioactive nucleus?
Why?
8. The alpha particle emitted by a particular nucleus are mono-energetic. Why is this
the case?
9. How does an alpha particle come out of the nucleus given that there is a potential
barrier for it to come out? What is responsible for the potential barrier?
10. How is the energy of the alpha particle related to the half life of the nuclide? Why?
11. What are beta particles and by what process are they produced?
12. Why does a nucleus decay by emitting beta particles?
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13. Given that the neutron in a laboratory decays with a half life of about 10 minutes,
why don’t all the neutrons in the nuclei decay?
14. Are beta particles mono-energetic? If not, why not?
15. How do we know that another particle apart from the beta particle must be coming
out in beta decay?
16. Why does positron emission happen in some nuclei and not in others?
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17. What is electron capture? What is the characteristic signal for electron capture?
18. Why is electron capture more likely in heavier atoms than in lighter atoms?
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19. What are gamma rays and what is the process by which they are produced?
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Chapter 3
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INTERACTION WITH MATTER
Learning Objectives
1. To understand the concept of cross section.
2. To study the interaction of a heavy charged particle with matter.
3. To derive the formula for energy loss by a heavy charged particle in its interaction
with matter.
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4. To study the interaction of electrons with matter.
5. To derive the energy loss formula for electrons moving through matter.
Introduction
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3.1
b
6. To study the different ways in which radiation interacts with matter.
The experiments that we perform in this laboratory are all concerned with the detection and measurement of radiation and particles, namely gamma rays and beta particles (We do not carry out experiments
with alpha particle emitters in this laboratory). Clearly, we need detectors for this purpose- in our laboratory, we use two kinds of detectors. Geiger-Muller counters (GM counters) and Scintillation counters.
We shall be studying in detail about the workings of these detectors in a later Chapter. In this Chapter,
we would like to understand in general how particles and radiation interact with matter. After all, if
radiation or a particle is to be detected, it must interact with the material of the detector. This might
be a gas, a liquid or even a solid. Before we do this, let us define some terms which we shall be using.
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Cross Section
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Cross section is a way to express the probability of interaction. Consider a beam of incident particles
impinging on a target material. We assume that each particle in the target material has a certain area,
which we call cross section. Any incident particle which passes within this area will interact with the
target particle. Clearly, the larger the area, the larger is the possibility of interaction. Of course, this
cross section which we think of as an area of influence in a way, depends on the nature of the process,
the nature and energy of the incident particle etc. In principle, it could be very different from the
geometric cross section.
We know that flux of a beam is defined as the number of particles crossing a unit area in unit time.
Suppose now we have a slab of some target material with area A and thickness ∆x. Let the number of
atoms per unit volume of the target be n, that is the total number of nuclei or atoms in the slab are
nA∆x. If each nuclei has a cross section of σ , then the aggregate cross section for the slab is σnA∆x.
Now consider a beam of incident particles, N of them hitting the slab. The number ∆N which will
interact with the nuclei in the slab will be
∆N
Aggregate cross section
nA∆xσ
=
=
= nσ∆x
N
Target area
A
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since one incident particle is assumed to interact only once with any nuclei, and get deflected, it is
removed from the beam. Then for a finite slab thickness, we have
N
dN
N
= −nσdx
dN
N
= −nσ
x
0
−σnx
N0
N = N0 e
(3.1)
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dx
We see that the number of surviving particles decreases exponentially with the slab thickness. Recall
that this is exactly what we see when we pass a beam of light through some absorber. The intensity
falls exponentially with thickness.
We can think of the cross section as in Figure (3.1) where an incident beam of flux F hitting a target.
dσ
The differential cross section is dΩ
(θ, φ).
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Figure 3.1: Differential Cross Section§
§(Source: ”ScatteringDiagram” by Original uploader was JabberWok at en.wikipedia - Transfered from en.wikipedia.
Licensed under CC BY-SA 3.0 via Wikimedia Commons
- https://commons.wikimedia.org/wiki/File:ScatteringDiagram.svg#/media/File:ScatteringDiagram.svg)
Then,
(3.2)
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1 dNs
dσ
(θ, φ) =
dΩ
F dΩ
dσ
dNs = F dΩ
(θ, φ)
dΩ
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We should think of this equation as the number of particles that scatter, Ns , into a portion of solid
angle per unit time is equal to the flux of incident particles per unit area per unit time multiplied by
the probability (represented by a cross section area) that would scatter into that portion of solid angle.
We can integrate Eq(3.2) over all solid angles to get the total cross section σ. The usual unit for cross
section is a barn which is defined as
1 barn = 10−28 m2
Please remember that this discussion of cross section, though in the context of scattering, is equally
applicable for absorption of incident particles or radiation by matter.
If Nt is the number density of the target particles, then the probability that a single interaction occurs
through a volume with thickness ∆x is simply Nt σ∆x. A convenient parameter in this discussion is the
mass thickness. This quantity is often more convenient to use instead of thickness. Thus
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R
mass thickness = ρ∆x gm cm−2
(3.3)
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This is convenient because it tells us directly the effect of the absorber on the incident beam. Thus,
for instance, a beam travelling through 2 gm cm−2 of air (of density ρ = 0.012 gm cm−3 ) has the same
effect as the beam passing through 2 gm cm−2 of water, even though it passes through 1.67 m of air
and just 2 cm of water. Essentially, what we have done is to factor out the density of the absorber and
encapsulate that in the mass thickness.
We are interested in the passage of alpha, beta and gamma radiation through matter. Thus we can
divide our discussion into two parts- interaction of charged particles with matter and interaction of
radiation with matter. For charged particles, once again, we need to consider two cases of passage of
heavy particles (like alpha particles) and the passage of electrons or beta particles.
3.2
Interaction of Charged Particles with Matter
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When a charged particle interacts with matter, two things happena) the particle loses energy traversing matter and
b) particle is deflected from its initial direction.
In general, there are two main processes which cause this. These are inelastic collisions with atomic
electrons in the material and also elastic scattering off the nuclei. There are some other processes which
contribute to the energy loss namely, Bremsstrahlung and nuclear reactions which are extremely rare.
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As mentioned above, the processes which are dominant causes of energy loss are different for light and
heavy charged particles. For heavy charged particles it is basically the inelastic collision with atomic
electrons which are responsible for energy loss.
3.2.1
Interaction of Heavy charged particle with matter
Consider a charged particle like an alpha particle entering a medium. The atoms of the absorbing
medium will have atomic electrons which will interact with the incoming charged particle. It is important to remember that the alpha particle interacts simultaneously with many atomic electrons. In
this process, some energy is transferred to the atomic electron and depending on the amount of energy
and the nature of the absorber, the atomic electron either gets into an excited state or in some cases,
may even become free, that is, the atom gets ionised. This energy transferred is of course exactly the
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energy that is lost by the incoming particle. However, as we shall see below, the maximum amount of
energy that is transferred in such a collision is very small compared to the kinetic energy of the incoming particle. This means that the incoming particle continuously loses small amounts of energy as it
passes through the absorber. However, the number of such encounters in any macroscopic length is large.
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Let us consider one such encounter in some detail. Consider a particle of mass M and velocity V
incident on another particle at rest of mass m, M m. This could be the case, for instance of an alpha
particle colliding with an electron. Classically, using non-relativistic energy momentum considerations,
we know that
1
1
1
M V 2 = M V12 + mv22
2
2
2
M V = M V1 + mv2
where V is the initial velocity of the incoming particle, V1 is the velocity of the alpha particle after
the collision and v2 is the velocity of the electron after the collision. Solving these two equations for the
velocity after collision of the incoming particle, V1 , we get
M −m
V
M +m
Thus, the loss of energy of the incoming particle is
V1 =
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1 4M mE
∆E = M V 2 − V12 =
2
(M + m)2
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where E is the initial kinetic energy of the incoming particle. Thus, we see that in the case when
M m, we get that the maximum energy transferred to the electron is 4m
E. This justifies our stateM
ment above that in the case of alpha particles (or protons) and electrons, each collision only diminishes
1
of the initial energy. It also
the incoming particle’s energy by a small amount, in this case about 2000
implies that the electron’s velocity after the collision can be at most 2V since the electron was at rest
initially and all this energy ∆E is the electron’s kinetic energy after collision.
Given that in any one collision, the energy transferred is small, we can also see that the momentum
P
transferred in any one collision, ∆p ∼ m(2V ) ∼ 2m M
which is again small since M m where P is
the initial momentum of the incoming heavy particle. Thus we can assume that the incoming particle
does not get deflected in any one collision and continues along the undeflected.
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Figure 3.2: Heavy particle in matter
Consider a heavy particle of mass M and charge ze entering a cylinder of absorber material with a
velocity v along the x direction. Consider a cylinder of radius b and length dx placed along the x axis as
shown in Fig 3.2. Let us look at the interaction of the heavy particle with a single atomic electron at a
distance b from the path of the incoming particle. This interaction is the Coulomb interaction between
the heavy particle and the electron and so we need to find out the field at the surface of the cylinder,
which is the location of the electron. The field is easily computed using Gauss’s Law as the symmetry
considerations tell us that the x component vanishes.
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ze
0
ze
Ey 2πb dx =
0
2ze
Ey dx =
4π0 b
k2ze
=
b
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b
Ey dA =
where k =
1
.
4π0
Consider the impulse produced on the electron as a result of this Coulomb force.
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I =
Fy dt
= e
Ey dt
= e
dx
v
2ze2 k
vb
(3.5)
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=
Ey
using Eq(3.4).
Now the energy gained by the electron (which is at a distance b from the incoming particle) is simply
I2
2me
4z 2 e4 k 2
=
2v 2 b2 me
2z 2 e4 k 2
= 2 2
b v me
∆E(b) =
(3.6)
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This is the energy loss to a single electron. We need to find out the energy loss of the incident
particle when it travels a distance dx in the absorber material. For this we need to add the energy lost
to all the electrons in the annular thickness of the cylinder between b and b + db. If Ne is the electron
density in the material, then this energy loss is given by
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b
−dE(b) = ∆E(b)Ne dV
= ∆E(b)Ne 2πb db dx
2z 2 e4 k 2
= 2 2 Ne 2πb db dx
b v me
4πz 2 e4 k 2
db
=
Ne
dx
2
me v
b
(3.7)
To find the total loss, we need to integrate over values of b and thus we get
dE
4πz 2 e4 k 2
−
=
Ne ln
dx
me v 2
bmax
bmin
(3.8)
Clearly, the values of bmax and bmin will be determined by physical considerations. These are easily obtained. Consider the maximum energy lost by the incoming particle (or the maximum energy
2
transferred to the electron). This, we have already seen is given by me (2v)
. This will happen when the
2
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impact parameter is the minimum. Thus we have, using Eq(3.6),
Tmax = 2me v 2 =
2z 2 e4 k 2
b2min v 2 me
(3.9)
where Tmax is the maximum energy that can be transferred.
bmin =
kze2
me v 2
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Now, for the maximum impact parameter, bmax , clearly, if the distance is very large and the energy
transfer is smaller than the ionization or excitation energy Ie , then no energy is transferred. Thus,
again using Eq(3.6), we get
2z 2 e4 k 2
b2max v 2 me
Ie =
Thus we have for the energy loss
R
dE
4πk 2 z 2 e4
=−
Ne ln
dx
me v 2
Tmax
Ie
4πk 2 z 2 e4
=−
Ne ln
me v 2
2me v 2
Ie
(3.10)
(3.11)
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since as we have seen Tmax = 2me v 2 . This is the classical formula for the energy loss of
a charged particle in an absorbing material as given by Bohr. Of course this formula
assumes that classical, non-relativistic, particle must be heavy compared to me , that the
interaction time is short thus the electrons are “stationary” and also does not account for
binding of atomic electrons. We can find the number density Ne of the electrons in the material as
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Ne =
ZρNA
AmN
where ρ is the density of the absorber material with atomic weight A and atomic number Z and NA
is the Avogadro’s number.
The classical formula of Bohr was extended by Bethe and Bloch to include relativity and other effects
like density effects and the effect of atomic electrons etc. When one uses relativity to find the maximum
energy transferred (and hence bmin ) by a particle with mass M interacting with a particle of mass m,
we get instead of Eq(3.9),
Tmax =
2mβ 2 γ 2
m
1 + 2γ M
+
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(3.12)
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where γ = √ 1
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1−β 2
and β = vc .
The basic result is the Bethe-Bloch formula
R
dE
4πk 2 e4 z 2 ρZNA
=−
B(v)
dx
me c2 β 2 A
where
B(v) = ln
2me v 2
Ie
− ln(1 − β 2 ) − β 2
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(3.13)
(3.14)
Clearly, as we have seen, for non-relativistic particles (that is β 1), only the first term in the factor
B(v) is significant .
This is the basic formula to estimate the energy loss by a charged particle in matter. We can rewrite
the Bethe-Block formula in another way. Remember that the classical electron radius re is defined as
re =
ke2
m e c2
In terms of re , we can rewrite Eq(3.13) as
R
z2 Z
1 dE
= 4πre2 me c2 NA 2 B(v)
ρ dx
β A
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−
(3.15)
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The quantity Ie in the Bethe-Bloch formula is the mean ionisation energy of the material. This is
usually determined empirically though one can find it out in principle by taking the average over all the
ionisation and excitation processes in the atom. Experimentally, it is usually found to be Ie ∼ 10Z eV.
Thus for instance, in air, Ie ∼ 85 eV while in aluminum, it is Ie ∼ 160 eV.
Let us see what this equation for the energy loss by a charged particle in matter tells us in general.
Z
∼ 12 except for hydrogen for which it is 1. Thus we can
Note that most materials have the same A
2
immediately say that apart from the corrections due to B(v), the energy loss depends on βz 2 since all
the other factors are constants. That is, for a given non-relativistic particle, the energy loss
varies inversely with particle energy since it varies inversely with v 2 . This can be easily
understood since if the particle has higher energy, it spends less time near any one electron
and hence the impulse on the electron is less and the energy transfer is lower. Slow moving particles lose more energy and as their momentum increases and thus their velocity
approaches c, we expect a flattening of the dE
curve. We can see this in the energy loss
dx
curves for various charged particles as shown in Fig 3.4. The second point is the variation with
z 2 . Energy loss depends directly on z 2 . Thus alpha particles with the greatest charge will have
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the highest energy transfer (among particles with the same energy or velocity). Finally,
when we compare different absorbers, the quantity which is entering the energy loss is the
product N Z. This obviously is a measure of the electron density in the absorber. Hence,
materials with high N Z or higher electron density will have a higher stopping power or
energy loss for a given beam.
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Another interesting thingabout the Bethe-Bloch formula is the logarithmic term. The first term in
2
the factor B(v) is ln 2mIeev which shows that there is a slow rise in the energy loss with increased
momentum or velocity. The reason for this is actually the behaviour of a relativistic particle’s electric
field as the velocity increases. We know that the electric field of the incoming charged particle becomes
more and more squashed as the velocity increases. The reason for this is that the transverse field, Ey , is
larger by γEy relative to the isotropic case of the charge at rest while the longitudinal field Ex decreases
by a factor of γ12 relative to the isotropic case as shown in Figure 3.3. The typical whiskbroom pattern
of the electric field of a moving charge is shown in the Figure 3.3. This leads to a slow increase in
ionization at farther and farther distances from the particle track.
Figure 3.3: Electric field of moving charge
§
§(Source: http:// sernam.ru/ lect f phis6.php? id=65 )
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From Figure 3.4, we see that the typical specific energy loss at the minimum ( for most particles
except the electron) is around 2 MeV cm2 gm−1 . Since most relativistic particles have very similar
behaviour (in terms of energy loss as a function of velocity or energy), we refer to these relativistic
particles (βγ > 3) as minimum ionizing particles.
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Figure 3.4: Specific energy loss in air versus energy of charged particles§
§(Source: Review of Modern Physics, Vol 24, page 273 )
b
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The Bethe-Bloch formula is a good guide to understand the energy loss of charged particles in matter.
However, there is another factor which comes into play as the charged particle moves through matter.
This is reduction in the effective charge. To understand this, consider Figure 3.5.
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Figure 3.5: Energy loss of Alpha particles of energy 5.49 MeV§
§(Source: Wikicommons)
This kind of plot of the energy loss of a particle in a medium is called a Bragg curve . There
are several things about this figure that one needs to understand. Firstly, as the alpha particles move
through matter, we expect from the Bethe-Bloch formula that the energy loss increases as E1 . We see
this in the initial part of the plot. However, as the particle keeps moving in the material, its velocity
and hence energy decreases and the energy loss increases to a maximum near the end of the track.
Just before the particle comes to a complete stop, the energy loss reaches a maximum which is called
a Bragg peak . At that point, we see that the energy loss sharply falls off. This is because the alpha
particles are now travelling slow enough that they can pick up electrons from the material and thus the
effective charge is reduced. We expect that charged particles with the maximum nuclear charge will
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pick up the electrons earlier than those with less nuclear charge. This is indeed seen as we can see in
Figure 3.6.
Figure 3.6: Energy loss for helium & hydrogen ions.§
§(Source: B.Wilken, T.A. Fritz, Nuclear Instrumentation Methods, 138, pp 331 (1976))
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Incidentally, the energy loss curve that we have discussed is for charged particles only. For photons,
for instance γ rays or light, the fall in intensity as we know is exponential. The energy loss curve is of
fundamental importance in medicine also. Proton beams of a fixed energy from a linear accelerator are
used to treat certain kinds of tumours. As we have seen, the maximum energy loss will occur at the end
of the proton trajectory (the Bragg peak) and it is a sharp maximum. To increase the target tumour
volume for the protons, the monoenergetic proton beam is widened to a spectrum of energies which
lead to a broadening of the Bragg peak thereby increasing the effect of the treatment on the tumour as
shown in Figure 3.7
Figure 3.7: Radiation dose produced by a monoenergetic proton beam and a modified beam.§
§(Source:CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=469545 )
Finally, we consider the phenomena of straggling. When we pass a beam of charged particles of a
fixed energy (instead of a single particle) through different materials and study the number transmitted,
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we find an interesting feature. We see that energy loss is not continuous but statistical or stochastic
in nature and some particles undergo less/more energy loss and their range will be larger/smaller than
the typical, expected range. This is known as straggling.
Figure 3.8: Energy Straggling§
§(Source: ‘DEVELOPMENT And IMPLEMENTAZIONE IN C++ OF ALGORITHMS FOR The CALCULATION OF PLANS
OF TREATMENT IN ADROTERAPIA’ Giovanni Nicco, Bachelor’s Thesis, University of Turin, 2000. )
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As we can see from Figure 3.8, the variation in the amount of collisions/energy loss is approximately
Gaussian (because of the central limit theorem). We define the mean range as the distance at which
50% of the particles are transmitted, as shown in Figure 3.8. We can find the range of transmission by
extrapolation to the distance at which we expect zero transmission as in the Figure.
Interaction with matter of electrons
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b
The discussion above is strictly valid for heavy charged particles and hence will be a good description
for the energy loss for a beam of alpha particles. However, things are different for beta particles.
In our discussion of the Bethe-Bloch formula, we had assumed that the incoming charged particle is
much heavier than the atomic electron in the absorber and therefore does not suffer any significant deviation from its straight line path. However, when one is interested in the interaction of beta particles
or electrons with matter, this assumption clearly is invalid. Here, the beam and the target particles are
identical and hence large deviations can be expected from the beam path. In addition, the interaction
between the beam and the nucleus of the absorber can lead to sudden changes in the path. The target
and the beam particles are identical and indistinguishable and this needs to be taken into account.
Finally, at the energies that we are interested in, namely nuclear energies (remember the beta particles
are in the MeV range), the electrons are always relativistic.
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The above differences relate to the considerations of energy loss by collisions with atomic electrons
as in the case of alpha particles. However, since the electrons are light and therefore suffer substantial
acceleration (or deceleration) because of their interaction with the absorber, they emit radiation as we
know from the well known Larmor formula in classical electrodynamics. The emission of radiation because of deflections (and hence acceleration) leads to energy loss. These are of two kinds: Cherenkov
radiation and Bremsstrahlung. To find out the energy loss of an electron beam in matter, we need
to take into account all these processes.
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To find out the energy loss, we need to modify the Bethe-Bloch formula to take into account all the
above mentioned differences between a heavy particle interaction and electron interaction. Recall the
Bethe-Bloch formula Eq (3.13) which we had for heavy particle interaction with matter,
dE
4πk 2 e4 z 2 ρZNA
=−
B(v)
dx
m e c2 β 2 A
with
B(v) = ln
2me v 2
Ie
− ln(1 − β 2 ) − β 2
This gets modified in the case of electrons to
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4πk 2 e4 z 2 ρZNA
dE
= −
B(v)
dx
me c2 β 2 A
2 p
p
me v 2 E
1
2
2
2
2
− (ln 2)(2 1 − β − 1 + β ) + (1 − β ) +
1− 1−β
B(v) = ln
(3.16)
2I 2 (1 − β 2 )
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This is a much more complicated formula for energy loss due to collisions of electrons in a beam
with atomic electrons in an absorbing material. Another interesting difference is in the straggling
for electrons as compared with heavy charged particles. Since the electron mass is small, there is a
significant fractional energy loss in each collision than for heavier particles. Hence we see a lot more
straggling for electrons, Figure 3.9.
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Figure 3.9: Energy Straggling for β particles§
§(Source: www.physics.queensu.ca/ ∼phys352/ )
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Apart from the collisions suffered by the beta particles, they also radiate and hence lose energy
because of the two processes mentioned above. Cherenkov radiation is the electromagnetic shock wave
generated by a particle in a medium when it moves faster than the velocity of light in that medium.
That is when the velocity of the particle, v > nc where n is the refractive index of the material. This radiation, which is the familiar blue radiation that one sees in pictures of nuclear reactor’s water shield, goes
as λ12 and thus peaks at small wavelengths or in the ultraviolet. In the case of electrons in a medium, the
loss due to Cherenkov radiation is small, around 1% and so we will ignore this. As it turns out, in the full
Bethe-Bloch formula for energy loss, this radiative loss due to Cherenkov radiation is already accounted.
b
The main radiative loss for an electron is due to Bremsstrahlung. This braking radiation is what
the electron emits when it is accelerated due to the strong nuclear electric field, though electron-electron
Bremsstrahlung is also possible. One can use the classical theory of radiation by an accelerated charge
(Larmor’s formula) to calculate the radiative loss due to this process.
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The treatment of the radiation loss in any collision process is easy to analyse in the case when the
particle is non-relativistic. This may be the case of an alpha particle, an ion or an electron. When
the particle passes near a target atom, it experiences an acceleration due to the field of the target.
We can think that the particle does not experience this field when it is far away from the target and
only when it is close to it. In effect, it receives an impulse or experiences a force for a short time.
(remember that impulse is the product of force and time that it acts). The time scale of this impulse, τ
can be estimated to be τ ≈ vb0 where b is the impact parameter and v0 is the incoming particle velocity.
Of course the direction of the force and hence the impulse will be determined by various factors but
one can assume for simplicity that on the average it is perpendicular to the initial velocity of the particle.
Now we know that an accelerated charge radiates energy and the power radiated from it in the
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non-relativistic case is given by Larmor’s formula
P =
q1 2a2
4π0 3c3
(3.17)
where q1 is the charge of the incoming particle, and a is its acceleration. To determine the acceleration, we can simply use the electrostatic force acting between the target atom and the incoming particle
as
F
q1 q 2
=
m1
4π0 m1 b2
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a=
(3.18)
where q2 is the charge of the target atom and m1 is the mass of the incoming particle and b is the
impact parameter. This allows us to get an estimate of the radiated energy since we know the time for
which this impulse is acting and therefore the charge is radiating , that is τ . This gives us
W ≈ Pτ ≈
q14 q22
2
q1 2a2 b
=
2
3
3
4π0 3c v0
(4π0 ) 3m1 v0 b3 c3
(3.19)
Clearly, this is the energy radiated by the incoming charge in one collision with impact parameter
b. To get the total energy radiated per unit length of the target, we need to integrate over the whole
range of impact parameters and multiply by the number density of target atoms, n2 .
bmax
= n2
q14 q22
2
2πb db
2
3
(4π0 ) 3m1 v0 b3 c3
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dW
dl
bmin
(3.20)
bmax
q14 q22
4π
1
= n2
2
3
3
(4π0 ) 3m1 v0 c b bmin
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Here it is important to point out the difference from the case of energy loss by collision that we
discussed in the alpha particle case, that is Bohr’s formula for the energy loss, Eq.3.11 where we could
not take the maximum limit of b, that is bmax to be ∞ because of the lograithmic dependence. In this
case, bmax can be taken to be infinity. On the other hand, for bmin we need to be careful and cannot
take it to 0. Instead, we use the minimum value of b to be where the wave nature of the incoming
particle becomes important, that is the de-Broglie wavelength of the incoming particle. Thus we take
bmin =
~
m1 v0
With this approximation, and taking bmax = ∞, we get the energy lost per unit length by radiation
as
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R
dW
q 4 q 2 4π 1
= n2 1 2 3 3
dl
(4π0 ) 3c m1 ~
(3.21)
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This non-relativistic expression allows us to say several things about radiation loss due to Bremsstrahlung
in collisions. First, notice that the incoming particle velocity cancels out but the mass of the incoming
particle enters in the denominator. This means that the loss for electrons is much more than that for
alpha particle or ions. Secondly, there is however the fact that if the incoming alpha particle collides
with a free electron, the free electron will be accelerated and radiate energy. The energy loss will be
exactly the same as in Eq 3.21 except that q1 and q2 will be interchanged and m1 will be
replaced by m2 .
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From Eq 3.21 it is also obvious that targets of heavy nuclei, that is with high Z will cause a higher
radiative loss because q2 will then be Ze and the loss is ∝ Z 2 e2 . Also, one might think that in case of
incoming electrons interacting with the target electrons, we would get radiation from both the incoming
and target electrons. It turns out electron-electron collisions do not produce significant bremsstrahlung
because both the incoming and target electrons experience equal accelerations in opposite directions
and thus produce radiation which is out of phase and thus interferes destructively to cancel each other.
This is the case for non-relativistic electrons. In the case of relativistic electrons, this is not the case
and one needs to take this into account. Also note that for the case of electron-positron collisions, this
cancellation does not occur and therefore even for the non-relativistic case, we need to take it into
account.
We can write Eq 3.21 more instructively in terms of constants like the fine structure constant α =
2
and the classical electron radius re = 4π0em2 c2 .
e2
4π0 ~c
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dW
4π
= n2 Z 2 me c2 α re2
(3.22)
dl
3
The important question that we need to address is about the relative importance of the energy lost
due to radiation versus the collisional energy loss that is given by the Bethe-Bloch formula. If we
take non-relativistic particles, we need to compare the loss due to bremsstrahlung to the loss due to
collisions. We know that the loss due to collisions is given by Eq 3.11
dE
4πk 2 z 2 e4
=−
Ne ln
dx
me v 2
2me v 2
Ie
(3.23)
In this expression, the dimensionless argument of the logarithm can be taken as Λ. Then we can
write Eq 3.23 as
dE
4πk 2 z 2 e4
=−
Ne ln Λ
dx
me v 2
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(3.24)
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To find the relative importance of these two energy losses, we need to take the ratio of the energy
loss per unit length due to the two processes. Thus
dW
me v02 1
= z 2 Za α
dE
m1 c2 3 ln Λ
(3.25)
where Za is the atomic number of the target nucleus, and therefore Ne = n2 . m1 is the mass of the
incoming particle and α is the fine structure constant.
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Eq 3.25 allows us to see that for non-relativistic case, bremsstrahlung is never important. In the
best case of electrons interacting with the heaviest nuclei, Za = 92, the factor Za α ≈ 0.67. However,
v02
1
and
are always less than one. In the case of heavy particles like alpha particles impinging on a
3 ln Λ
c2
target, the mass of the incoming particle m1 in the denominator makes the ratio very small and thus
we can conclude that the energy loss due to radiation for alpha particles is very small. We might think
that the heavy alpha particles could accelerate the atomic electrons which could radiate. However,
note that this loss can never exceed the total energy transferred by the alpha particle to the electrons
since that energy transfer is shared by collissional loss and bremsstrahlung. This can also be seen if we
take the energy loss by bremsstrahlung expression (Eq 3.21) and change m1 and m2 and q1 and q2 and
take the ratio of this energy loss and the collisional energy loss. We see that the denominator in the
equation corresponding to Eq 3.25 will be me and thus will cancel the me in the numerator. Thus we
can conclude that in the non-relativistic case, energy loss by radiation of a charged particle
passing through matter is negligible compared to the loss due to collisions.
This situation changes in the relativistic case. The calculation for the energy loss in the relativistic
case is quite complicated. It turns out that the energy loss for the relativistic case is given by
dE
dx
r
N EZ 2 e4 α
=
m2 c4
4
2E
−
4 ln
mc2 3
(3.26)
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Clearly then, the total energy loss is the sum of the energy loss due to collisions (Eq(3.16)) and that
due to Bremsstrahlung (Eq (3.26)). That is
dE
=
dx
dE
dx
+
c
dE
dx
(3.27)
r
If we compare the two expressions for energy loss of beta particles, Eq(3.16) and Eq(3.26), we notice
some important features. Firstly, the collisional energy loss increases logrithmically with the energy E
and linearly with the atomic number of the absorber, Z while the energy loss due to radiation increases
linearly with the energy E and as the square of the atomic number Z. The radiative energy loss also
has a mass factor in the denominator which implies that the radiative energy losses are most for lighter
particles (like beta particles) than for heavier particle (like alpha particles). Also radiative energy losses
are most for high energy particles and in materials of high Z.
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This also allows us to define a critical energy, Ecrit for which
dE
dx
=
c
dE
dx
r
An approximate formula for Ecrit given by Bethe & Heitler is
1600mc2
Z
The value of the critical energy for various materials is given in Table 3.1.
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Ecrit ≈
Material
Ecrit (MeV)
Cu
24.8
Pb
9.51
air (STP)
102
plastic
100
water
92
Table 3.1: Critical energy for various materials
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As we have discussed above, the electron being of the same mass as the scattering particles (atomic
electrons), can suffer large deviations from its original path. Thus, if one has a beam of mono-energetic
electrons from a source and we pass it through an absorber, even a thin absorber can lead to a loss of
electrons from the detector. This scattering therefore means that the intensity of the transmitted beam
drops immediately and goes to zero as the absorber thickness is increased.
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A corollary to this random motion of the electron is therefore that the range of the electron is hard
to define since it might have travelled a much larger distance within the absorber than simply the
thickness of the absorber. What is normally done is to extrapolate the range from the linear portion of
the graph.
To summarise then, we can say that the energy loss of electrons is much smaller than that of heavy
charged particles of the same energy. This means that they have a much larger range. What is observed
experimentally is that for a wide variety of absorber materials, the product of the range and the density
of the absorber is a constant for any particular electron energy.
The situation is very different for the beta particles emitted by a radioactive source. This is because,
as we have seen in Section 2.2.2 in Fig 2.2, the energy spectrum of the beta particles is continuous.
What is seen therefore is that the beta particles at the lower end of the spectrum are absorbed even
with a very thin absorber. However, for the most part of the spectrum, the transmission of the beta
particles shows an exponential decrease with thickness. This is an experimental fact which cannot be
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derived easily from fundamental physics. What we see is that the counting rate (or intensity) falls
off exponentially with an attenuation coefficient which depends on the end point energy of the beta
particle.
C = C0 e−nd
(3.28)
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where C is the counting rate with the absorber material, C0 is the counting rate without the absorber
and d is the mass thickness in units of mass per unit area. The coefficient n is the attenuation coefficient.
This behaviour is shown in Fig 3.10.
Figure 3.10: Range-Energy for 1.17 MeV beta particles from
210
Bi in Al absorbers§
3.2.3
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§(Source: ‘Radiation Protection’, course at Oregon State University Extended Campus )
Interaction of gamma rays with matter
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The interaction of electromagnetic radiation like gamma rays with matter is fundamentally different
than what we have studied so far. The interaction in this case is between radiation and the charged
particles like atomic electrons unlike between two charged particles that we have been investigating
so far. The interaction of electromagnetic radiation with matter depends on its frequency (and hence
energy). Gamma rays, as we have seen in Section 2.2.3 have typical energies in the MeV range since their
origin is in the de-excitation of the nucleus. In general, the interaction of electromagnetic radiation with
matter can be of four kinds depending on the energy of the radiation. These are Rayleigh Scattering,
Photoelectric Absorption, Compton Scattering and Pair Production. The relevant photon
energies for which these processes are operative are given in Table 3.2 ( EI is the ionization energy of
the atom).
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Rayleigh Scattering
Photoelectric Absorption
Compton Scattering
Pair Production
hν < EI
hν > EI
hν ∼ me c2
hν > 2me c2
∼ eV
∼ keV
∼ MeV
≥ MeV
Visible
X-rays
γ rays
Hard γ rays
Table 3.2: Interaction of Photons with matter
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The fact that the three processes listed above are dominant in different energy regimes can be clearly
seen if we plot the variation of the cross section for the three processes with the gamma ray energy.
This is shown in Fig 3.11.
Figure 3.11: Photon interaction cross sections for uranium as a function of photon energy. Solid line, PE absorption; dashed line, Compton scatter; dash-dotted line, pair production; dotted line, total attenuation§
§(http://rsif.royalsocietypublishing.org/content/7/45/603 )
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To understand these processes of course one needs to develop a fully relativistic theory of the interaction of photons and electrons, which is known as quantum electrodynamics. However, for our purposes,
a simple classical description is sufficient to provide some insights into the mechanisms. We think of
the atom as a dipole with the electron in the atom attached to the nucleus by a spring. The electron
is oscillating with a frequency ω0 around its mean position. This ‘natural’ frequency of oscillation can
be modelled by a linear restoring force −kx on the electron, with the ‘spring constant’ k and ω0 being
related as
ω02 =
k
me
Of course, k is related to the attractive electric force between the electron and the nucleus and is
related to the binding energy of the electron. Now when the photon or electromagnetic wave is incident
upon such an oscillating dipole, the electric field of the wave exerts an added force on the electron,
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−eE(t) where E(t) is the oscillating electric field of the electromagnetic wave given by E(t) = E0 sin ωt
where ω is the frequency of electromagnetic wave. The equation of motion for the electron is then
me ẍ = −kx − eE(t)
e
ẍ + ω02 x = − E(t)
me
(3.29)
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The harmonic solutions to this equation of a forced oscillator are well known as x(t) = A sin ωt which
we can substitute in Eq(3.29) and get
e
E0
me
1
e
E0
A =
2
ω 2 − ω0 me
1
e
x(t) =
E0 sin ωt
2
2
ω − ω0 me
(ω02 − ω 2 )A = −
(3.30)
We know from classical electromagnetic theory that an accelerated charge radiates energy. The
power radiated is given by Larmor’s formula as
2 e2 2
a
3 c3
where a is the acceleration. In our case, the time averaged acceleration squared is simply (the only
time dependent quantity is
sinωt which time averages (over a period) to 21 .
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P =
2
a
=
ω2
e
E0
2
2
ω − ω0 me
2
1
2
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which gives us the radiated power as
1
P =
3
e2
me c2
2
ω4
cE02
2 2
2
(ω − ω0 )
(3.31)
Now we can think of the power radiated
P = σI
where σ is the cross section of the interaction and I is the intensity of the incoming radiation. We
know that
I=
cE02
8π
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since I = uc with u =
E02
2
being the energy density of the radiation. Thus we get
P =σ
cE02
2
or
8π
σ=
3
e2
me c2
2
ω4
(ω 2 − ω02 )2
(3.32)
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This classical cross section can be expressed in terms of the classical radius of the electron re =
as
R
σ=
2
4πre2
3
ω2
(ω 2 − ω02 )
e2
me c2
2
(3.33)
This is the expression which describes the interaction of an electromagnetic wave with an electron.
Please remember that this expression is classical ( no quantum) and is non-relativistic. We can use this
to get some idea about the various processes which take place when a photon interacts with an atom
in matter.
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Consider Rayleigh Scattering first. This, as we have seen happens when the incoming energy is
much less than the binding energy of the electron. Now the incoming energy of the photon is related
to the frequency of incoming radiation, that is ω. What about the binding energy? The electron is
bound to the nucleus by the electrostatic force which in this model is what provides the restoring force
with a force constant k. This leads to a ”natural” frequency of the dipole which is ω0 . Thus, Rayleigh
scattering regine is when ω ω0 . In this limit,
ω2
(ω 2 − ω02 )
2
ω4
∼ 4
ω0
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and the cross section is simply
R
σRay
8πre2 ω 4
=
3 ω04
(3.34)
This is the famous Rayleigh scattering cross section where we can see the dependence on the
wavelength of the electromagnetic wave. This is what gives us that blue colour of the sky where blue
light is scattered the most.
Rayleigh Scattering occurs when the electromagnetic wave has a much smaller energy than the
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binding energy. What about if the electromagnetic radiation is energetic enough to ionize the atom but
not energetic enough for it to accelerate it to relativistic speeds? This is the domain ~ω0 ~ω me c2 .
This kind of scattering is called Thomoson Scattering In this limit,
ω2
(ω 2 − ω02 )
∼
1
ω02
ω2
∼ −1
−1
Thus the cross section for Thomson Scattering becomes
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R
σT =
8πre2
3
(3.35)
This is a remarkable expression since it is totally independent of the frequency of the incoming radiation provided the condition above is met. In fact, it has a fixed value of about σT ∼ 32 barn. In fact,
Thompson scattering is basically the low energy limit of Compton scattering which we shall explore a
little later. In Thompson scattering, the photon is scattered elastically while Compton scattering is the
inelastic scattering of the photon from an electron. (Recall that in elastic scattering of particles, the
kinetic energy of the particle is conserved in the center of mass frame though not in the lab frame.)
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Though we have discussed Rayleigh and Thomson scattering, we know that these processes do not
play an important role in the interaction of gamma rays with matter. This is simply because the energies what we have encountered above are clearly too low for the gamma rays that we are interested in
since we have seen that the gamma ray energies are typically in the MeV range (Figure 2.4 for instance)
while the binding and ionisation energies of atoms is in the eV range and so Rayleigh scattering is
unimportant for gamma rays. Similarly, for Thomson scattering, though the energy of the photon is
more than the ionisation energy, it is less than the rest mass energy of the electron since the assumption
is that the electron is non-relativistic. Thus, the energy of the photon is in the keV range. Thus,
neither of these two processes of elastic collision of photon and electron are important for gamma rays.
For gamma rays, there are three processes which play a significant role in the energy loss in matter.
These are Photoelectric Absorption, Compton Effect and Pair Production. It is important to
note that unlike the case of alpha and beta particles, where the energy of the incoming particle is lost
gradually due to continuous interaction with the electrons in the material, in the case of gamma rays,
the photon can simply disappear or get scattered significantly abruptly.
Photoelectric Absorption
The classical theory that we have described above of the interaction of the photon and the electron
breaks down when ω ≈ ω0 since the denominator blows up. This is the resonance condition where
the theory outlined above is not valid. In this case, the photon interacting with the atomic electron
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transfers all its energy to the electron and disappears. If the photon energy was more than Eb , the
binding energy of the electron it interacts with, then of course the balance energy manifests itself as
the kinetic energy of the free electron, Ee . That is
Eγ = hν = Ee + Eb
R
σP E
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It turns out that to properly understand the process, one needs quantum mechanics to calculate the
cross section for photoelectric absorption. If a proper quantum mechanical calculation is done, we see
that the cross section σPE ∼ Z 5 where Z is the atomic number of the absorber. Thus we can see that
if we want an absorber to absorb gamma rays of the relevant energies, then we need to use high Z
materials like lead etc.
One can use the classical electromagnetic theory together with the correspondence principle to get
a fairly accurate expression for the cross section for photoelectric absorption. If one does this for both
the K shell electrons, for the non-relativistic case, one obtains (for the cross section per atom)
√
= 4 2Z 5 α4
me c2
hν
3.5
σT
(3.36)
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where Z is the atomic number of the atom, ν is the frequency of the incoming photon, and σT =
is the Thomson scattering cross section.
8π 2
r
3 e
For the strongly relativistic case, that is Eγ me c2 , the corresponding expression is
R
5
σP E = 1.5Z α
4
me c2
hν
σT
(3.37)
R
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An alternate formula which is applicable when the energy of the incoming photon is higher than the
binding energy of the K shell electrons is given by
σP E
4π
= √ σT α3 Z 4
3
m e c2
hν
3
(3.38)
There are several things to note about these expressions. One is that the cross section shows an
absorption edge- when the energy of the photon is below the K shell binding energy, photoionization of
the K shell electrons cannot happen. Thus the cross section would drop at that energy. However, since
the L shell electrons require less energy to be ionized ( 14 of the K shell energy), they would be ionized
till the energy drops to below that level which again gives rise to an absorption edge. Below that, the
M shell electrons would be ionized and so on. Also note that the cross section goes as ∼ (hν)−3 and
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therefore the lower the energy, the higher is the cross section.
When a photoelectron is ejected, typically from an inner shell, it creates a vacancy in that shell.
This vacancy is filled either by a free electron which is captured or by an outer electron falling into the
vacant position. This results in the production of characteristic X-rays which it turns out are mostly
absorbed in the material itself.
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Although we will study Compton Scattering next, it is instructive to point out the differences between
the two processes.
1. In Compton scattering, the combined momentum of the electron and the photon is conserved while
in photo absorption, momentum is transferred to the nucleus also.
2. In Compton scattering, the outgoing photon carries some energy while in photo absorption, all
the energy of the incoming photon is transferred to the electron as binding and kinetic energy.
3. Finally, Compton scattering becomes important only when the photon energy is atleast equal to
the rest mass energy of the electron while for photo absorption, the cross section increases sharply
at low energies and for energies less than ∼ 100 keV, it dominates the total cross section as is clear
from Fig 3.11.
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To conclude, photoelectric absorption is the main process by which gamma rays of relatively low
energy interact with matter (the cross section in Eq(3.36) varies inversely with energy and so at high
energies is small). The cross section depends strongly on the atomic number of the absorber and also
at the binding energies of the various shells in the absorber atoms.
Compton Scattering
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For gamma rays of energies that we typically encounter in radioactive decay, the dominant process
for interaction is Compton Scattering. This, as we know, is the interaction of an atomic electron with
the incoming photon, in which the electron acquires enough energy to become free and also relativistic. The scattering is therefore inelastic ( as opposed to Rayleigh and Thomson scattering which were
elastic). The photon transfers a part of its energy to the atomic electron and thereby loses energy
to the recoil electron. The process can be analysed using relativistic energy momentum conservation
as follows. We perform the calculation for the case of a free, unbound electron for simplicity.
Let the initial 4-momentum of the electron be Q̃i .
Q̃i = (me , 0)
.
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Note that the electron is assumed to be at rest and hence its energy is simply the rest mass energy.
Here and in the following treatment, we take c = 1. Let the final 4-momentum of the electron
after being hit by the photon be Q̃f . We know that the sqaured of the four momentum is an invariant
quantity. Hence
Q̃2i = Q̃2f = m2e
.
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Let P̃i and P̃f be the initial and final four momenta of the photon. Then
P̃i = (pi , pi n̂i )
and
P̃f = (pf , pf n̂f )
where n̂i and n̂f are the unit vectors in the direction of the initial and final photon 3-momentum
respectively.
For the photon, we know that
P̃i2 = P̃f2 = E 2 − p2 c2 = 0
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Now energy momentum conservation implies that
Q̃i + P̃i = Q̃f + P˜f
Q̃f = Q̃i + P̃i − P̃f
Q̃2f = P̃i2 + P̃f2 + Q̃2i − 2P̃i · P̃f + 2Q̃i · (P̃i − P̃f )
P̃i · P̃f = Q̃i · (P̃i − P̃f )
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pi pf − pi pf cos θ = me (pi − pf )
1
θ
1
2 sin2 ( ) = me
−
2
pf
pi
θ
1
1 2 sin2 ( ) = me c2
−
2
hνf
hνi
(3.39)
where θ is the scattering angle of the photon and in the final expression, we have not taken c = 1. In
the derivation above, we have used the fact that Q̃2i = Q̃2f , P̃i2 = P̃f2 = 0 and the fact that Q̃i = (me , 0)
and therefore Q̃i · (P̃i − P̃f ) = me (pi − pf ).
We can use the above expressions to get the energy of the scattered photon as
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R
hνf =
hνi
hνi
1 + me c2 (1 − cos θ)
(3.40)
or, the recoiling electron energy which is simply the kinetic energy
R
 
(1 − cos θ)

Ee = hνi − hνf = hνi 
hνi
1 + me c2 (1 − cos θ)
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hνi
me c2
(3.41)
The calculation of the cross section for Compton Scattering once again needs to take into account
relativistic and quantum mechanical effects. The result is a very complicated formula called the KleinNishina formula which is
2
dσ
1
2 1 + cos θ
= re
dΩ
2
[1 + hνi (1 − cos θ)]2
1+
hνi2 (1 − cos θ)2
(1 + cos2 θ)[1 + hνi (1 − cos θ)]
(3.42)
This is of course a very complicated expression. A useful way to remember this is to approximate it
by
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R
σComp ≈ σT
m e c2
hν
(3.43)
Thus we expect the cross section to decrease at high energies.
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It is instructive to consider two extreme cases for Compton scattering- one in which the photon is
hardly scattered, that is θ ∼ 0 and the other of back scattering, that is θ ∼ π. For θ ∼ 0, we can see
from Eq(3.40) and Eq(3.41), that
hνf ∼
= hνi
(3.44)
Ee ∼ 0
(3.45)
and
There is hardly any energy transfer as is expected in such a case.
For θ ∼ π, the photon is back scattered and
hνf =
hνi
1 + 2 mhνe ci2
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(3.46)
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and
Ee = hνi
1
2hνi
m e c2
2hνi
+m
2
ec
!
(3.47)
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This is the case for the maximum energy transfer from the photon to the electron. Figure 3.12 shows
how the energies of the scattered photon and recoil electron vary.
Figure 3.12: Compton Effect energies as a function of scattering angle§
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§(Source: Wikipedia)
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Of course, when gamma rays pass through any absorber, scattering will occur at all angles and thus if
one was to measure the energy spectrum of the recoiling electron, we would find a continuous spectrum
till the maximum energy given by Eq(3.47). If one takes a monoenergetic gamma ray, then the energy
spectrum will look like that in Figure 3.13.
Figure 3.13: Compton Effect- Energy spectrum of recoiling electron§
§(Source:By flutefreek (Flutefreek) (Own work) [GFDL (http://www.gnu.org/copyleft/fdl.html), CC-BY-SA-3.0
(http://creativecommons.org/licenses/by-sa/3.0/)
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or CC BY-SA 2.5-2.0-1.0 (http://creativecommons.org/licenses/by-sa/2.5-2.0-1.0)],
via Wikimedia Commons)
In this figure, the difference between the Compton edge and the full energy peak is Ec and is given
by
Ec = hνi − hνi
1
2hνi
me c2
2hνi
+m
2
ec
!
=
hνi
2hνi
1+ m
2
ec
(3.48)
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or for large gamma ray energies (hνi me c2 /2), we get
Ec ≈
me c2
= 0.256 MeV
2
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Pair Production
For high energy gamma rays, another process can result in the loss of gamma ray photons. This
is the production of an electron-positron pair by the photon in the presence of a nucleus. An isolated
photon cannot produce a particle-antiparticle pair because of conservation of energy and momentum.
To see this, just go to the center of mass frame in which the electron and positron fly back to back
with equal momentum. The final momentum in this frame is zero. However, the photon, in any frame
always moves with the velocity c and has a momentum equal to the energy of the photon (in units
where c = 1). Thus it is not possible. If there is another nucleus or another particle then the situation
is completely different and that nucleus or particle can recoil and balance the momentum.
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By energy considerations, the minimum energy of the photon to be able to produce a pair of electron
and positron would be equal to twice the mass of the electron, that is 1.02 MeV. Of course, this is
the threshold energy and it is only when the photon energies are of several MeV that pair production
becomes significant. The excess energy of the photon, above 1.02 MeV is shared by the electron-positron
pair as kinetic energy.
The pair production cross section can only be obtained using relativistic quantum mechanics. However, the cross section turns out to be, in the range of photon energy, 2me c2 hν me c2 Z −1/3 α−1 ,
σPP ≈ Z
2
αre2
28
ln
9
2hν
me c2
218
−
27
(3.49)
This is of course a very complicated expression. However, note the weak dependence on the photon
energy in this range. Of course, for hν < 2me c2 , the cross section goes to zero as it should. The
shielding of the nucleus by the atomic electrons basically for energies hν me c2 Z −1/3 α−1 means that
the cross section becomes constant beyond these energies.
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We also see that the cross section in the field of an atomic nucleus varies as Z 2 and so absorbers with
high atomic number will give an increased energy loss due to this process. Further, the expression Eq
3.49 tells us that the cross section at low energies goes as the logarithm of the photon energy. Finally,
since we have seen above that the Compton Scattering cross section varies as ∼ ω1 , pair production is
significant at higher energies when the the Compton effect cross section falls off.
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In practice, when a high energy gamma ray passes through an absorber and pair production takes
place, what we see are two photons emerging as secondary products of the interaction. This is because
the positron which is produced in the pair production process, typically interacts with an electron soon
after being produced and produces two photons (again, it cannot produce one photon due to energy
momentum conservation).
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Thus far, we have studied five processes which are operative when electromagnetic radiation is incident on a material. We have already seen that for gamma rays produced in radioactive decays, Rayleigh
and Thomson scattering are unimportant. The three process of importance are the photoelectric absorption, Compton Effect and Pair Production. All of these are operative but depending on the energy
of the incoming gamma rays, their cross sections vary. This is shown in Figure 3.14.
Figure 3.14: Relative importance of interactions of gamma rays§
We now have some understanding of how
alpha, beta and gamma rays interact with matter. We next turn to a study of the Geiger-Muller counter
which is the mainstay of our laboratory. We will study its working and see how it can be used to detect
these radioactive emissions.
§(Source: Encyclopedia of Occupational Health & Safety, www.icocis.org)
3.3
References
1. “Radiation Detection & Measurement”, Glenn F. Knoll, Wiley India (2009).
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Lab Manual for Nuclear Physics
Questions
1. What is mass thickness? Why is it important and what does it tell us about the
absorber?
2. What is the mechanism for an alpha particle to lose energy while interacting with
matter?
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3. When an alpha particle travels through matter, why does it not get deflected from its
initial path?
4. How does the energy loss by an alpha particle depend on its energy?
5. What is a Bragg peak and why does it occur?
6. How is the energy loss of an electron in matter different from that of alpha particles?
Why?
7. What are the various processes by which an electron can lose energy in matter?
8. What is Bremsstrahlung? Do alpha particle travelling through matter also lose energy
by Bremsstrahlung? If not, why not?
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9. In the case of electrons, what is the relative importance of energy lost by collisions
and that by radiation in the non-relativistic and relativistic case?
10. What is the difference in the range of a beam of mono-energetic electrons and that
of the beta particles emitted from a radioactive nucleus? What is the difference in
change of intensity with distance in both the cases?
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11. What are the different processes by which electromagnetic radiation interacts with
matter? What determines which process is dominant?
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12. What is the difference between Rayleigh scattering and Thomson scattering? Are
these two processes important for the interaction of gamma rays with matter? If not,
why not?
13. What are the processes which are dominant in the energy loss by interaction of gamma
rays with matter? How does the relative dominance of each of these processes vary
with the energy of the gamma rays?
14. How does the Compton effect cross section vary with the energy of the incoming
gamma rays?
15. Can a gamma ray photon of energy 5 MeV in space produce an electron-positron pair?
If not, why not?
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G-M COUNTER
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Chapter 4
Learning Objectives
1. To study various kinds of detector models.
2. To understand the concept of ionisation of gases and Townsend avalanche.
4.1
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3. To study the working of a GM counter.
Introduction
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In our laboratory, we use the Geiger-Muller counter (GM Counter) to study the radiation from
radioactive sources. The GM counter is a kind of counter or detector which uses the ionization produced
in a gas by the radiation to detect and study the radiation from the sources. It is a very convenient
detector or counter but can not be used for studying the energy characteristics of the radiation and is
only used to detect and count.
Before we study the GM counter, it is instructive to understand the general principles of ionisation
of a gas by radiation since these are used in the GM counter as well. In addition to the GM counter,
other kinds of detectors also use this phenomenon. Thus, ion chambers and proportional counters are
also based on this process. We first look at a generalised model of a detector and then investigate the
process of ionisation by radiation in gas filled detectors before studying the GM counter in detail.
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Detector Models
Fundamentally, all detectors of nuclear radiation work on the principle of the radiation (alpha, beta
or gamma rays) interacting with matter as discussed in Chapter 3 and depositing its energy in some
form in the material. This could lead to an ionisation of the material, or the production of photons etc.
Typically we detect and measure this change in the detector material and from that infer the properties
of the incoming radiation.
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In most detectors of interest to us, the net result is the production and collection of charge in the
detector. The charge Q produced by the interaction at some time t = 0 is collected by the presence of the
field which separates the two kinds of charges produced in the detector and collects them. Depending
on the detector, the time taken to collect the charge will obviously vary. The flow of this charge will
lead to a current which lasts till the time the charge is collected and whose integral over the time from
t = 0 to the charge collection time tc will give us the amount of charge deposited and collected.
tc
Q=
I(t)dt
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0
Figure 4.1: Current versus time
A pulse produced in the detector is due to one single interaction and we normally assume that the
rate of the ionising radiation entering the detector is low enough that we can distinguish between various
pulses as shown in Figure 4.2. Clearly, the size and the duration of each pulse depends on the actual
interaction taking place between the detector and the ionising radiation.
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Figure 4.2: Current versus time
We can distinguish between three different modes of the operation of detectors:
1. Current Mode
2. Pulse Mode
3. Mean Square Voltage Mode
¯ = 1
I(t)
T
t
I(t0 )dt0
t−T
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In the Current Mode, one basically measures a detector signal as a current measurement. In this
mode, a current meter is connected to the detector output. Since the current level is in picoAmperes
or nanoAmperes, a precise meter is required. Given that the current meter has a response time T , the
observed current from a sequence of events at time t will be
The response time is usually longer than the time between individual detection events, so that an
average current is recorded at a time t. The current mode is used when event rates are very
high, which makes a stable current.
In the Pulse Mode, the information on energy and timing of individual events, i.e. the
information on the signal amplitude and time of occurrence is usually recorded. In this
mode, the detector records the charge from each individual ionising event. This mode is used when
we need to get the information on the timing and the amplitude of individual pulses. Thus, given this
property, this mode is not suitable for high count rates since then the time between neighbouring events
may not allow for getting that information. Because the information on the charge collected in each
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individual event is recorded, this mode is used for energy measurements and spectrum.
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The signal shape from a radiation detector depends on the electronics to which the detector is
connected as well as the detector response. In most cases, the the input stage of the electronics is
an RC circuit where the resistance is the total input resistance and the capacitance C is the total
capacitance of the detector, cables, electronics etc as shown in Figure 4.3. Here V (t) is the time
dependent voltage across the load R and is the signal produced.
Figure 4.3: Detector Circuit
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It is easy to see that the time dependent voltage V (t) is given by
V (t) = V0 (1 − e−t/RC )
(4.1)
where τ = RC is the time constant of the RC circuit.
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We can distinguish two kinds of operations : When the time constant τ = RC tc , the charge
collection time. In this case, the current through R is the instantaneous value in the detector and
V (t) = I(t)R
In this case, the detector can collect charge from a single event with time tc .
In the second case, we have τ = RC tc . Now we can see that very little current flows through
R during the time tc and the current from the detector gets integrated in the capacitor which charges
to Vmax = Q
. If the time between the two events is long, this charge will discharge through R. In this
C
case, note that for a fixed C (which is determined by the electronics and the geometry and construction
of the detector etc.), Vmax is directly proportional to the charge Q deposited by the event. Thus,
measuring the height of the pulse that is Vmax , we can determine the charge deposited in
the detector and hence the energy produced by the incoming radiation in its interaction
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with the detector material. Furthermore, a measurement of the pulse rate gives us a rate for the
interactions of the incoming radiation with the detector material. The two cases are shown in Fig 4.4
Figure 4.4:
Ionisation of Gases
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Finally, the Mean Square Voltage Mode (MSV) Mode is used for certain kinds of measurements where we need to measure radiation from sources which produce charges which very different
from each other. Basically, what we do is that in a current mode operation, the varying current which
can be regarded as the superposition of a steady current and a varying component, we block the steady
component. The varying component is then squared and the signal is thus proportional to the square
of the charge that is created by the incident radiation. This enhances the differences between the two
kinds of radiation entering the detector. We shall not discuss this much since it is not used in our lab.
The basic idea in any ionisation based detector is that of gas multiplication. In its most rudimentary
form, consider a cylindrical enclosure filled with some gas. The enclosure has an electric field which
increases from the center to the walls of the cylinder. A radiation in the form of alpha, beta or gamma
rays enters the gas and interacts with the atoms or molecules of the gas, ionizing the atom and producing a positive ion and a free electron. The main characteristic of any gaseous ionisation detector is to
amplify this initial ionization event. Let us see how this happens.
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Lab Manual for Nuclear Physics
Townsend Avalanche
The ionizing radiation upon interacting with the gas produces an ion and an electron. If there was no
field present, then of course the positive ion and the electron would just drift till they lose all their
energy. However, in the presence of the field, they move towards the electrodes, the ion towards the
cathode and the electron towards the anode.
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The motion of the electrons and the ions is of course very different because of their different masses.
Both the ions and the electrons at any given temperature are always in some random thermal motion.
In the presence of the field, there is a force on the charges and they also get a drift velocity. (This is
very similar to the motion of electrons in a conductor when a voltage is applied between the two ends
of the conductor). It is seen that in a gas, the drift velocity v in the presence of a field is given by
E
P
E
v = µ
P
v ∝
(4.2)
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where E is the electric field and P is the pressure. The proportionality constant µ is called the
mobility and it depends on the nature of the gas and it turns out to be fairly independent of the value
of E and P . For ions, typical value of the mobility is around ∼ 10−4 m2 s−1 atm V−1 , For electrons,
because of their much smaller mass, the mobility is typically higher by a factor of 103 than the ions.
The variation of the drift velocity with E
for some gases is shown in Figure 4.5.
P
Figure 4.5: Variation of drift velocity of electrons in various gases with
E§
P .
§(Source: http:// encyclopedia2.thefreedictionary.com/ Mobility+of+Ions+and+Electrons)
Now suppose we increase the field. The ions being heavier move somewhat faster than before but
still have low mobility. The electrons on the other hand could gain significant kinetic energy. In their
path towards the anode, the electron would collide with other neutral atoms and if it transfers more
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than the ionisation energy to the electron in a neutral atom, it will ionise it. Of course, exactly when
this secondary ionisation caused by the electrons in motion actually happens, depends on the density
and pressure of the gas as well as the electric field strength. It is seen that in most gases at atmospheric
pressure, the threshold value of the field required for this is around 106 V m−1 .
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So after the first ionisation by the radiation, we now have a second ionisation produced by the ionizing
electron from the first ionisation. The two electrons now again move towards the anode because of the
electric field and gain kinetic energy. When this energy is large enough, they could collide with two other
neutral atoms and ionise them thereby producing 4 electrons now. These four electrons move towards
the anode and after gaining enough energy could ionise four neutral atoms and thereby producing 4 more
electrons and so on. This process continues as a chain reaction or a cascade as depicted schematically
in Figure 4.6 and is known as the Townsend Avalanche.
Figure 4.6: Schematic Diagram of Townsend Avalanche
§
§(Source: “Electron avalanche” by Dougsim - Own work. Licensed under CC BY-SA 3.0 via Wikimedia Commons http:
b
//commons.wikimedia.org/wiki/File:Electron avalanche.gif#/media/File:Electron avalanche.gif)
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We can quantify this process of cascade ionisation by the Townsend Equation which gives us the
increase in the number of electrons per unit length as
dN
= αdx
N
(4.3)
where α is the first Townsend coefficient which depends on the field strength and the nature as well
as the density and pressure of the gas. It is clearly zero for fields below the threshold value of the field
referred to above. The solution to this equation is as expected, an exponential increase in the number
of electrons
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R
N (x) = N (0)eαx
(4.4)
4.3.2
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The charges are finally deposited on the electrodes and this leads to a pulse in the associated electronics and whose characteristics give us some information on the nature and properties of the original
ionizing radiation.
Kinds of Detectors & Detector Regions
We mentioned that there are several kinds of detectors which use the principle of ionisation. Proportional Counters and GM counters are two examples which though using the same basic principle of
ionisation, work in different regions as we shall see below. Both these work in the pulse mode.
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Consider a gas filled detector where, as we have seen above, we have a chamber filled with a gas
and an electric field. Suppose in such a detector, we have one ionisation event due to the interaction
of radiation. If we increase the applied voltage to the electrodes, hence increasing the electric field
between them and note the amplitude of the pulse generated (which is a measure as we have seen of
the charge which has been generated by the ionisation and is collected by the electrodes), we can see
the differences in the response of the detector as shown in Figure 4.7.
Figure 4.7: Detector Regions
§
§(Source: “Radiation Detection & Measurement”, G. Knoll, Wiley India, 2009 )
Starting from very low values of the voltage, we see that the electric field experienced by the charges
is so small that some of the original ions can recombine and the charge that is collected by the detector
is less than that produced by the original radiation. This region is of course of no use for detection of
radiation. As we increase the applied voltage, we reach a saturation. This is when the charge collected
by the electrodes is equal to the charge created by the original radiation. This is the region where
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another class of detectors called ion chambers operate.
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When the voltage is increased even further, at some point the field becomes strong enough that the
threshold is reached for gas multiplication which as we have seen above, then produces secondary ions
and electrons by collisions. The charge collected by the detector increases and is proportional to the
energy of the ionising radiation. As the voltage increases, the collected charge increases and we see an
increase in the pulse amplitude. Initially, for a range of applied voltage, the gas multiplication is linear
and we see that the charge collected is proportional to the original number of ion pairs (and hence to
the energy of the incoming radiation). This region is called the region of true proportionality and is
the region in which proportional counters operate. We can achieve a multiplication of about 106 in
this region.
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If we increase the field further, then this proportionality breaks down. The reason for this is the
accumulation of a sheath of positive charges formed from the ions which are produced in the primary
and secondary ionisation. The positive ions being much heavier have smaller mobility and hence take a
much longer time to move towards the cathode. The space charge that is formed, distorts the electric
field experienced by the electrons in the gas and this leads to some nonlinearity. Thus in this region of
the applied voltage, we will not see a strictly linear increase of pulse amplitude with voltage though there
is still an increase. This region is known as the region of limited proportionality. We can say that
in this region, the pulse amplitude becomes more dependent on the voltage than on the initial ionisation.
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As the field is increased even more, we reach what is called the Geiger Region. Now the conditions
are there for an avalanche to occur in the gas and a maximum number of electrons are produced in the
gas due to the cascade effect as described above in Section 4.3.1. This is the region where the ion pair
count is now independent of the original ionisation and the detector cannot distinguish between different
ionising radiations or their energies. However, the detector efficiency is very good in this region. What
we see because of this independence is that the pulse amplitude is independent of the applied voltage
over a range of voltage and we get a plateau in the graph.
Finally, when the voltage increases beyond a certain value ( which depends on the nature and the
pressure of the gas as well as the geometry of the detector), the gas breaks down into a plasma and we
get a continuous discharge. This is bad for the detector and can damage the detector.
4.4
GM Counter
We have already seen how in general the detectors which are based on ionisation of gases work. The
GM counter, which is what we use in the laboratory is a kind of ionisation based detector which has
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certain special characteristics which we shall examine now.
Figure 4.8: GM Counter Schematic Diagram
4.4.1
Geiger Discharge
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The Townsend avalanche which we discussed above (Section 4.3.1) consists of secondary ions and electrons forming after the initial ionisation event. However, the electrons produced in the avalanche also
sometimes produce excited gas molecules, that is when the energy transferred to the molecule is less
than that required for ionisation. These excited molecules deexcite with a characteristic time of around
10−9 seconds and in doing so, emit a photon. This photon could be in the visible or the ultraviolet,
depending on the excitation state of the molecule. Now when this photon interacts (by photoelectric
absorption as we discussed in Section 3.2.3) with a less tightly bound electron somewhere else in the gas,
the electron might be released. This electron will move towards the anode and once again be a source
of a new avalanche. The photon could also hit the walls of the tube and release an electron when it is
absorbed. Typically, in a GM counter, the gas multiplication factor is very large, ∼ 109 − 1010 , and thus
we have an increasing number of avalanches created after a single ionisation event. The avalanches begin
when the electron is close to the anode wire and thus the time needed for producing all the ions and
electrons is less than a microsecond. The ultraviolet photons produce more electrons near the original
avalanche and these electrons also move towards the anode and create further avalanches. Thus, what
we see is that the discharge grows along the central anode wire in both directions from the position of
the original ionising event. The speed of this growth along the anode is very high, ∼ 2 − 4 cm (µ s)−1 .
This process is shown in Figure 4.9. The electrons collected by the anode constitute a pulse that is an
increased voltage across the external load resistor and this is counted as an event or a count.
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Figure 4.9: Geiger Discharge
§
§(Source: “Spread of avalanches in G-M tube” by Dougsim - Own work. Licensed under CC BY-SA 3.0 via Wikimedia Commons
- http://commons.wikimedia.org/wiki/File:Spread of avalanches in G-M tube.jpg#/media/File:Spread of avalanches in G-M tube.
jpg)
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So we have a scenario where a single ionizing event has in a very short time, created many avalanches
throughout the tube. How does this process end? To understand this, we need to realise that whenever
a free electron is created in the avalanche or otherwise, a positive ion is also created. The positive ions
have a much lower mobility because of their larger mass. Thus they hardly move while all the electrons
created in the various avalanches are collected by the anode. Clearly, this concentration of positive
charges near the anode causes the electric field to reduce since we know that in a cylindrical tube the
electric field at a distance r from the anode will be
E(r) =
V
r ln ab
(4.5)
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where V is the constant voltage between the cathode and the anode and b and a are the inner radius
of the cathode and the radius of the anode wire respectively. (To see this, think of the positive space
charge acting as a sheath around the anode, thereby increasing the effective radius of the anode wire , a.
Now obviously, with an increase in a, the field at a fixed r will increase. However, and this is important,
the r where the field is relevant now for ionisation, increases because of the space charge. This effect is
much larger and so overall the field will decrease. To demonstrate this, let V = 100 V, b = 10 cm and
a = 1 cm. Then the field at r = 1 cm, that is just above the anode wire, will be 43.4 V cm−1 . Now let
a = 2 cm, then the field at r = 2 cm, that is just above the effective anode radius now is 31.1 V cm−1 .
) With the decrease in the electric field below some threshold value, no further avalanches are possible
and the Geiger discharge ends. For a particular tube (that is a particular geometry and composition of
the gases), with a fixed external voltage the termination of the Geiger discharge is always after a fixed
number of avalanches or, what is the same thing, a fixed amount of total charge. This does not depend
on the original ionizing event at all. To put it another way, the size of the pulse is always the same no
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matter what the nature of the incoming radiation which caused the initial ionizing event is.
4.4.2
Quenching
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What we have seen till now is that incoming radiation generates an ionisation induced Geiger discharge
in the GM counter. We saw that the mechanism for this is a Townsend avalanche which, in a GM tube
generates multiple avalanches till the discharge ends because of the space charge due to the positive
ions. In all of this, it is clear that the role of the gas in the GM tube is very important. Firstly, we must
make sure that the gas used in the counter, called the fill gas is such that there is no possibility of it
forming negative ions. This is obvious since the whole operation depends on the generation of electrons
and positive ions. Thus, we need the fill gas to have a low electron attachment coefficient like hydrogen
or the rare gases and not a high electron attachment coefficient like oxygen. Secondly, the properties
of the fill gas should be such that the energy gained by the free electrons is enough for the multiple
avalanches to be generated. Recall that the drift velocity and hence the average energy depends on the
ratio E
(Eq(4.2)). Thus we try to have a gas mixture which gives the maximum of this ratio so that
P
we can generate the minimum electron energy to generate multiple avalanches. Typically, we use argon
in the GM counters used in our labs.
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The discussion above on Geiger discharge began with the single ionising event creating a cascade of
avalanches and then the avalanches travel down the anode wire very quickly and a pulse is created by
the charge collected at the anode which then essentially gets stored in the capacitance of the circuit and
discharges through the load resistor. We also discussed how the discharge ends because of the positive
space charge formed by the low mobility positive ions which reduces the electric field near the anode as
we discussed above (Eq(4.5)). Now let us consider what happens to the positive ions. These ions, say of
argon or some other inert gas with a high ionisation potential, are much heavier and therefore have low
mobility. Their motion towards the cathode is slow but they at some point will arrive at the cathode.
In this motion, they will gain kinetic energy. Now when they hit the cathode they will neutralise by
picking up an electron from the cathode. But suppose their energy was larger than twice the work
function of the cathode, that is the energy required to liberate an electron from the surface of the
cathode. In that case, the positive ion could, with some probability, liberate more than one electron
from the cathode. If the total number of positive ions is large (which it will be as we saw above since in
a GM tube, a large number of ion-electron pairs are produced) there could be at least one extra electron
liberated from the cathode. This electron, like the other electrons in the tube, would accelerate towards
the anode and can cause another Geiger discharge producing more positive ions and the process can go
on endlessly producing a continuous pulse.
There are several ways of dealing with this problem of multiple pulses. In the initial days of the
GM counter, a method known as External Quenching was used. In this, the electronics of the device
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was arranged in such a way so as to reduce the voltage for some time after a pulse. The lower voltage
is such that no gas multiplication can take place and therefore no Geiger discharge happens after the
initial pulse. The time that the voltage needs to be reduced obviously depends on the time that the
positive ions take to travel to the cathode, typically ∼ 10− 1 milliseconds. One way to do this is to
choose a large enough R in the RC circuit external to the tube. Then, as we have seen, (Eq(4.1)), the
time constant of the RC circuit becomes much larger than the charge collection time. Clearly, this also
means that the GM counter will only be able to record events which are separated by more than a few
milliseconds. This method therefore can only be used with low count rates.
In the GM counters used in our lab, we use Internal Quenching. In this method, a small percentage, typically 5 − 10% of another gas, called the quench gas is added to the tube. The quench gas
has the property that it has a lower ionisation potential than the fill gas and is also a more complex
molecular structure. Popular quench gases are ethyl alcohol and halogens, both of which satisfy these
conditions compared to argon, a typical fill gas. Now as the positive argon ions are drifting towards the
cathode, they will collide with the halogen molecules and since these have a lower ionisation potential,
the argon ions will transfer their energy to the quench molecules and neutralise. The quench gas ions
will now start drifting towards the cathode. When they strike the cathode, they will neutralise by
picking up an electron but, and this is the crucial point, the extra energy instead of liberating another
electron, will go towards dissociating the quench gas molecules.
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As mentioned above, both halogen gases and complex organic gases like ethyl alcohol are used as
quench gases. However, the organic gases are generally consumed and so the tubes which use them
have a finite lifetime. Halogens on the other hand can recombine spontaneously and so can be used
since they are self replenishing.
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Dead Time & Recovery Time
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4.4.3
After a Geiger discharge takes place, as mentioned above, we still have the positive ions of the fill gas
drifting away from the anode towards the cathode. Initially of course, when they are closer to the
anode, they reduce the electric field experienced by the electrons to below the level required for creating
another avalanche. At this time, if another particle of radiation enters the GM tube, and causes a single
ionising event, that will not lead to a Geiger discharge because the field is lower than that required by
the ionising electron to create an avalanche. This time is called the dead time and if another ionising
event occurs in this period, it will not be recorded. However, slowly the positive ions drift towards
the cathode and so the field near the anode rises. In this time, there could be some avalanches formed
and therefore some pulses may be observed depending on the sensitivity of the counting circuit. These
pulses would not be of the same amplitude as the original Geiger discharge pulse though. Finally, all
the positive ions do reach the cathode and at this point the electric field near the anode becomes strong
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enough again for the Geiger discharge of the same amplitude to take place. The behaviour is shown in
Figure 4.10
Figure 4.10: Time behaviour of a GM tube & Dead Time
§
§(Source: “Dead time of geiger muller tube” by Dougsim - Own work. Licensed under CC BY-SA 3.0 via Wikimedia Commons http://commons.wikimedia.org/wiki/File:Dead time of geiger muller tube.png#/media/File:Dead time of geiger muller tube.png)
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In most tubes, the dead time is around ∼ 50 − 100µs. As mentioned above, the tube takes a longer
time than the dead time to return to its original configuration, that is when it can produce a second
pulse of the same magnitude as the first one. This time is called recovery time.
Clearly, in cases where the count rates are high, determination of dead time is of critical importance.
There are several methods of determining the dead time. Here we discuss the commonly used two
source method.
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Let us define the following variables:
τ , the dead time of the counter
tr , the real or actual time the counter is in operation. This is the total time that we take to measure
the counts. It is clearly something which does not depend on τ .
tl , the live time of the detector, or the time that the detector is able to record the counts. This obviously
depends on τ .
N , the total number of counts recorded by us during tr .
.
m, the measured counting rate, that is N
tr
n, the true counting rate, that is Ntl .
So we have
m=
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n=
N
tl
Thus
m
tl
=
n
tr
(4.6)
So we have
tl = tr − N τ
(4.7)
tl
N
= 1 − τ = 1 − mτ
tr
tr
(4.8)
Using Eq(4.6) and Eq(4.7), we see that
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The interpretation of this is clear- the fraction of the counts we record is equal to the ratio of the live
time to the real time or to put it another way, it is the fraction of the time that the detector can actually
record the counts. The fraction of the time that the detector cannot record the counts is given by mτ .
It turns out that to a very good approximation, the live time is simply the real time minus N τ . This is
easy to see since the detector is unable to record any event during the total counting time tr is simply N τ .
m
tl
=
tr
n
= 1 − mτ
m
n =
1 − mτ
(4.9)
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Eq(4.9) tells us the relationship between the measured count rate m , the dead time τ and the true
count rate n. Note that n is always greater than m as it should be. Further, for small values of mτ , the
dead time is not important since the difference between n and m is small as a percentage. Finally, note
that the parameter of importance is not just τ but mτ . So, we can keep the product small by either a
small value of τ or that of m.
To find the dead time of the counter, we use the fact that the count rate from two sources individually
do not add up to the count rate from both of them together. This is because when the count rate is
high, the counter is dead for a longer period and therefore the counts dont add up. If m1 and m2 are
the measured counts for the two sources and m12 is the measured count rate for both of them together,
m12 6= m1 + m2
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Of course, the true count rates do add up, that is
n12 = n1 + n2
In carrying out any counting experiment with a source, we also need to take into account the measured
background count rate, that is the count rate caused by ambient sources of activity. Let the true count
rate for the background be nb and the measured one be mb . Then we have
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n12 − nb = n1 − nb + n2 − nb
n12 + nb = n1 + n2
(4.10)
Using Eq(4.9) for each of these true count rates n12 , n1 and n2 , we get
m12
mb
m1
m2
+
=
+
1 − m12 τ
1 − mb τ
1 − m1 τ
1 − m2 τ
(4.11)
In this equation all the quantities except τ is a measured quantity and therefore we can use this
equation and the measured values of the count rates for the sources individually, together and the
background to get the value of τ . Solving for τ , we get
τ =
A(1 −
√
1 − B)
(4.12)
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C
A = m1 m2 − mb m12
C = m1 m2 (m12 + mb ) − mb m12 (m1 + m2 )
C(m1 + m2 − m12 − mb )
B =
A2
This is obviously a very complicated expression. An approximate solution to the equation (Eq(4.11))
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is
τ≈
m1 + m2 − m12
2m1 m2
(4.13)
The above expression, in the presence of significant background counts is replaced by
R
τ≈
m1 + m2 − m12 − mb
2(m1 − mb )(m2 − mb )
(4.14)
Similarly, many other approximations are sometimes used instead of the more complicated expression Eq(4.12).The experiment is usually carried out by measuring the counts from source 1, placing the
second source 2 close by to count the combined counts and finally removing source 1 to get the counts
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4.4.4
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for source 2. To take care of absorption and scattering, when measuring counts from individual sources
1 or 2, a dummy source without activity is placed in place of the other source. One needs to be sure
that the positions of the sources does not change when measuring them together. We also know that
since there is not much difference between m12 and m1 + m2 , the count rates need to be measured very
accurately. We already know from Chapter 1 that the error in counting experiments goes as ∼ √1N
where N is the count rate. Thus to get good results we need to have the fractional dead time m12 τ in
Eq(4.9) to be around 15 − 20%.
Geiger Counting Plateau & Operating Voltage
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Now that we have discussed the general working of a GM counter, we need to understand in detail as
to how exactly one needs to use the GM counter in our experiments. The first thing to determine is the
Operating Voltage of the tube. From the discussion above, we know that the GM tube will only work
if the electric field inside it is above a certain value. If the electric field (or the applied voltage) is too
low, there will no counts recorded because the field cannot produce any pulse. As we raise the applied
voltage above the starting voltage , which is defined as the voltage at which the GM counter just
begins to count, we start observing counts. As the voltage is increased further, we observe a very rapid
increase in the counting rate which is almost like a step function. This rising region is called the knee
because of its resemblance to the knee. The counting rate then rises till we reach a threshold voltage
Vt . Above this voltage, all ionising events produce the same output pulses. The threshold voltage depends on the GM tube and the circuit components of the counter. Above the threshold value, the graph
levels off for a broad range of applied voltage. This region, after the knee or threshold value, where
the counting rate is level is called the plateau region. As we increase the voltage, the counting rate
remains essentially constant, that is the shape of the counting rate versus applied voltage is a straight
line almost parallel to the x axis. This continues till the voltage is high enough that a continuous
discharge takes place in the tube and there is a breakdown. The reason for the continuous discharge are
clearly related to the inability of the quenching mechanism to stop runaway avalanches because of the
high energy gained by the positive ions. The operating voltage is a value of the voltage in the middle
region of the plateau, roughly equidistant from the knee and the point where continuous discharge starts.
To illustrate this, let us consider a GM counter where the starting voltage is Vs . If the applied voltage
is less than Vs , then the pulse is not recorded as we have seen. As the voltage is increased to beyond
Vs , to reach the plateau region, the pulse can be measured. However, if the voltage is somewhat close
to the knee, then the low amplitude tail of any pulse will cause a slight rise in the slope of the plateau
region. The plateau region could also have a finite slope because of the quenching mechanism failing
sometimes.
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Figure 4.11: Counting Plateau of GM tube
The operating voltage of the GM tube depends therefore on the fill gas and the quenching gas used.
For instance, the typical operating voltage for an argon filled tube with alcohol as the quenching gas is
around 1000 − 1200 V while those with argon as a fill gas and bromine as a quenching agent is around
200 − 400 V.
Counting Efficiency
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Now that we have looked at how the GM counter works in theory, let us see what are the design considerations that go into the making of a GM counter. Firstly, we need a gas filled enclosure with a
window. The window and the walls of the enclosure need to be strong enough to withstand the pressure
difference between the inside and outside the tube since most GM tubes operate below the atmospheric
pressure. The window while being strong enough must be thin enough so as not to have significant
effect on the flux of radiation which one is trying to measure. The gas, as we have seen is typically a
rare gas along with a quenching agent.
The operation of the tube depends crucially on the existence of a high enough field near the anode.
This to a large extent determines the geometry of the counter. Thus, for instance, a parallel plate kind
of arrangement of electrodes will be less efficient than a cylindrical geometry. This is evident from
the expressions for the fields in the two geometries. For a parallel plate geometry, the field would be
uniform and inversely proportional to the distance between the electrodes. On the other hand, the field
for a cylindrical geometry, for the same voltage would be inversely proportional to the distance from
the anode (Eq(4.5)) and so near the anode, where we need a large field to ensure the avalanche, we can
use a much smaller external voltage.
With the above mentioned design of the GM tube, when a charged particle like an alpha or a beta
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particle enters the tube, we can detect it. The reason for this, as we have seen is that in a GM tube,
a single ionising event can cause a pulse. So if the charged particle enters the tube, the efficiency in
its detection will be close to 100%. However, we know that alpha particles have a low penetrating
power and so if we need to detect alpha particles, we need to ensure that the window is extremely thin.
For beta particles, we use a thicker window though here too some particles could be reflected or back
scattered from the window material also.
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The situation with gamma rays is very different from that for charged particles. The gamma ray
photons do not cause direct ionisation of the fill gas atoms. Instead, the photons interact with the
walls of the detector to produce a secondary electron. If the electron is able to penetrate the material of the wall (this depends on the thickness and the material used for the walls), and enter the gas
in the tube, it will essentially behave like an ionising electron from an ionising event produced by a
charged particle in the GM counter and cause a Geiger discharge as discussed above. Thus we need
to make the walls of the tube thinner than the range of the electrons which are produced by the interaction of the gamma ray photons with the wall material. Further, to increase the cross section of
this interaction, material with a higher atomic number needs to be used. Even with all this, for high
energy gamma rays, the typical GM counter efficiency is a few percent. However, for low energy gamma
rays, the probability of a direct interaction with the fill gas atoms increases and if we use appropriate gasses of high atomic number (Xenon for instance) at high pressures, the efficiency can be very high.
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Gamma sources are also are typically isotropic, that is the radiation is emitted isotropically in all
directions. Since the GM counter has a window of a finite area, only a fraction of these emitted gamma
rays will hit the counter window. Geometrically, it is clear that if we take the counter window to be
a circle of radius r and place the counter at a distance d from the gamma source, then the fraction of
gamma rays hitting the counter window will be the ratio of the area of the spherical cap of radius r
and the total area of the sphere of radius R. The area of the spherical cap with height x can be easily
seen to be 2πRx using double integrals or by any other method.
Area of a spherical cap:
Consider a sphere of radius r. We are interested in finding the surface area of a portion of the
sphere with height h as shown in Fig 4.12.
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Figure 4.12: Area of a spherical cap
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We divide the cap into segments each with an arc length of ds. Now since ds is infinitesimal, we
have
ds2 = dx2 + dy 2
where x and y are the two coordinates. The area of segment is obviously 2πxds. The total area
of the cap is obtained by adding all the segments or integrating. Thus
S=
2πxds
But
ds =
p
s
dx2
+
dy 2
= dy
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We also know that
x2 = r 2 − y 2
Thus
dx
y
y
= −p
=−
2
2
dy
x
r −y
So we have
S = 2π
r
p
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r
y2
x 1 + 2 = 2π
x
x2 + y 2 dy = 2πr
dy = 2πrh
r−h
Incidentally, this problem was first solved by Archimedes, of course without any calculus. Take
a sphere of radius r. Now take a cylinder of height 2r and radius r. the sphere will just fit into
the cylinder and will touch the cylinder along a circle. Imagine a slice of the cylinder of height or
thickness z. This slice will subtend an arc on the sphere. Let the length of the arc be w.If we take
the height z to be small enough and the arc to be a straight line, we can see that
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x
z
= sin θ =
r
w
The area of the slice of the cylinder would be
Ac = 2πrz
and that of the spherical slice would be
As = 2πxw
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But xw = rz and hence these two areas are the same. We could now repeat this exercise by
slicing the spherical cap into small spherical slices and adding the areas. Each spherical slice would
have an area equal to the corresponding cylindrical slice. Thus the area of the spherical cap would
be
A = 2πr
X
z = 2πRrh
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Figure 4.13: Area of a spherical cap
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Thus we have
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fraction of gamma rays incident on counter window =
2πRx
x
R−d
1
d
=
=
= − √
(4.15)
2
2
4πR
2R
2R
2 2 r + d2
This expression for the fraction of gamma rays incident on the counter window is valid for all
distances of the source from the counter. We can intuitively see it to be true since when the distance
of the source is 0, that is the source is next to the window, we expect one half of the emitted gamma
rays to enter the counter, which is what we get. On the other hand, when the source is very far away,
then d r and we get no gamma rays entering the counter.The geometry of the arrangement is shown
in the Figure 4.14.
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Figure 4.14: Isotropic source with GM counter
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We can define two kinds of efficiency of the counter/detector in principle. Absolute Efficiency and
Intrinsic Efficiency . Absolute efficiency, Abs can be thought of as the number of particles detected
as a fraction of the total number of particles emitted. That is
R
Abs =
Number of particles detected
Number of particles emitted
(4.16)
R
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Clearly, the absolute efficiency will depend on the counter geometry and the distance from the source.
Intrinsic efficiency int factors in these parameters and gives us a measure of efficiency for that particular
detector
int =
Number of particles detected
Number of particles incident on the detector
(4.17)
We are now familiar with the mechanism by which a GM counter works as well as its operational
details. We also have learnt about the various parameters associated with the GM counter like operating voltage, dead time, recovery time and its efficiency. With this knowledge, we can now proceed to
use the GM counter for various experiments.
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4.5
Lab Manual for Nuclear Physics
References
1. “Radiation Detection & Measurement”, Glenn F. Knoll, Wiley India (2009).
4.6
Questions
1. What is a Townsend avalanche and how is it produced?
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2. How does the number of electrons in a Townsend avalanche increase with time?
3. What are the various regions in which detectors operate?
4. What are the differences between the proportional region and the Geiger plateau?
5. Explain what happens when a beta particle enters a GM counter.
6. How does the avalanche once started in a GM tube end?
7. Explain the process of quenching in a GM tube.
8. What is dead time and how is it different from recovery time of a GM counter?
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9. Are the counting efficiencies for beta and gamma rays different in a GM counter? If
yes, why?
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10. Consider a very energetic beta particle and a very energetic gamma ray photon. What
are the differences between the way in which these two interact with a GM counter?
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Chapter 5
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Experiment: GM Characteristics
Things to know before you do the experiment
1. Familiarity with radioactivity (types, properties, safety and precautions, etc.) and
radioactive sources (Natural and artificial) and proper handling of the radioactive
sources. For safety precautions, please see Section 5.2.
2. Concept of half life, activity, decay rate, decay law of the radioactive sources.
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3. Familiarity with the various sources of background radiation.
4. The different process by which radiation interacts with matter.
5. Basic interactions and their properties.
6. Properties of Inert gases and molecular gases with regard to ionisation.
b
7. Electric field distribution for various configurations like a cylinder, sphere etc.
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8. Concept of a Solid angle.
9. Knowledge of statistics. Familiarity with probability, Central Limit theorem, various kind of basic probability distribution functions (Binomial, Poisson, Gaussian),
concept of mean, average, variance and standard deviation of a distribution.
10. Curve fitting techniques like least square fit and χ2 fit.
11. The meaning of p-value in statistics.
12. Difference between interpolation and curve fitting.
13. Concept of dead time, recovery time, RC circuit, time constants, etc.
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5.1
Lab Manual for Nuclear Physics
Introduction
5.2
Precautions
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Now that we have discussed the theoretical basis for the experiments in the laboratory, let us look at
the actual experiments that we perform in this laboratory. The basic equipment that we use in the
laboratory is the GM counter. We use it to find out the count rates from various nuclear sources and
thus try and understand the characteristics of the radiation coming out from them. For this purpose,
we first need to study the operational characteristics of the GM counter itself. This experiment
investigates the working of a GM counter and finding its parameters like the optimum
operating voltage, dead time etc. In addition, it also demonstrates the statistical nature
of radioactive processes.
Since all the experiments in this laboratory deal with radioactive sources, it is important to follow certain guidelines and precautions for working in this laboratory. Radioactive sources pose a substantial
health hazard and so it is essential that you follow these guidelines very carefully.
5.2.1
Health Effects of Radiation
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As we have already seen in earlier chapters, radioactive materials decay spontaneously to produce ionizing radiation like alpha, beta and gamma rays. These radiations have sufficient energy to strip away
electrons from atoms (as we saw in Sections 3.2.1, 3.2.2 & 3.2.3) or to break some chemical bonds. Any
living tissue in the human body can be damaged by ionizing radiation in a unique manner. The body
attempts to repair the damage, but sometimes the damage is of a nature that cannot be repaired or it
is too severe or widespread to be repaired. Also mistakes made in the natural repair process can lead to
cancerous cells. All types of radiation to which the person is exposed and the pathway by which they
are exposed influence health effects. Different types of radiation vary in their ability to damage different
kinds of tissue. Radiation and radiation emitters (radionuclides) can expose the whole body (direct
exposure) or expose tissues inside the body when inhaled or ingested. All kinds of ionizing radiation
can cause cancer and other health effects. The main difference in the ability of alpha and beta particles
and gamma and x-rays to cause health effects is the amount of energy they can deposit in a given space.
Their energy determines how far they can penetrate into tissue. It also determines how much energy
they are able to transmit directly or indirectly to tissues and the resulting damage. Although an alpha
particle and a gamma ray may have the same amount of energy, inside the body the alpha particle will
deposit all of its energy in a very small volume of tissue. The gamma radiation will spread energy over
a much larger volume.
There are two kinds of health effects that radioactive sources can cause - Stochastic and Non-
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stochastic.
Stochastic health effects are those that are associated with long term and low level exposure to radiation. Increased levels of exposure make these health effects more likely to occur, but do not influence
the type or severity of the effect. Cancer is considered by most people the primary health effect from
radiation exposure. Simply put, cancer is the uncontrolled growth of cells. Ordinarily, natural processes
control the rate at which cells grow and replace themselves. They also control the body’s processes
for repairing or replacing damaged tissue. Damage occurring at the cellular or molecular level, can
disrupt the control processes, permitting the uncontrolled growth of cells cancer This is why ionizing
radiation’s ability to break chemical bonds in atoms and molecules makes it such a potent carcinogen.
Other stochastic effects also occur. Radiation can cause changes in DNA, the “blueprints” that ensure
cell repair and replacement produces a perfect copy of the original cell. Changes in DNA are called
mutations. Sometimes the body fails to repair these mutations or even creates mutations during repair.
The mutations can be teratogenic or genetic. Teratogenic mutations are caused by exposure of the
fetus in the uterus and affect only the individual who was exposed. Genetic mutations are passed on
to offspring.
Non-stochastic health effects appear in cases of exposure to high levels of radiation, and become more
severe as the exposure increases. Short-term, high-level exposure is referred to as ‘acute’ exposure.
Many non-cancerous health effects of radiation are non-stochastic. Unlike cancer, health effects from
‘acute’ exposure to radiation usually appear quickly. Acute health effects include burns and radiation
sickness. Radiation sickness is also called ‘radiation poisoning.’ It can cause premature aging or even
death. If the dose is fatal, death usually occurs within two months. The symptoms of radiation sickness include: nausea, weakness, hair loss, skin burns or diminished organ function. Thus, for instance,
some medical patients receiving radiation treatments often experience acute effects, because they are
receiving relatively high “bursts” of radiation during treatment.
The unit of effective or equivalent dose or radiation is the rem or Roentgen Equivalent in Man. This
is the older though much used unit of the radiation dose exposure. A more modern unit is the Sievert
which is defined as 100 rem. A rem is a very large unit and so typically millirem is used for ordinary
exposures. 1 rem or 0.01 sievert translates to typically a 0.055% chance of the development of cancer
eventually. Typically,the exposure from background sources is around 0.01 millisieverts (mSv) per day
while the exposure to a chest x-ray is 0.06 mSv during the exposure to the x-ray. For a CT scan,
the exposure can be as high as 2 mSv. To compare, during the Fukushima nuclear accident, near the
accident site, a level of 1000 mSv hour−1 has been reported.
General Precautions
1. Handle the radioactive sources with utmost care and respect. Don’t bend or try to break them.
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2. Although the sources in the laboratory are always in their holders, it is in general important
to never touch the source using bare hands. Always use forceps to handle sources.
3. Do not eat or drink during the lab. Please do not keep any edible material or even drinking
water on your work table. Keep your bags with the food and water on the table on the side.
4. When bringing or returning the source to the source room, please be extremely careful to not
let it fall.
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5. Do not leave the source lying around the work table. Always keep it carefully while using it
and return it promptly after you are finished.
6. Do not keep the source in your pocket or in close contact with your body.
7. Wash your hands after the experiment and before eating or drinking anything.
5.3
5.3.1
Experiment
Purpose
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The purpose of this experiment is to study the GM counter and determine its characteristics. Specifically, we will be plotting the counts versus the external voltage curve to determine the plateau region
(Section 4.4.4) and hence the operating voltage. We will also be determining the dead time (Section
4.4.3) of the counter. Finally, we will investigate the statistical nature of the radioactive process. For
this purpose we would need a GM counter (which comprises of the GM tube, associated electronics for
counting purposes and an external high voltage source), two radioactive sources (which in our case are
60
60
27Co) and a dummy source. The decay scheme for 27Co is given in Figure 2.4. The reactions are
5.3.2
60
27Co
→
60
∗∗
28Ni
60
∗∗
28Ni
→
60
∗
28Ni
60
∗
28Ni
→
60
28Ni
+ e− + ν̄ e
+γ
+γ
Method
1. GM Characteristics:
To determine the characteristics of the GM counter, we first set the voltage to zero. Place the
source provided near the counter window and set the preset time to 50 seconds. This is the
time during which the counter will measure the counts. We increase the voltage gradually, initially
increasing it by a larger amount (say 50 V) till the counts start. This voltage at which the counts
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start is the threshold voltage. Once the counts start, the voltage should be increased in smaller
steps (say 10 V). You will note that after a sharp sudden rise in the counts (the knee of the characteristic), the counts remain constant with an increase in the voltage. This is the plateau region
discussed in Section 4.4.4. Finally, at some point, the count rate starts to increase. This is the
region where the continuous discharge starts and one should not increase the voltage beyond this
to avoid damage to the counter.
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2. Measurement of Dead Time:
We use the two source method to determine the dead time of the GM counter (Section 4.4.3). For
this we need two sources. First, we set the voltage in the GM counter at the operating voltage.
This, as we have seen is the value at which the GM counter should be operated and is the value
in the middle of the plateau region of the GM characteristic. After setting the voltage, we set the
preset time. For this experiment now, we set the preset time to a large value, say around 500−1000
seconds. Recall that for determining the dead time, we need to find the counts from two sources
separately and then together and also need to know the background counts (Section 4.4.3, Eq
(4.14)). First we remove all sources and measure the background count rate only with the dummy
source. Then we place one source and the dummy source (to ensure that all the absorption and
other effects from the source holder are equalised) and measure the counts. We repeat the same
with the second source and a dummy. Finally, we take both the sources together and determine
the combined count rate for the two sources.
5.3.3
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3. Counting Statistics:
To see the statistical nature of the counts, we remove all sources. We set the operating
voltage and then for various preset times (say 5, 10, 20 seconds), we note the number of
counts to get the background reading. We can repeat the measurement for 150 − 200 times
to get a viable statistical sample.
Sample Data
1. GM Characteristics:
Given below is the sample data for determining the GM characteristics.
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Error
0
4
11
20
23
28
31
30
31
30
32
31
32
31
31
31
32
31
32
32
32
31
31
32
31
32
31
32
33
31
32
32
32
33
33
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Voltage (V) Counts (N)
320
0
324
16
326
111
328
411
330
539
332
795
334
930
336
909
338
950
340
912
345
998
350
933
360
1005
370
964
390
947
400
983
420
1009
430
980
440
993
450
1035
460
1037
470
971
480
991
490
1054
500
938
510
1029
520
987
530
1043
540
1085
550
952
560
998
570
1026
580
1049
590
1058
600
1069
Table 5.1: Sample Data for GM Characteristics
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The first two columns give us the voltage (V) in Volts and the number of counts N . Since this is
a counting experiment with a fixed time interval (the preset time) and the events are random (radioactive decay is a random event), the conditions of the distribution being a Poisson distribution
are satisfied as we saw in Section 1.4.2. Therefore, square root rule (Section 1.5) tells us that the
√
error in each measurement of the counts is N . This is tabulated in column 3. As expected, the
estimated error is lower for lower count rates.
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Notice how the errors are reported. The voltage that we use is assumed to be accurate
and hence by assumption has no error. The counts N , that we measure at any voltage,
√
as discussed earlier have an error which is N . But we have already seen that the
error or uncertainty should be reported to 1 or 2 significant figures with appropriate
rounding off (See Section 1.2.4). Hence, for instance, the square root of N = 1058 is
32.52 and that of N = 1049 is 32.38, and so we report the uncertainty in these as 33 and
32 respectively. Further, though in this case it is not relevant, the data itself should
only be reported to a precision which is the same as the uncertainty. This is very
important since your calculators will give you a precision of 8 or more digits which is
meaningless in our experiment set up.
We can plot the date as shown in Fig 5.1.
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GM Characteristic
1200
1000
La
b
Counts
800
600
400
200
0
300
350
400
450
500
550
600
Voltage
Figure 5.1: Sample GM Characteristics
We see that the graph looks very similar to what we expect theoretically as in Figure 4.11. The
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m=
2. Dead Time:
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sudden rise after the threshold voltage, the knee, the plateau are all clearly visible in this sample
data. The plateau region is not a line of zero slope as expected in an ideal GM tube but has some
finite slope. Nevertheless, the constancy of the count rate beyond the knee of the characteristic is
clearly visible in the graph. The error bars represent our estimate of the error in each measurement
of N . We can draw a smooth line through the points and we obtain the characteristic. Alternatively,
we can assume that the plateau region is a straight line and fit the points to a straight line using
Method of Least Squares (Section 1.7, Eq(1.73) & Eq(1.74)). We can use the linear part of the
curve to find the operating voltage of our GM tube. For this, recall that the operating voltage is
typically taken to be in the middle of the plateau region. We can take two values near the extreme
ends of the straight line in the plateau region and compute the middle point. In this case, the
operating voltage turns our to be ∼ 450 V. The slope of the plateau region should ideally be 0.
However, we can determine the slope from the data given as
∆y
952 − 933
19
=
=
≈ 0.1
∆x
550 − 350
200
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Preset time is set at 600 seconds. The observed counts for source 1 + dummy (N1 ), source 2 +
dummy (N2 ), source 1 + source 2 (N12 ) and the background (NB ) are are follows.
Table 5.2: Dead Time Determination
La
b
N1
N2
N12
NB
28401
23926
51957
268
28589
23970
51827
236
N1 = 28495
N2 = 23948
N12 = 51892
NB = 252
p
p
p
p
∆N1 = N1 = 168 ∆N2 = N2 = 154 ∆N12 = N12 = 227 ∆NB = NB = 15
Once again, note the reporting of the uncertainty in the count rates N1 , · · · , NB . We are again taking
the square root of the counts and reporting it to 3 significant figures here. We can and should report
these to two significant figures as per our convention. However, in this case we have retained more
than 2 significant figures because as we will see below, we dont actually use these numbers to calculate
the error in the dead time.
We can calculate the dead time using Eq(4.14)
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τ≈
m1 + m2 − m12 − mb
2(m1 − mb )(m2 − mb )
To get the proper dead time,we need to multiply τ with the preset time since the counts m1 , m2 , · · ·
are the counts per unit time. In our case N1 , N2 , · · · are the total number of counts in the preset time
interval.
R
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Putting in the numbers we get,
τ ≈ 1.34 × 10−4 seconds
(5.1)
The error in τ can also be calculated. We know that τ is a derived quantity from the measured
quantities N1 , N2 , N12 and Nb . We can use the error propagation equation (Eq(1.48) to calculate the
error in τ if the errors in the measured quantities are known. But the measured quantities are simply
the counts in a fixed time interval and the events are random, the distribution will be Poissononian
and hence the error in the measurements will go as the square root of the counts.
To calculate the error in the dead time, we use the error propagation expressions. We know that
u
m1 + m2 − m12 − mb
≡
2(m1 − mb )(m2 − mb )
v
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τ≈
u = m1 + m2 − m12 − mb
v = 2(m1 − mb )(m2 − mb )
But
La
b
Then by Eq(1.56), we have
σu2 σv2
στ2
=
+ 2
τ2
u2
v
2
2
2
2
σu2 = σm
+ σm
+ σm
+ σm
1
2
12
b
using Eq(1.53).
Similarly using Eq(1.55) and Eq(1.53) on the denominator, we get
2
2
2
2
σm
+ σm
σm
+ σm
σv2
1
2
b
b
=
+
2
2
v
(m1 − mb )
(m2 − mb )2
Putting it all together, we get the expression for the error in the dead time as
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στ2
=τ
2
2
2
2
2
2
2
2
2
σm
+ σm
+ σm
+ σm
σm
+ σm
σm
+ σm
1
2
12
1
2
b
b
b
+
+
(m1 + m2 − m12 − mb )2 (m1 − mb )2 (m2 − mb )2
In this expression, we know all the quantities. One way to think about the errors in m1 , m2 , m12 , mb or
equivalently in N1 , N2 , N12 , NB would be that they are given by
q
σm1 =
N1
q
N2
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σm2 =
q
σm12 = N12
q
σmb = NB
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However, there is a subtle point here that one needs to understand. Let us see what we are doing- we
are taking the total number of background counts in 600 seconds and reporting it as a number. Then
we are repeating the same procedure 2 times. Note that in each of the values of N1 , N2 , N12 &NB ,
there is an error. This is the inherent statistical error that is associated with the event which as we
know is a result of a Poisson distribution. We can therefore think of each value of N1 , N2 , N12 &NB
as a mean of a Poisson distribution. The associated error with each value of N1 , N2 , N12 &NB is thus
√
√
N1 , N2 , · · · , NB . Therefore, the correct procedure to exhibit this inherent statistical error would
be to take the expression for the mean value of N1 , N2 , · · · NB , that is N1 , N2 , N12 , NB and apply the
appropriate error prorogation equation to it. In the case of N1 , for instance, this means
2
P
2
σN
=
1
La
b
N1 =
=
=
σN1 =
=
(N1 )i
i=1
2
1 2
2
σ(N1 )1 + σ(N
1 )2
4
1
[28401 + 28589]
4
56990
√4
56990
2
119.4 ' 120
Similarly we can calculate the errors in N2 , N12 , NB . They are
√
σN2 =
47896
= 109.4 ' 110
2
163
(5.2)
Shobhit Mahajan
Lab Manual for Nuclear Physics
√
103784
= 161.1 ' 160
2
√
504
σNB =
= 11.2 ∼ eq11
2
With these values of the errors in the counts, we can calculate the error in the deadtime.To get the
actual dead time, the above expression will need to be multiplied by the preset time.
σN12 =
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3. Counting Statistics:
Preset Time: 5 seconds
Operating Voltage: 460 Volts
In Table 5.3, xi is the number of counts in the preset interval and fi is the number of times out of a
P
total of
fi = 200 that we get xi counts.
0
1
2
3
4
5
6
fi
xi f i
27
0
55
55
60
120
31
93
16
64
06
30
05
30
P
P
fi = 200
xi fi = 387
(xi − x) (xi − x)2
-1.94
-0.94
0.06
1.06
2.06
3.06
4.06
3.76
0.880
0.086
1.12
4.24
9.86
16.48
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xi
pe =
Pfi
fi
(xi − x̄)2 p
0.135
0.507
0.275
0.242
0.300
0.010
0.155
0.173
0.080
0.339
0.030
0.280
0.025
0.412
P
P
2
pe = 1 σE = (xi − x̄)2 p = 1.96
La
b
Table 5.3: Counting Statistics for Preset time = 5 seconds
We note that the sum of the probabilities
P
pe = 1 as it should be.
We can easily calculate the sample mean as in Eq(1.5) as
P
x i fi
387
x̄ = P
=
= 1.94
fi
200
The corresponding sample variance can be calculated from the expectation value of the square of the
deviations from the sample mean
σE2 =
X
(xi − x̄)2 p = 1.96
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or
σE = 1.40
Our theoretical expectation is that the distribution of the counts would follow a Poisson distribution.
However, we do not know the mean of the distribution. We can estimate the mean of the parent
distribution by taking it to be the sample mean given above. This means that
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µ = x̄
Then we expect that the probability distribution function to be (Eq(1.25))
PPoisson ≡ P (xi , µ) =
We can tabulate these values also
PPoisson
0.144
0.279
0.271
0.175
0.084
0.030
0.011
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xi
0
1
2
3
4
5
6
µxi −µ
e
xi !
pe
0.135
0.275
0.300
0.155
0.080
0.030
0.025
Table 5.4: Experimental and Poisson probabilities: Preset Time 5 seconds
La
b
The standard deviation of the Poisson distribution is simply
σPoisson =
√
µ = 1.39
The graph of the experimental data and the theoretical Poisson distribution with the same mean is
shown in Figure 5.2.
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GM Counting Statistics: Preset time 5 seconds
0.35
Poisson
Data
0.3
0.25
0.15
0.1
0.05
0
0
1
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p
0.2
2
3
4
5
6
x
Figure 5.2: Counting Statistics : Preset 5 seconds
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We can see that the theoretically expected Poisson distribution and the experimental distribution are
fairly similar though there are deviations. However, the error bars on the experimental values of the
derived probabilities are such that the theoretical distribution falls within the range. The error bars
are the error on pe . However, we know that pe is actually a derived quantity, since
fi
pe = P
fi
(5.3)
La
b
√
P
Now
fi = 200 and obviously there is no error in this. We take the error in fi as fi and get the
√
fi
error bars by taking the error in pe to be 200
.
We also saw in Chapter 1, Section 1.8 that there is another way to determine if the experimental data
conforms to our assumed statistical model. The way to do this through the χ2 test. Recall that we
define a quantity χ2 as (Eq.(1.81))
N
1X
χ =
(xi − x̄)2
x̄ i=1
2
Here x̄ is simply the sample mean. The numerator in the above expression can be easily related to the
sample variance (Eq(1.9)) as
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(N − 1)s2
x̄
We can also define another quantity, the reduced Chi squared (Eq(1.84)) as
χ2 =
χ̃2 =
χ2
ν
where ν is simply the degrees of freedom defined as
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ν = N − Nc
Here N is the number of observations and Nc is the number of constraints.
The number of constraints is 1 , coming from the fact that we have already used the data to determine
x̄ . Therefore
ν =N −1
This implies
s2
x̄
If our data was a true Poisson distribution, then we know that
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χ̃2 =
s2 = x̄
b
and therefore, for a true Poisson distribution, the value of χ̃2 will simply be 1. If this value of χ̃2 turns
out to be too different from 1 then we can say that either our experimental data is not quite correct or
our underlying statistical model is faulty.
La
In our case, we have already calculated both the sample mean and the sample variance above. We thus
get
s2
σE2
1.96
2
χ̃ =
=
=
= 1.01
x̄
x̄
1.94
The value of the χ2 is then simply
χ2 = 1.01 × ν = 1.01 × (200 − 1) = 201.05
We can look up tables of χ2 (for instance at en.wikibooks.org/wiki/Engineering Tables/
Chi-Squared Distibution or www.pd.infn.it/∼lunardon/didattica/docsper2/TavoleChi2.pd) to find
out the probability associated with this value of χ̃2 . To use these tables, one first determines the
degrees of freedom in the experiment. In our case, the number of observations is 200 and so the degrees
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of freedom are 199. Then one looks in the row corresponding to the number of degrees of freedom (199
in our case) and finds the column where the number is the closest (next smallest) to the value of χ2
obtained from your data. The value of p corresponding to that column is the probability that a random
sample chosen from a Poisson distribution would have a larger value of χ2 than that shown in the
table. In our case, we see that this probability is less than 0.5 which is a very good match. A perfect fit
to the Poisson distribution should yield a value of p = 0.5 exactly. In our case, it is slightly less than 0.5.
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In fact, one can get the p-value directly from MS Excel by using the function CHISQ.DIST.RT(χ2 , ν).
If we use this we get a p-value of 0.445989. If we plot the distribution for 199 degrees of freedom with
this observed value of χ2 , we see that the graph will look like
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Figure 5.3: p-value for ν = 199 and χ2 = 201.05
Preset Time: 10 seconds
Operating Voltage: 460 Volts
x i fi
06
0
18
18
33
66
38
114
36
134
27
135
18
108
13
91
05
40
04
36
02
20
P
P
fi = 200
xi fi = 762
b
0
1
2
3
4
5
6
7
8
9
10
fi
La
xi
(xi − x)
(xi − x)2
-3.81
-2.81
-1.81
-0.81
0.19
1.19
2.19
3.19
4.19
5.19
6.19
14.52
7.890
3.280
0.660
0.036
1.420
4.790
10.180
17.560
26.940
38.320
pe =
Pfi
fi
(xi − x̄)2 p
0.030
0.435
0.090
0.710
0.165
0.541
0.190
0.125
0.180
0.006
0.135
0.191
0.090
0.431
0.0650
0.661
0.0250
0.439
0.020
0.538
0.010
0.383
P
P
2
pe = 1 σE = (xi − x̄)2 p = 4.46
Table 5.5: Counting Statistics for Preset time = 10 seconds
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We can easily calculate the sample mean as in Eq(1.5) as
P
x i fi
762
x̄ = P
=
= 3.81
fi
200
The corresponding sample variance can be calculated from the expectation value of the square of the
deviations from the sample mean
σE2 =
(xi − x̄)2 p = 4.46
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or
X
σE = 2.11
Our theoretical expectation is that the distribution of the counts would follow a Poisson distribution.
However, we do not know the mean of the distribution. We can estimate the mean of the parent
distribution by taking it to be the sample mean given above. This mean that
µ = x̄
Then we expect that the probability distribution function to be (Eq(1.25))
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PPoisson ≡ P (xi , µ) =
µxi −µ
e
xi !
La
b
We can tabulate these values also
xi PPoisson
0
0.022
1
0.083
2
0.159
3
0.202
4
0.193
5
0.147
6
0.093
7
0.050
8
0.024
9
0.010
10 0.003
pe
0.030
0.090
0.165
0.190
0.180
0.135
0.090
0.065
0.025
0.020
0.010
Table 5.6: Experimental and Poisson probabilities: Preset Time 10 seconds
The standard deviation of the Poisson distribution is simply
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σPoisson =
√
µ = 1.95
The graph of the experimental data and the theoretical Poisson distribution with the same mean is
shown in Figure 5.4.
GM Counting Statistics: Preset time 10 seconds
0.25
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Poisson
Data
0.2
p
0.15
0.1
0.05
0
0
2
4
6
8
10
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x
Figure 5.4: Counting Statistics : Preset 10 seconds
La
b
We can see that the theoretically expected Poisson distribution and the experimental distribution are
fairly similar though there are deviations. However, the error bars on the experimental values of the
derived probabilities are such that the theoretical distribution falls within the range.
The reduced χ2 can again be calculated as
χ̃2 =
σE2
4.46
=
= 1.17
x̄
3.81
and the χ2 as
χ2 = 232.95
Again, referring to the tables for χ2 for 199 degrees of freedom, we get a value between 0.05 < p < 0.1
indicating a reasonable though not very good fit to our theoretical model. The exact value from Excel
is 0.0492.
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Shobhit Mahajan
Figure 5.5: p-value for ν = 199 and χ2 = 232.95
Preset Time: 20 seconds
Operating Voltage: 460 Volts
xi f i
(xi − x)
(xi − x)2
-6.155
-5.155
-4.155
-3.155
-2.155
-1.155
-0.155
0.845
1.845
2.845
3.845
4.845
5.845
6.845
7.845
8.845
37.88
26.57
17.26
9.95
4.64
1.33
0.024
0.714
3.4
8.09
14.78
23.47
34.16
46.55
61.54
78.23
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5
10
5
15
15
60
12
60
23
138
25
175
21
168
30
270
26
260
15
165
09
108
05
65
02
28
04
60
02
32
01
17
P
P
fi = 200
xi fi = 1631
b
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
fi
La
xi
pe =
Pfi
fi
0.025
0.947
0.025
0.664
0.075
1.294
0.060
0.597
0.115
0.533
0.125
0.166
0.105
0.002
0.150
0.107
0.130
0.442
0.075
606
0.045
0.665
0.025
0.586
0.010
0.341
0.020
0.937
0.010
0.615
0.005
0.391
P
P
2
pe = 1 σE = (xi − x̄)2 p = 8.893
Table 5.7: Counting Statistics for Preset time = 20 seconds
We can easily calculate the sample mean as in Eq(1.5) as
171
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Lab Manual for Nuclear Physics
P
x i fi
1631
x̄ = P
= 8.155
=
200
fi
The corresponding sample variance can be calculated from the expectation value of the square of the
deviations from the sample mean
σE2 =
X
(xi − x̄)2 p = 8.893
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or
σE = 2.98
Our theoretical expectation is that the distribution of the counts would follow a Poisson distribution.
However, we do not know the mean of the distribution. We can estimate the mean of the parent
distribution by taking it to be the sample mean given above. This mean that
µ = x̄
Then we expect that the probability distribution function to be (Eq(1.25))
PPoisson ≡ P (xi , µ) =
La
b
M
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We can tabulate these values also
xi PPoisson
2
0.009
3
0.025
4
0.052
5
0.080
6
0.117
7
0.136
8
0.139
9
0.130
10 0.100
11 0.076
12 0.051
13 0.032
14 0.010
15 0.010
16 0.010
17 0.003
172
µxi −µ
e
xi !
pe
0.025
0.025
0.075
0.060
0.115
0.125
0.105
0.150
0.130
0.075
0.045
0.025
0.010
0.020
0.010
0.005
Shobhit Mahajan
Lab Manual for Nuclear Physics
Table 5.8: Experimental and Poisson probabilities: Preset Time 20 seconds
The standard deviation of the Poisson distribution is simply
σPoisson =
√
µ = 2.85
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The graph of the experimental data and the theoretical Poisson distribution with the same mean is
shown in Figure 5.6.
GM Counting Statistics: Preset Time 20 seconds
0.18
Poisson
data
Gaussian
0.16
0.14
0.12
p
0.1
0.08
0.06
0.04
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0.02
0
-0.02
2
4
6
8
10
12
14
16
18
x
b
Figure 5.6: Counting Statistics : Preset 20 seconds
La
The value of the reduced χ2 is simply
χ̃2 =
8.893
= 1.09
8.155
and of χ2 is
χ2 = 217
The tables tell us that the value of p is around 0.2 which is a reasonably good fit to the assumed
theoretical model. The exact value is 0.182.
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Shobhit Mahajan
Figure 5.7: p-value for ν = 199 and χ2 = 217
5.4
Questions
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We can see that the theoretically expected Poisson distribution and the experimental distribution are
fairly similar though there are deviations. From our theoretical considerations (Section 1.4.3), we know
that when the mean of the Poisson distribution is large, the distribution tends towards a Gaussian or
normal distribution. In Figure 5.6, we have also plotted the Gaussian distribution with the sample
mean as the mean and the sample standard deviation as the standard deviation. We can see the
similarities or differences between the experimental distribution and the normal distribution. Indeed
one sees that the normal distribution approximates the experimental distribution much better than the
Poisson distribution.
1. Describe the working of a GM counter.
2. What are the various regions of a GM characteristic curve?
b
3. What is the meaning of threshold region, knee region, plateau region and breakdown
region? Explain the reasons behind their formation.
La
4. With reference to the experiment, explain the concept of threshold voltage, knee
voltage, Operating voltage and breakdown voltage.
5. What does one mean by primary and secondary ionisation?
6. Explain the formation of a Townsend avalanche?
7. What is quenching (internal as well as external) and how does it work?
8. What are statistical fluctuations and what is their relationship to the number of
events?
9. What are the two kinds of errors encountered in any experiment? What is the importance of statistical errors on the measurements?
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Chapter 6
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Experiment: GM Counter: Counting
Efficiency for β & γ rays
Things to know before you do the experiment
1. All the concepts in Chapter 5 .
2. Concept and importance of efficiency and its types (intrinsic and extrinsic).
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3. Concept of solid angle.
4. Error propagation and estimating errors on the derived quantities.
Introduction
b
6.1
La
We have already studied the GM counter and recorded its characteristics. These include finding the
operating voltage for the GM counter by plotting its characteristic and taking the middle point of the
flat plateau region. We have also determined its dead time. In this experiment we will study the
efficiency of the GM counter for detecting gamma and beta radiation. As already discussed
in Section 4.4.5, alpha particles are difficult to detect with the GM counters available to us because
of the thickness of the window (which needs to be thick for structural reasons to be able to withstand
the difference in pressure inside and outside the tube). Beta particles on the other hand have a larger
penetrating power and so can penetrate the window and being charged can easily cause an ionising
event in the tube which leads to a Gieger discharge. Gamma rays on the other hand, rarely interact
directly with the fill gas in the tube but instead cause the emission of a photoelectron from the inner
walls of the tube which leads to an avalanche as explained in Section 4.4.5.
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Shobhit Mahajan
6.2
6.2.1
Lab Manual for Nuclear Physics
Experiment
Purpose
The purpose to this Experiment is To estimate the efficiency of the GM counter for the detection of β and γ rays. We determine the efficiencies for a set of various β and γ rays and study the
dependence of efficiency as a function of distance between the source and the counter.
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To determine the efficiency of the detector, we clearly need to measure the number of particles
emitted by the source and the number detected. The efficiency can be defined in two ways, Absolute
Efficiency, Abs which does not take into account geometric factors like the distance from the source
etc. (Eq(4.16)) and Intrinsic efficiency , int which does take into the account the number of particles which actually enter the counter, Eq(4.17). For gamma ray sources, since the emission is roughly
isotropic, we need to take into account the number of particles actually striking the counter window
while for beta rays, we assume that all the particles emitted by a source are actually striking the window.
For this experiment, we would need a GM counter setup, a range of beta and gamma sources, which
in our case are 60Co, 57Co, 133Ba, 204Tl and 147Pm. We would also need some aluminum sheets.
La
b
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The decay schemes for these sources are given below.
Figure 6.1: Decay scheme for
176
60
Co
Lab Manual for Nuclear Physics
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Shobhit Mahajan
Figure 6.2: Decay scheme for
Co
133
Ba
La
b
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Figure 6.3: Decay scheme for
57
Figure 6.4: Decay scheme for
177
204
Tl
Lab Manual for Nuclear Physics
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Shobhit Mahajan
Figure 6.5: Decay scheme for
6.2.2
Method
147
Pm
As mentioned above, we need to basically know three numbers- one, the total number of particles
emitted by the source; the number of particles detected by the source and finally, the number of
particles entering the detector (which, as discussed in Section 4.4.5 may be different from the number
emitted.). The total number of particles emitted by the source depends on the activity of the source.
The activity of a source is defined as the rate of decay of the radionuclides in a source and is of course
a time dependent quantity as we know from Section 2.1.2. The activity follows an exponential law
(Eq(2.2))
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R(t) = R(0)e−λt
where λ is the decay constant. The decay constant is related to the half life T 1 by
2
λ=
ln 2
T1
2
La
b
For the sources mentioned above, we are given the activity of the source at the time of its fabrication,
R(0). The time of fabrication ti is also given. From this, we can determine the time elapsed since the
source with the given activity R(0) was fabricated. This time is t. Knowing the source, we know the
half life T 1 of the particular radioisotope. This allows us to calculate λ, the decay constant. Finally,
2
knowing R(0), t and λ, allows us to calculate the present activity R(t).
For the purely gamma sources, 133Ba, 57Co, we need to take into account the geometric factor also
as discussed in Section 4.4.5 since the sources are isotropic and the number of particles entering the
counter window will depend on the distance from the source and the radius of the counter window.
In addition to the pure gamma sources, we can also use 60Co which is a gamma and beta source. We
place an aluminum sheet between the source and the detector window which effectively blocks all the
beta particles, its thickness being more than the range of the 0.514 MeV beta particle but is not thick
enough to significantly attenuate the gamma rays.
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Lab Manual for Nuclear Physics
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For the pure beta sources, 204Tl, 147Pm, the geometry is not a significant issue since we assume all the
beta particles emitted enter the window. This is because the source is placed in a holder which has only
a small aluminum window through which the beta particles can escape. The source is otherwise enclosed
by thick plastic casing which absorbs the beta particles. Nevertheless, we will use the geometric
factor for determining the number of particles striking the window of the counter, instead
of taking it to be equal to the number emitted. Of course as we increase the distance between
the source and the counter, there is some absorption of the beta particles in the air.
We start by determining the operating voltage of the GM tube as discussed in Chapter 5 and choosing
an appropriate preset time, say 180 seconds. The radius of the GM counter window r is known to be
7.5 mm.We also determine the background count rate NB in the absence of all sources. With the GM
counter set up at the appropriate voltage, we then use the pure gamma sources at various distances d,
and measure the count rate N . We repeat this for various distances for the gamma sources and beta
sources. The detected count rate per second has to be decreased by the background count rate per
second to get the net counts detected per second (cps). The counts emitted per second or the
number of decays per second (dps) which are hitting the counter window are obtained from the
calculation of the activity today R(t). The geometric factor is what we obtained in Equation 4.15 and
so we get for the number of particles hitting the window as
dps = R(t) ×
1
d
− √
2
2 2 r + d2
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where the geometric factor is important for the gamma sources but not so important for the beta
sources. The efficiency is then simply
=
Sample Data
La
b
6.2.3
cps
× 100%
dps
(6.1)
Operating Voltage: 425 V
Preset Time: 180 seconds
nB = 85 counts
85
= 0.47 counts per second
180
Error estimation and the correct method to determine the error in the efficiency, is
discussed in the next section.
NB ( average) =
γ sources:
179
Shobhit Mahajan
1.
Lab Manual for Nuclear Physics
133
Ba
R(0)
ti
t
T1
111 KBq
August 2013
1.833 years
10.5 years
λ
R(t)
r
0.066 years−1
98.35 × 103 Bq
0.75 cm
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2
Table 6.1: Data for
R(t) Bq
d (cm)
98.35 × 103
98.35 × 103
98.35 × 103
2.7
4.7
7.0
dps = R(t)
1
2
−
√ d
2 r2 +d2
133
Ba
N (counts)
1794
614
279
3045
1201
629
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Table 6.2: Count rate and efficiency data for
57
b
Co
La
2.
R(0)
ti
t
T1
76.2 KBq
August 2013
1.833 years
0.74 years
λ
R(t)
r
0.936
13.64 × 103 Bq
0.75 cm
2
cps =
Table 6.3: Data for
180
57
Co
N
180
− NB
16.45
6.20
3.02
133
Ba
=
cps
dps
× 100% ± σ
0.91 ± 0.02
1.01 ± 0.03
1.08 ± 0.04
Shobhit Mahajan
Lab Manual for Nuclear Physics
R(t) Bq
d (cm)
13.64 × 103
13.64 × 103
13.64 × 103
2.7
4.7
7.0
dps = R(t)
1
2
−
√ d
2 r2 +d2
N (counts)
240.1
83
38
cps =
239
117
101
Co with Aluminum sheet
R(0)
ti
t
T1
17 KBq
August 2013
1.833 years
5.27 years
λ
R(t)
r
0.131
13.37 × 103 Bq
0.75 cm
2
13.37 × 103
13.37 × 103
13.37 × 103
2.7
4.7
7.0
dps = R(t)
1
2
−
√ d
2 r2 +d2
60
235.3
81.5
37.4
1010
426
277
La
Table 6.6: Count rate and efficiency data for
β sources:
1.
204
57
cps
dps
× 100% ± σ
0.35 ± 0.02
0.21 ± 0.02
0.23 ± 0.02
Co
Co
N (counts)
b
d (cm)
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Table 6.5: Data for
R(t) Bq
=
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60
− NB
0.86
0.18
0.091
Table 6.4: Count rate and efficiency data for
3.
N
180
Tl
181
60
cps =
N
180
− NB
=
5.14
1.89
1.07
Co with aluminum sheets
cps
dps
× 100% ± σ
2.11 ± 0.07
2.23 ± 0.11
2.81 ± 0.17
Shobhit Mahajan
Lab Manual for Nuclear Physics
R(0) 0.11µ Ci = 0.11 × 3.7 × 104 = 4070 Bq
ti
August 2011
t
3.89 years
T1
3.9 years
2
λ
R(t)
r
0.178
2036.48 Bq
0.75 cm
R(t) Bq
d (cm)
2036.48
2036.48
2036.48
2.7
4.7
7.0
dps = R(t)
1
2
−
204
Tl
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Table 6.7: Data for
√ d
2 r2 +d2
35.84
12.42
5.70
N (counts)
cps =
1972
579
406
147
=
204
cps
dps
× 100% ± σ
28.22 ± 0.64
21.59 ± 0.89
30.81 ± 2.16
Tl
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Pm
− NB
10.48
2.74
1.78
Table 6.8: Count rate and efficiency data for
R(0)
13 KBq
ti
Decemeber 2014
t
0.493 years
T1
2.6 years
b
2
λ
R(t)
r
La
2.
N
180
0.266
11.40 × 103 Bq
0.75 cm
Table 6.9: Data for
R(t) Bq
d (cm)
11.40 × 103
11.40 × 103
11.40 × 103
2.7
4.7
7.0
dps = R(t)
1
2
−
√ d
2 r2 +d2
147
Pm
N (counts)
200.6
69.5
31.9
3238
907
341
182
cps =
N
180
− NB
17.52
4.566
1.420
=
cps
dps
× 100% ± σ
8.42 ± 0.15
6.42 ± 0.21
4.39 ± 0.24
Shobhit Mahajan
Lab Manual for Nuclear Physics
Table 6.10: Count rate and efficiency data for
147
Pm
6.2.4
Error Estimation
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From the data and the results obtained above, we see that the efficiency of the counter for γ rays is
indeed very small as expected while that for the β particles is much higher. Theoretically, we expect an
almost 100% efficiency for beta particles. There could be several reasons why this was not seen in this
experiment. One, the efficiency depends on the dead time and that might have reduced the efficiency for
some source. This would be particularly noticeable for high activity sources where the time between the
particles entering the detector is small. Secondly, as the distance increases, the probability of the beta
particles being absorbed in the air or being scattered off increases. This also might be a contributing
factor which our experiment does not take care of. It only takes care of the distance via the geometric
factor.
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The experiment aims to determine the efficiency of the GM counter in measuring the counts for beta
and gamma rays. We calculate the efficiency by finding the counts per second and the detected counts
per second as in Eq(6.1). Clearly, both these quantities are derived quantities and so to estimate the
errors in these, we need to estimate the errors in the measured quantities from which they are derived
and then use the error propagation equation Eq(1.49).
We have
La
or
cps
dps
b
=
e=
c
dp
Thus
2
σe2
σc2 σdp
= 2 + 2
e2
c
dp
Let us consider the numerator first.
N − nB
180
Each of these quantities is a measured quantity, that is the total number of counts in 180 seconds
with the source and without the source (with the background only). We have seen already in Chap 1
c=
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that the errors will be simply the square root of the counts. Thus
p
2
2
σN
+ (σnB
)
σc =
180
We assume that there is no uncertainty in the time measurement. Then
√
N
√
= nB
σN =
Therefore we get
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σnB
σc2
and
σc2
=
c2
p
2
2
σN
+ σnB
=
1802
N +nB
1802
(N −nB )2
1802
=
2
+ σn2 B
σN
(N − nB )2
What about the denominator? Here again, the quantity is a derived quantity and we need to estimate
the error in each of the measured quantities. We have
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1
d
dps = dp = R(t)
− √
2 2 d2 + r2
where
La
b
The error calculation for dps is a little more complicated. Let us use our knowledge of error
propagation and see how it works. If we were to use the standard formulae for error propagation in
products and quotients etc. as discussed in the discussion around Eq(1.55) and Eq(1.56), then we
will need to first write
dps = R(t)B
1
d
− √
B=
2
2 2 d + r2
Now by Eq(1.55), we have
2
σdp
σR2
σB2
=
+
dp2
R2 B 2
But we know that R(t), which we have calculated from the original activity and the time elapsed
since the fabrication of the source does not have any error since we assume we know these quantities
184
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precisely. Thus
2
σdp
σB2
= 2
dp2
B
Now
d
1 G
1
− √
= −
B=
2 2 d2 + r 2
2
2
Using Eq(1.51), we have
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1 2
σB2 = σG
4
or
2
2
σdp
σG
=
dp2
4B 2
2
, we have
To calculate σG
d
F
G= √
=
J
d2 + r 2
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where
F = d, J =
√
d2 + r 2
Using Eq(1.56), we have
2
σF2
σJ2
σG
=
+
G2
F 2 J2
La
b
But F = d and so
σF2 = σd2
2
σG
σd2 σJ2
=
+
G2
d2 J 2
What about J? Here we have
J=
√
d2 + r 2 =
√
with
M = d2 + r 2
185
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Shobhit Mahajan
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Using Eq(1.57), we have
σJ
σM
=
J
2M
or
2
σJ2
σM
=
J2
4M 2
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Writing
M = d2 + r 2 = P + S
Now r, the radius of the GM counter window also is specified by the manufacturer and so we
assume no error in it. Thus
σM = σP
But
P = d2
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Thus, using Eq(1.57), we have
2σd
σP
=
P
d
or
σP2 =
La
b
So
4P 2 σd2
= 4d2 σd2
d2
2
σM
= σP2 = 4d2 σd2
But
2
σJ2
σM
4d2 σd2
d2 σd2
=
=
=
J2
4M 2
4(d2 + r2 )2
(d2 + r2 )2
and so
2
σG
σd2 σJ2
=
+
G2
d2 J 2
gives us
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2
σG
σd2
d2 σd2
=
+
G2
d2 (d2 + r2 )2
But
2
2
σdp
σG
=
dp2
4B 2
2
Putting in the value of σG
, we get
or
2
σdp
σd2
=
dp2
4B 2
Putting in the value of B, we get
2
σdp
σd2
=
dp2
4
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2
2
σdp
σG
G2 σd2
d2 σd2
=
=
+
dp2
4B 2
4B 2 d2 (d2 + r2 )2
d2
d2 + r 2
d2
d2 + r 2
d2
1
+
d2 (d2 + r2 )2
d2
1
+
h
d2 (d2 + r2 )2
1
2
1
−
√ d
2 d2 +r2
i2
Finally, we get the expression for the error in the efficiency as
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2
σe2
σc2 σdp
=
+
e2
c2
dp2
or
N + nB
σd2
σe2
=
+
e2
(N − nB )2
4
d2
d2 + r2
d2
1
+
h
d2 (d2 + r2 )2
1
2
1
−
√ d
2 d2 +r2
i2
La
b
However, there is a problem with this method in this case. The issue is basically
that whenever we have expressions like the one above, involving the same variable
in the numerator and the denominator, then the above method of calculating error
misses out on the cancellation of errors between the numerator and the denominator.
To see this, consider a simple case of a function of three variables, u, v, x as
u+v
u+x
Now suppose all of these variables are positive numbers. Now suppose we overestimate the error
in one of them, say u. Now this overestimate will affect BOTH the numerator and denominator
and these overestimates will cancel each other to a large extent. Similarly if we underestimate the
f (u, v, x) =
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Shobhit Mahajan
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error in u, it will affect both the numerator and denominator and these underestimates will cancel
each other largely. This kind of cancellation is missed when we do the error calculation by the above
method of splitting the function f (u, v, x) by repeated use of different error formulae Instead, what
we need to do is to use the basic error propagation equation to get the correct error.
Let us see how this is done.
We know that
Thus
c
dp
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e=
2
σe2
σc2 σdp
= 2 + 2
e2
c
dp
The error in the numerator c is not a problem and we have already computed the error in c. This
is simply
σc2
=
c2
N +nB
1802
(N −nB )2
1802
=
N + nB
(N − nB )2
The problem is with the error in the denominator. For the error in dp, we have
1
d
dp = R(t)
− √
2
2 2 d + r2
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or
dps = R(t)B
where
1
d
B=
− √
2
2 2 d + r2
La
b
We can use the error formula for the product (Eq(1.55)) here since again, the two functions B
and R are independent. There is no error in R(t). So we have
2
σdp
σB2
=
dp2
B2
1
d
1 G
B=
− √
= −
2 2 d2 + r 2
2
2
For finding the error in B, again there is no problem in using the expression for weighted sums
(Eq(1.53)). Using this, we have
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1 2
σB2 = σG
4
or
2
2
σdp
σG
=
dp2
4B 2
where
d
+ r2
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G= √
d2
To find the error in G, we might think of using the error formula for division (Eq(1.56)). But as
noted above, this will give us the wrong result since both the numerator and the denominator contain
the variable d and doing this, we will miss out on the errors cancelling out in both the numerator
and denominator. To calculate σG , we use the basic error propagation equation Eq(1.49). We know
that there is no error in r. Thus
Then
2
σG
But
∂G
∂d
∂G
∂d
2
d2
1
− 2
= √
d2 + r2 (d + r2 )3/2
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=
σd2
Thus
2
σG
La
b
or
σB2
=
σd2
d2
1
√
−
d2 + r2 (d2 + r2 )3/2
2
2
1 2
1
d2
= σd √
−
4
d2 + r2 (d2 + r2 )3/2
Finally
2
2
σdp
σB2
1 2
1
d2
= 2 =
σ √
−
dp2
B
4B 2 d
d2 + r2 (d2 + r2 )3/2
or
2
σe2
σc2 σdp
=
+
e2
c2
dp2
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Shobhit Mahajan
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that is
σe2
e2
=
N +nB
(N −nB )2
+
1
σ2
4B 2 d
h
√ 1
d2 +r2
−
d2
(d2 +r2 )3/2
i2
(6.2)
where
d
1
− √
B =
2
2 2 d + r2
2
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2
The only estimation we need therefore to determine the error in the efficiency is the error in d.
We assume that the error is equal to the least count of the scale used to measure the distance, namely
0.1 cm. As an example, consider the measurement for the beta source 204Tl. We have
N = 1972
nB = 85
r = 0.75 cm
d = 2.7 cm
σd = 0.1 cm
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= 0.2822
Then a calculation of σe2 gives us
σe2 = 0.00004
La
b
or
σe = 0.0064
Thus we have for this case
R
= 28.22 ± 0.64%
(6.3)
A similar analysis can be carried out for all the other readings to get an estimate of the error involved
in our experiment.
It is easy to write a program for instance in C language to do the calculation for the efficiency and
the error in the efficiency. A sample program is given below
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Shobhit Mahajan
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#include <s t d i o . h>
#include <math . h>
main ( )
{
f l o a t se , e , n , nb , sd =0.1 , d , r =0.75 , se1 , se2 , x , y , z ;
f l o a t r t , d1 , c , b ;
p r i n t f ( ” i n p u t N” ) ;
s c a n f ( ”%f ” ,&n ) ;
nb = 0 . 4 7 ;
s c a n f ( ”%f ” ,&d ) ;
p r i n t f ( ” i n p u t R” ) ;
s c a n f ( ”%f ” ,& r t ) ;
c=(n / 1 8 0 . 0 ) − nb ;
y= d / ( 2 . 0 ∗ s q r t ( d∗d+r ∗ r ) ) ;
d1=0.5−y ;
d=r t ∗ d1 ;
e=c /d ;
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p r i n t f ( ” i n p u t d” ) ;
p r i n t f ( ”N=%f \ t NB= %f \ t d=%f \ t R( t ) = %f \ t c= %f \ t e = %f \n” , n , nb , d , r t , c , e ) ;
p r i n t f ( ” input e p s i l o n ” ) ;
s c a n f ( ”%f ” ,& e ) ;
x = ( n+nb ) / ( ( n−nb ) ∗ ( n−nb ) ) ;
z =1/( s q r t ( d∗d+r ∗ r ) ) ;
s e 1=x+(sd ∗ sd / ( 4 ∗ d1 ∗ d1 ) ) ∗ ( z−d∗d∗ z ∗ z ∗ z ) ∗ ( z−d∗d∗ z ∗ z ∗ z ) ;
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s e 2=s e 1 ∗ e ∗ e ;
s e=s q r t ( s e 2 ) ;
p r i n t f ( ” e = %f \ t s e=
}
Questions
b
6.3
%f ” , e , s e ) ;
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1. How does one define efficiency and what is meant by intrinsic and extrinsic efficiency.
Which of these efficiencies is a more practical quantity?
2. Why are the efficiencies of a given GM counter for β and γ rays not the same?
3. How does the efficiency of a GM counter depend on the distance between the source
and detector?
4. What are the sources of systematic errors in this experiment?
5. How is the error in the efficiency evaluated given that it is a derived quantity?
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Chapter 7
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Experiment: Absorption of γ rays in Iron
Things to know before you do the experiment
1. All the concepts in Chapter 5 .
2. The interaction of γ rays with matter and the dependence of these interaction
processes on the energy of the γ ray.
3. The concept of mass thickness.
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4. Knowledge and use of semi-log graphs.
5. Concept of intensity of radiation and probability of interaction.
Introduction
b
7.1
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We have already seen how gamma rays interact with matter in Chapter 3, Section 3.2.3. We saw there
that at energies of ∼ MeV, the three processes of Photoelectric absorption, Compton Effect and Pair
Production are operative. The relative importance of these depends on the energy of the gamma rays
(Figure 3.14). We also saw that when gamma rays pass through an absorber, there is an exponential
attenuation of the incoming beam (Section 3.1.1). The number of transmitted particles after crossing
a distance ∆x is (Eq(3.1))
N = N0 e−σn∆x = N0 e−µ∆x
(7.1)
where µ is called the linear attenuation coefficient. Clearly, this attenuation coefficient, being
a product of the cross section σ and n, the number of scatterers per unit volume, will depend on the
density ρ of the scattering material. A more convenient measure to quantify absorption is the mass
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attenuation coefficient which is simply µρ . Consequently, we replace Eq(7.1)by
N = N0 e−
µ∆xρ
ρ
(7.2)
where ρ∆x is the mass thickness of Eq(3.3). This, as we have seen is much easier to measure and
also much more informative in comparing different absorbers.
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We can define a quantity called the half thickness, or the mass thickness where the intensity of the
incoming gamma rays is reduced to one half the initial value. Thus
N0
= N0 e−µm d1/2
(7.3)
2
where µm is the mass attenuation coefficient and d1/2 is the half thickness of that particular
absorber for the particular energy gamma rays used. Clearly
N=
d1/2 =
7.2
7.2.1
Experiment
Purpose
0.693
µm
(7.4)
7.2.2
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The aim of the experiment is to find the half thickness of monoenergetic gamma ray source 137Cs, which
emits gamma rays with energy 0.6617 MeV, using iron plates of varying thickness.
The experiment uses a GM counter, a gamma ray source, 137Cs and iron plates of varying thickness.
Method
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As with all experiments using a GM counter, the first step is always to find the operating voltage of
the counter by plotting its characteristic as in Chapter 5. We also obtain the background counts for a
fixed preset time, say 180 seconds.
We then put the source in the lead stand and note down the number of counts during the preset time.
The lead absorber stand is then placed between the source and the counter window and an increasing
number of iron plates of known thickness are placed in the absorber stand and the counts noted. A
plot of the logarithm of the net number of counts versus the mass thickness of the absorbers gives us a
straight line whose slope allows us to calculate µm and d1/2 .
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7.2.3
Lab Manual for Nuclear Physics
Sample Data
Operating Voltage = 425 V
Preset Time = 180 seconds
Background counts NB = 97 in 180 seconds
Density of Iron = 7.86 gm cm−3
Atomic Number of Iron = 56
√
Error ( Ni + NB )
20.83
19.23
19.41
18.62
18.62
17.69
17.00
16.40
16.43
16.24
16.61
15.42
15.13
14.62
15.09
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Counts (Ni ) Net Counts (Ni − NB )
337
240
273
176
280
183
250
153
250
153
216
119
192
95
172
75
173
76
167
70
179
82
141
44
132
35
117
20
131
34
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Mass Thickness (ti = xi ρ) gm cm−2
0
2.28
3.76
4.60
6.08
8.36
9.86
11.41
13.92
15.26
16.20
19.12
23.72
25.29
27.61
137
Cs with iron plates
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Table 7.1: Count rate and thickness data for
We need to plot the graph between ln(N − NB ) and t = xρ. This, as we expect will be straight line.
However, given the statistical nature of the data, we would need to find the best fit straight line using
the Method of Least Squares (Section 1.7, Chapter 1).
To determine the slope and the intercept of the best fit straight line, we need to use Eq(1.75) and
P
P P
P
P
P 2
P
Eq(1.76) and therefore need
xi =
ti , yi =
ln(Ni − NB ), x2i =
ti and
xi y i =
P
ti ln(Ni − NB ). We can calculate them as below.
xi = ti
0
x2i = t2i
0
yi = ln(Ni − NB )
5.48
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√
xi yi = ti ln(Ni − NB ) ln( N + NB )
0
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Lab Manual for Nuclear Physics
5.17
5.21
5.03
5.03
4.78
4.55
4.32
4.33
4.25
4.41
3.78
3.46
2.99
3.53
P
yi = 66.34
11.79
19.59
23.14
30.58
39.96
44.86
49.29
60.83
64.09
71.44
72.27
82.07
75.62
97.46
P
xi yi = 742.99
2.58
2.60
2.51
2.51
2.39
2.28
2.16
2.17
2.12
2.20
1.89
1.78
1.50
1.76
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2.28
5.10
3.76
14.13
4.60
21.16
6.08
36.97
8.36
69.89
9.86
97.22
11.41
130.19
13.92
193.77
15.26
232.83
16.20
262.44
19.12
365.57
23.72
562.64
25.29
639.58
27.61
762.31
P
P 2
xi = 187.47
xi = 3393.80
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Table 7.2: Least Square Fitting for graph of ln(N − NB ) vs t
The number of observations, N = 15.
With these values, we can compute the best fit straight line using Method of Least Squares. The results
are
b
m = −0.0819, c = 240
La
The graph of N − NB vs t is plotted on a semi-log scale with the error bars as indicated. We have also
drawn the best fit line with the above mentioned slope and intercept.
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Gamma Ray absorption
1000
data
best fit
10
1
0
5
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N-NB
100
10
15
20
25
30
t (gm cm-2)
Figure 7.1: γ ray absorption
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With the given slope, we can calculate the mass attenuation coefficient µm , which is precisely
the slope calculated as −0.0819 gm cm−2 . The half thickness can be calculated from Eq(7.4) as
= 8.55 gm cm−2 .
d1/2 = 0.693
µm
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What about the error in our calculation? The error in N − NB is clear. We know that a weighted sum
√
√
or difference has an error given by Eq(1.53). The error in N is simply N and that in NB is NB .
√
These two quantities are uncorrelated and hence the overall error in N − NB will be N + NB . The
√
error bars are then simply ln( N + NB ). But what about the errors in the slope and the intercept of
the graph which is what we are using to determine the desired derived quantity, d1/2 ? As you would
recall from Section 1.7, the uncertainty in the slope determined by this method can be calculated using
Eq(1.79). Since we have the values of m and c, for each xi = ti , we calculate y = mxi + c . We also
P
P
have the observed yi as well as the values of ∆ = N x2i − ( xi )2 . Putting it all together, we then
know the uncertainty in y which is given by
rP
(yi − mxi − c)2
(7.5)
σy =
N −2
With this uncertainty, we can calculate the uncertainty in the slope determined by the Least Square
method as
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r
σm = σy
N
= σy
∆
s
N
P
x2i
N
P
− ( xi )2
(7.6)
The values for the uncertainty in the slope turns out to be
σm = 0.00018
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Thus, our value for the mass attenuation coefficient is
µm = −0.0819 ± 0.0002
The corresponding error in the half thickness can also be calculated using Section 1.5.1 is
σd = −d1/2
Thus we can quote the half thickness as
R
7.3
σm
= 0.017
m
d1/2 = 8.55 ± 0.02 gm cm−2
Questions
(7.7)
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1. What is half thickness and how it is measured?
2. How does the half thickness depend upon the energy of the incoming γ rays, intensity
of the γ rays, nature of the absorber and nature of the source?
3. What are the various ways in which radiation can interact with matter?
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4. How does the relative dominance of each of these processes vary with the energy of
the gamma rays?
5. How does the Compton effect cross section vary with the energy of the incoming
gamma rays?
6. Can a gamma ray photon of energy 5 MeV in space produce an electron-positron pair?
If not, why not?
7. How do we estimate the errors on the measurements of derived quantities (error
propagation).
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Chapter 8
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Experiment: Verification of the Inverse
Square Law for γ rays
Things to know before you do the experiment
1. All the concepts in Chapter 5 .
2. Concept of solid angle.
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3. Error propagation and estimating errors on the derived quantities.
4. The inverse square law for radiation.
Introduction
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8.1
La
The intensity of radiation from a point source emitting isotropically is known to fall of as the square of
the distance from the source. We know that this is a purely geometric effect since the area of the surface
increases as d2 and the same energy therefore spreads over four times the area leading to a decrease in
intensity by a factor of four. This can be clearly seen in Figure 8.1.
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Shobhit Mahajan
Figure 8.1: Inverse Square Law for radiation
§
§(Source: http:// hyperphysics.phy-astr.gsu.edu/ hbase/ forces/ isq.html )
We expect that gamma rays being electromagnetic radiation will exhibit similar behaviour.
8.2
8.2.1
Experiment
Purpose
8.2.2
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The objective of this experiment is to determine the relationship between the intensity
of gamma radiation and the distance from the source to the detector. We expect that the
inverse square law would be valid.
Method
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If we have a point source of gamma radiation, which we assume emits isotropically at a rate of N0
particles per second, and we observe the intensity, I at a distance r from the source, we expect a
relationship like
N0
(8.1)
4πr2
If we use a long lived source such that N0 is constant through the duration of the experiment, we then
will see that as we increase r, the intensity will go down as the r12 . That is a graph of I vs r12 will be
a straight line with a slope N4π0 . Alternatively, a graph of log I vs log r will give us a straight line with
slope −2.
I=
We use a variety of long lived gamma sources and determine the relationship of the intensity and
distance.
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137
Cs,
60
Co,
22
Na and
133
Ba. The decay schemes for these are given below.
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The sources we use are
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Figure 8.2: Decay scheme for
60
Cs
Co
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Figure 8.3: Decay scheme for
137
Figure 8.4: Decay scheme for
200
133
Ba
Lab Manual for Nuclear Physics
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Shobhit Mahajan
Figure 8.5: Decay scheme for
22
Na
We also need a GM counter to measure the intensity at various distances as well as a scale to measure
the distance from the source to the counter window.
8.2.3
Sample Data
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As with all experiments which use a GM counter, we first need to determine the GM characteristics
and find the operating voltage. We then choose a preset time and determine the background radiation
rate so that it can be subtracted from the count rate with the sources to get the net or true count
rate. Finally, with each of the sources placed at different distances, the count rate is measured and
tabulated to see if the inverse square law is valid or not.
Operating Voltage = 425 V
Preset Time = 60 seconds
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Background Count
S.No.
1
2
3
4
5
NB
(NB − NB )
18
-0.8
17
-1.8
20
1.2
19
0.2
20
1.2
NB = 18.8
(NB − NB )2
0.64
3.24
1.44
0.04
1.44
P
(NB − NB )2 = 6.80
Table 8.1: Background Count rate for 60 seconds
With these 5 values of the background counts, we determine the mean to be 18.8 counts in 60 seconds.
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What about the error in the background count rate? Once again, we could think of using the standard
deviation Eq(1.65) to determine the error in the mean. For the σ of the parent distribution, we can use
the sample standard deviation s given by
rP
s=
(x − x)2
N −1
In our case, we get s = 1.30. From this, we can estimate the error in the mean (from Eq(1.65)) to be
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s
σµ = √
N
or
1.3
σNB = √ = 0.58
5
Thus one would think that one should report the result as
NB = 18.80 ± 0.58
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However, there is a subtle point here that one needs to understand. Let us see what we are
doing- we are taking the total number of background counts in 60 seconds and reporting
it as a number. Then we are repeating the same procedure 5 times. Note that in each of
the values of NB , there is an error. This is the inherent statistical error that is associated
with the event which as we know is a result of a Poisson distribution. We can therefore
think of each value of NB as a mean of a Poisson distribution. The associated error with
√
each value of NB is thus NB . Therefore, the correct procedure to exhibit this inherent
statistical error would be to take the expression for the mean value of NB , that is NB and
apply the appropriate error prorogation equation to it. In our case this means
5
P
NB =
2
σN
=
B
=
=
σNB =
=
(NB )i
i=1
5
1 2
2
2
σ(NB )1 + σ(N
+ · · · + σ(N
B )2
B )5
25
1
[18 + 17 + 20 + 19 + 20]
25
94
25
√
94
5
1.93
Thus our background count should be reported as
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(8.2)
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NB = 18.8 ± 1.9
137
Cs source with Activity ∼ 3.1 µ Ci
Counts (N)
8728
4858
2976
1969
1418
1061
867
711
572
485
407
N − NB
8709.2
4839.2
2957.2
1950.2
1399.2
1042.2
839.2
692.2
553.2
466.2
388.2
Error=
1.55
1.16
0.91
0.74
0.63
0.55
0.50
0.45
0.41
0.37
0.34
N − NB
8709.2
4839.2
2957.2
1950.2
1399.2
1042.2
839.2
692.2
553.2
466.2
388.2
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Counts (N)
8728
4858
2976
1969
1418
1061
867
711
572
485
407
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Table 8.2: Measured Count rates at various distances from
d (cm)
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
NB
60
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d (cm)
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
q
N +σ 2
B
yi = R = N −N
60
145.15
80.65
49.28
32.50
23.32
17.37
13.98
11.53
9.22
7.77
6.47
P
yi = 397.29
137
Cs with error estimates
xi = d12 m−2
x2i
2500
625 × 104
1600
256 × 104
1111
123.4321 × 104
816
66.5856 × 104
625
39.0626 × 104
493
24.3049 × 104
400
16 × 104
330
10.89 × 104
278
7.7284 × 104
236
5.5696 × 104
204
4.1616 × 104
P
P 2
xi = 8593
xi = 1178.74 × 104
Table 8.3: Count rates at various distances from
203
137
Cs & Least Square fitting
xi y i
362875
129040
54751
26520
14575
8563
5592
3937
2563
1834
1320
P
xi yi = 611571
Shobhit Mahajan
Lab Manual for Nuclear Physics
We can plot the curves for R vs d as shown in Figure 8.6.
Rate vs distance
160
data
best fit
140
120
80
60
40
20
0
2
3
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Rate
100
4
5
6
7
d (cm)
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Figure 8.6: Graph of R vs d
We can also plot a graph of R vs
we get
1
d2
using the Method of Least Squares. Using Eq(1.75) and Eq(1.76),
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m = 0.0593, c = −10.250
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Rate vs 1/d^2
160
data
best fit
140
100
80
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Rate counts/sec
120
60
40
20
0
0
500
1000
1500
2000
2500
1/d^2 (m)^-2
1
d2
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Figure 8.7: Graph of R vs
Finally, we need to compute the error in our estimation of the slope of this graph. In Section 1.7, we
saw that the uncertainty in the slope determined by this method can be calculated using Eq(1.79).
Since we have the values of m and c, for each xi , we calculate y = mxi + c . We also have the observed
P
P
yi as well as the values of ∆ = N x2i − ( xi )2 . Putting it all together, we then know the uncertainty
in y which is given by
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rP
(yi − mxi − c)2
σy =
(8.3)
N −2
With this uncertainty, we can calculate the uncertainty in the slope determined by the Least Square
method as
r
σm = σy
N
= σy
∆
s
N
P
x2i
N
P
− ( xi )2
Putting in the values, we get a value of
σm = 0.002
We can thus quote the result for the slope of the best fit line as
205
(8.4)
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m = 0.0593 ± 0.002
A similar exercise would need to be done with all the other 3 sources and similar graphs would be
obtained.
Questions
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8.3
1. Why do γ rays follow the inverse square law?
2. Do β rays follow the inverse square law? If not, why not?
3. Do all kinds of radiation follows the inverse square law?
4. Is the inverse square law valid in vacuum only or in matter also?
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5. What could be the sources of systematic errors in this experiment?
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Chapter 9
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Experiment: To Determine the Range of β
rays in Aluminum and to determine the
End Point Energy
Things to know before you do the experiment
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1. All the concepts in Chapter 5 .
2. Concept of solid angle.
3. The β decay process and its description with the help of Fermi theory of β decay.
4. The concept of continuous energy spectrum of β particles and the end point energy.
9.1
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5. Error propagation and estimating errors on the derived quantities.
Introduction
In Chapter 3, we saw how electrons, both mono-energetic and those which are produced in nuclear
beta decay interact with matter. Section 3.2.2, we saw that for a beam of mono-energetic electrons the
energy loss of electrons is much smaller than that of heavy charged particles of the same energy. This
means that they have a much larger range. What is observed experimentally is that for a wide variety
of absorber materials, the product of the range and the density of the absorber is a constant for any
particular electron energy.
The situation is very different for the beta particles emitted by a radioactive source. This is because, as
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we have seen in Section 2.2.2 in Fig 2.2, the energy spectrum of the beta particles is continuous. What
is seen therefore is that the beta particles at the lower end of the spectrum are absorbed even with
a very thin absorber. However, for most part of the spectrum, the transmission of the beta particles
follows an exponential curve with thickness. This is an experimental fact which cannot be easily derived
from fundamental physics. What we see is that the counting rate (or intensity) falls off exponentially
with an attenuation coefficient which depends on the end point energy of the beta particle.
(9.1)
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C = C0 e−µd
where C is the counting rate with the absorber material, C0 is the counting rate without the absorber
and d is the mass thickness in units of mass per unit area. The coefficient µ is the attenuation
coefficient. Thus, a graph between ln CC0 and d would give us a straight line whose slope will be
the attenuation coefficient. This behaviour is shown in Figure 3.10 where the flat part of the curve
corresponds to the count rate going to the constant background value.
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We also know from Chapter 2, that the beta particle spectrum is a continuous one with a maximum
energy determined by the Q value of the nuclear reaction producing the beta particle. Thus, we can infer
that beta particles from a radioactive source will have different penetrating power and the maximum
penetration depth, the range will be for particles with the maximum or end-point energy. It turns out
that for aluminum absorbers, a single range energy relationship for both monoenergetic electrons and
beta particles with energy range 0.01 ≤ E ≤ 2.5 MeV exists as was shown empirically by Katz and
Penfold (Reviews of Modern Physics, 24, page 28, 1952). They found that the relationship is given by
R = 412E0n
n = 1.265 − 0.0954 ln E0
(9.2)
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Here E0 is in MeV and the range R is in units of mass thickness, that is mg cm−2 . Eq(9.2) allows
us to determine the end point energy of the beta rays from a radioactive source. If we can find the
maximum range R0 experimentally, then that will give us the value of Emax which will be the end point
energy since the maximum energy will correspond to the maximum range.
9.2
9.2.1
Experiment
Purpose
In this experiment, we will study the absorption of β rays in aluminum and investigate the exponential
attenuation of Eq(9.1). We will also determine the end point energy of beta rays from 90Sr using the
range-energy relationship Eq(9.2).
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9.2.2
Lab Manual for Nuclear Physics
Method
For this experiment, we would need a GM counter, a radioactive source ( 90Sr), and aluminum absorber
foils of varying thickness.
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We first need to determine the operating voltage of the GM counter as in Chapter 5. We then determine
the background counts for a preset time of say 120 seconds. Next we take a 90Sr source with known
activity. The decay scheme for the source is given in Figure 9.1.
Figure 9.1: Decay scheme for
90
Sr
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We see that the dominant β emission is with a maximum energy of around 2.28 MeV and so we can
assume that the range energy relation given above will be valid.
We first find the count rate without any absorber and then use different aluminum foils to block the
beta particles and measure the count rate with the varying thickness of the absorber which is aluminum.
We continue this process till the number of counts reaches a constant which is the background value
obtained earlier.
9.2.3
Sample Data
Operating Voltage = 420 V
Preset Time = 120 seconds
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90
Sr source with Activity ∼ 3.7 KBq
Background Count
NB
(NB − NB )
45
-1.8
47
-0.2
47
-0.2
49
1.8
48
0.8
NB = 47.2
(NB − NB )2
3.24
0.04
0.04
3.24
0.64
P
(NB − NB )2 = 7.16
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S.No.
1
2
3
4
5
Table 9.1: Background Count rate for 120 seconds
With these 5 values of the background counts, we determine the mean to be 47.2 counts in 120 seconds.
To determine the error in this, we can think of using use Eq(1.65) to determine the error in the mean.
For the σ of the parent distribution, we can use the sample standard deviation s given by
rP
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s=
(x − x)2
N −1
In our case, we get s = 1.33. From this, we can estimate the error in the mean (from Eq(1.65)) to be
s
σµ = √
N
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or
1.33
σNB = √ = 0.59
5
Thus one would think that one should report the result as
NB = 47.20 ± 0.59
Here again, as in the discussion around Eq(8.2), there is a subtle point that one needs to understand.
Let us see what we are doing- we are taking the total number of background counts in 120 seconds and
reporting it as a number. Then we are repeating the same procedure 5 times. Note that in each of the
values of NB , there is an error. This is the inherent statistical error that is associated with the event
which as we know is a result of a Poisson distribution. We can therefore think of each value of NB as
√
a mean of a Poisson distribution. The associated error with each value of NB is thus NB . Therefore,
the correct procedure to exhibit this inherent statistical error would be to take the expression for the
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mean value of NB , that is NB and apply the appropriate error prorogation equation to it. In our case
this means
5
P
2
σN
=
B
=
=
σNB =
=
5
1 2
2
2
σ(NB )1 + σ(N
+ · · · + σ(N
B )2
B )5
25
1
[45 + 47 + 47 + 49 + 48]
25
236
25
√
236
5
3.072
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NB =
(NB )i
i=1
Thus our background count should be reported as
NB = 47.20 ± 3.07
Next we take several values of the counts with the source and without any absorbers.
N0
(N0 − N0 )
(N0 − N0 )2
8082
47.8
2284.84
7975
-59.2
3504.64
8010
-24.2
585.64
8188
153.8
23654.54
7916
-118.2
13971.24
P
N0 = 8034.2
(N0 − N0 )2 = 44000.9
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S.No.
1
2
3
4
5
Table 9.2: Count rates without absorbers for
The sample standard deviation is therefore
rP
r
(x − x)2
44000.9
s=
=
= 104.9
N −1
4
Thus the error in the mean is
s
104.9
σµ = √ = √ = 46.9
5
N
Thus our count rate is
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90
Sr
(9.3)
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N0 = 8034.2 ± 46.9
Again, here too the error calculation given above is not quite correct for exactly the same reason as
we discussed for the error in the background count. We follow the same procedure for calculating the
actual error in N0 .
N0 =
2
σN
=
0
=
=
σN0 =
=
(N0 )i
i=1
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5
P
5
1 2
2
2
σ(N0 )1 + σ(N
+
·
·
·
+
σ
(N0 )5
0 )2
25
1
[8082 + 7975 + 8010 + 8188 + 7196]
25
40171
√25
40171
5
40.08
(9.4)
Thus we should report the value of the counts in the given preset time without any absorbers as
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N0 = 8034.2 ± 40.08
The data for various thicknesses of aluminum plates is as below.
2.7 gm cm−3 .
Thickness(cm) t ( mg cm−2 )
b
0
100
207
256
307
332
356
432
443
461
539
543
La
0
0.037
0.077
0.095
0.114
0.123
0.132
0.160
0.164
0.172
0.200
0.201
No. of Counts N N − NB
8034.2
4396
2768
2346
1938
1946
1796
1188
1041
910
556
586
7987.2
4349
2721
2299
1891
1899
1749
1141
994
863
509
539
212
The density of aluminum is
Error Transmission Coefficient
47.00
66.37
52.70
48.53
44.13
44.22
42.49
34.60
32.41
30.32
23.77
24.40
1
0.5445
0.3407
0.2878
0.2368
0.2378
0.2190
0.1428
0.1244
0.1081
0.0640
0.0670
N −NB
N0 −NB
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551
588
650
699
707
758
775
807
883
888
914
963
988
994
1039
1151
1258
1331
513
402
276
212
199
136
110
103
62
72
65
59
51
50
51
53
51
52
466
355
229
165
152
89
63
56
15
25
18
12
4
3
4
6
4
5
22.86
20.28
16.89
14.88
14.44
12.06
10.93
10.60
8.45
9.02
8.63
8.27
7.77
7.71
7.77
7.90
7.77
7.84
0.0580
0.0440
0.0290
0.0210
0.0190
0.0110
0.0080
0.0070
0.0019
0.0031
0.0022
0.0015
0.0005
0.0004
0.0005
0.0006
0.0005
0.0005
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0.204
0.218
0.241
0.259
0.262
0.281
0.287
0.299
0.327
0.329
0.339
0.357
0.366
0.368
0.385
0.426
0.466
0.493
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Table 9.3: Count rates with absorbers from
90
Sr, Error Estimation & Transmission Coefficient
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b
We can plot the graph of “N − NB vs t” on a semilog scale. The error bars for N − NB are obtained as
discussed in Section 1.5.1. It is important however to be careful when calculating the error in N − NB .
√
√
We know that the error in N is simply N . But the error in NB is NOT NB as discussed above.
Instead, it is 3.07. So to calculate the error in N − NB we need to add the errors in quadrature. Thus,
p
for instance when N = 4396, the error in N − NB is 4396 + (3.07)2 = 66.37. We see that as the net
count rate becomes small, the errors increase drastically and at very small net count rates, the errors
are much more than the count rate itself. We also plot the transmission coefficient (multiplied by a
√
100, to get a percentage) against t. We also plot the graph for N vs t with the error bars given by N .
The graphs are shown in Figure 9.2.
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Beta Rays Range
N-N_B: data
N-N_B data
N-N_B: best fit
Tramsission Data
Transmission Best Fit
N: data
N: data with errorbars
1000
100
10
1
0.1
0.01
0
200
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N, N-N_B, transmission coeff
10000
400
600
800
1000
1200
1400
t( mg cm^-2)
Figure 9.2: Graph of N − NB vs t
La
R
b
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We can see that the net count rate approaches a constant value which is close to zero while the count
rate approaches the background value after a point. This is the point where essentially all the beta
particles have been stopped by the absorber and thus gives us the range for the beta particles from
this particular 90Sr source. To determine the range, we take the N vs t graph and determine the point
where it turns to become the constant background value. The error in the determination of the range
√
is then the difference in the values between the range obtained from N and that from N ± N . For
this sample data, we obtain
R = 900 ± 15 mg cm−2
(9.5)
Using this range, we can now calculate the end point energy for the beta rays from this source, using
Eq(9.2). Thus
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R = 412E0n
n = 1.265 − 0.0954 ln E0
ln R = ln 412 + n ln E0
= ln 412 + (1.265 − 0.954 ln E0 ) ln E0
R
412
= 0
1.265 ± 1.141
0.1908
= 1.92, 299539
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0.0954(ln E0 )2 − 1.265 ln E0 + ln
ln E0 =
E0
E0 = 1.92 MeV
(9.6)
The equation to determine ln E0 is a quadratic and the two solutions are given above. Clearly, the only
reasonable value is E0 = 1.92 MeV.
We can also determine the lower and upper limits of the end point energy corresponding to the error
in the determination of the Range. Thus
R = 900 − 15 mg cm−2 , E0 = 1.88 MeV
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R = 900 + 15 mg cm−2 , E0 = 1.94 MeV
Thus
R
E0 = (1.92 + 0.02 − 0.04) MeV
(9.7)
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The theoretical value of the end point energy for this source is known to be 2.28 MeV (Figure 9.1).
Thus the percentage error in our determination is simply
9.3
% error =
2.28 − 1.92
= 15.8%
2.28
Questions
1. Why do β rays have a continuous energy spectrum?
2. Is the emission of β particles from a radioactive source isotropic?
3. Can one say something about the energy spectrum of the ν̄ which is emitted in the β
decay process?
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4. What is meant by the range or maximum range of β particles?
5. How does the range of the β particles depend on their energy?
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6. How do we derive the range from the graph while minimizing systematic errors?
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Chapter 10
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Experiment: Gamma Ray spectrum using
a Scintillation Counter
Things to know before you do the experiment
1. All the concepts in Chapter 2 .
2. Interaction of Gamma Rays with matter.
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3. Least Square Fitting, Section 1.7.
4. Goodness of Fit- Section 1.8.
Introduction
b
10.1
La
In all the previous experiments, we have been using a GM counter to detect and measure radiation
from radioactive nuclides. In this experiment we will use a different kind of detector, the Scintillation
Counter. This works on a very different principle than a GM counter which is a gas filled detector
working on the principle of an ionising particle or radiation causing an avalanche. We shall study the
working of this counter as well as its method of operation. The basic purpose of the experiment is
to detect and study the energy characteristics of gamma rays. Recall that the GM counter
does not allow us to measure the energy of the incoming radiation. The scintillation
counter on the other hand allows us to measure the energy as we shall see. In this
experiment, we shall be studying the energy spectrum of gamma rays. For this purpose,we
shall be using a scintillation counter and several gamma sources.
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Lab Manual for Nuclear Physics
Theory
The Scintillation counter basically can be thought of a scintillating material which emits light when
radiation or an ionising particle interacts with it, and a mechanism for collecting and measuring the
light which is produced. We shall study these separately. The incoming particle loses its energy and it
is this energy which is ultimately converted into light.
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Actually the use of light produced in certain kinds of materials when radiation or a particle interacts
with it is one of the oldest ways of detection of such radiation. In the historic Rutherford experiment for
instance, the scattered α particles were detected by the light flashes they produced on a Zinc Sulphide
screen. In fact, the first detection of X-rays by Roentgen also used scintillation- the platino-barium
cyanide crystals began to glow when the rays from his apparatus interacted with them. So how is this
light produced and what are the properties that we want the material of the detector to have in order
for it to be useful?
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The emission of light by a material which is excited can be of several kinds. The most common one,
fluorescence is what we get when a material which has been excited, emits light immediately after
excitation and the light is in the visible region of the electromagnetic spectrum. Phosphorescence
is basically the same as fluorescence but here the light emitted is of a much longer wavelength and is
emitted usually on a time scale much longer than fluorescence.
We can define some terms which would be useful later on:
1. Luminescence is the process of exciting a material (not thermally) and the subsequent emission
of light.
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b
2. How the material is excited determines the type of luminescence (e.g. photoluminescence, chemiluminescence, triboluminescence)
3. Fluorescence is photoluminescence or scintillation (i.e. excitation produced by ionizing radiation)
that has a fast decay time (nanoseconds or µs)
4. Phosphorescence is the same as fluorescence, but with a much slower decay time (milliseconds
to seconds)
There are basically two types of scintillating materials- inorganic and organic . These two kinds of
materials have very different mechanisms of production of light. In our laboratory, we use inorganic
scintillators. We shall study these two kinds separately.
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Lab Manual for Nuclear Physics
Inorganic Scintillators
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To understand the phenomenon of scintillation in inorganic materials, we first need to understand the
energy structure of crystalline materials since the mechanism for scintillation depends crucially on the
structure of the crystal. Recall that an atom has discrete energy levels or orbitals. When several
atoms form a molecule we get molecular orbitals which as molecules aggregate into a solid, combine
and become more and more dense. Finally, when a large number of molecules ‘combine’ to form a
solid, the energy levels become so close to each other that they can be considered to form a continuum
which is called an energy band. The width of energy bands depends on the atomic orbitals which are
superposed to form the band. It also happens that there are some energies where there is no overlap at
all and we then get band gaps.The width of the bands of course can vary and depends on the overlap
between the underlying atomic orbitals.
Typically, a solid has an infinite number of allowed bands but most of the them have very high
energies to be of any relevance. It turns out that the electronic properties of solids depends on
the bands which are near the Fermi level. Fermi level (NOT Fermi energy) in the band theory
is a hypothetical level such that at equilibrium, it has a 50% probability of being filled. It is not
necessarily an actual energy level. In the language of Fermi-Dirac statistics, it corresponds to the total
chemical potential of the electrons. The closest band above the Fermi level is called the conduction
band and the one closest below is called the valence band. The band gap is large in insulating
materials, somewhat smaller in semiconductor materials and very small in conductors, as seen in Figure
10.1. The electrons will never have energy in the forbidden region or the band gap. Electrons in a
lower energy state are in the valence band and these are tightly bound to lattice sites. On the other
hand, electrons in the conduction band have a higher energy and are more mobile throughout the crystal.
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Figure 10.1: Band gap in different class of materials
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Now let us consider what happens when radiation (we shall use the term radiation to denote both
ionising radiation and charged particles which can ionise the material), hits a pure inorganic crystal
like Sodium Iodide. The radiation can deposit its energy in the crystal and one of the electrons in the
valence band can gain enough energy to move to the conduction band, leaving a hole in the valence
band. When this excited electron returns to the valence band, it will emit a photon though in a pure
crystal, this is a very inefficient process. Furthermore, the energy gap between the conduction and
valence band is typically so large that the photon emitted is of short wavelength and not in the visible
region.
La
b
To get the crystal to emit light in the visible range, we obviously need to decrease the energy gap that
the electron experiences when it falls or de-excites from the conduction band. This is usually done by
adding a trace amount of impurity to the pure crystal. These impurities are called activators. The
activator sites modify the energy band structure of the pure crystal locally while the overall energy
band structure remains unmodified. The net effect of the activator sites is to create energy levels in
the forbidden region (the region between the conduction and the valence band, where ordinarily in a
pure crystal, the electrons cannot be). This is shown in Figure 10.2.
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Figure 10.2: Energy Band Structure of Pure crystal & Activated Crystal
Now let us consider what happens when radiation or a charged particle enters the crystal with activator
sites. The incoming particle deposits its energy and creates an electron hole pair in the valence band.
This primary electron-hole pair, through a cascade effect creates many secondary electron-hole pairs.
When the energy of the electronic excitations becomes below the ionization energy, thermalization
takes place. At the end of this stage, all the electrons are at the bottom of the conduction band and
the holes at the top of the valence band. This whole process takes place on a time scale of a picosecond.
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b
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After the thermalization stage, the free electron hole pairs migrate through the material. The hole will
migrate to the activator site and ionise it. The electron in the conduction band continues migrating
till it meets one of the ionised activator site and neutralises it. Now we have a neutral activator, but
one which depending on the energy of the electron will be an excited state (of the neutral activator).
This excited state will de-excite to its ground state and in the process give out a photon which, now
since the energy difference between the activator excited and ground state is smaller than the original
band gap of the crystal, will be in the visible region. This process takes place on a time scale of around
10−10 − 10−11 seconds. This is essentially then the time scale of the scintillation which one observes.
Sometimes, the electron in the conduction band when it encounters the ionised activator or impurity
site, neutralises it but goes into an excited state from where the transition to the ground state is
forbidden. In that case, the electron typically gains more energy from thermal motion and moves to
a higher excited state from which it can de-excite to the ground state emitting light. Obviously, this
process is on a much longer time scale and sometimes therefore we get an after glow in scintillating
materials. This component of light emitted, as we saw above is phosphorescence.
It can also happen that the de-excitation of the electron from the activator excited state to its ground
state is such that no visible light is emitted. This process is called quenching.
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The processes we have outlined above depend on the dynamics of an electron-hole pair created in the
valence band, which are essentially independent of each other. In semiconductors, there is another kind
of process which can take place. Instead of the electron and hole moving independently, sometimes it
happens that the electron and the hole form a loosely bound state called an exciton. The exciton
band is typically just below the conduction band and therefore the energy of the electron in an exciton
is somewhat lower than that in the conduction band. Here again, the electron hole exciton moves
together to the activator site and the hole gets neutralised while the electron in the exciton band
neutralises the ionised activator and when it de-excites to the ground state, can emit visible light. It
should be noted that the electron hole pair in an exciton can also recombine at the site of impurities or
traps in a crystal without any external doping of the kind used to create activator sites. The time scale
for the exciton recombination is typically much faster than that of the electron-hole pairs which move
independently since here the electron and hole move together. So we have two kinds of components
in most materials- the fast component which is caused by the recombination of excitons and the slow
component which is when the electrons in the conduction band and holes in the valence band are
captured successively by the activator sites. The fast and slow components can be resolved for most
scintillating materials.
10.2.2
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What we have thus seen is how an incoming radiation interacting with a crystal which has been doped
with an activator produces light through luminescence. One can do an elementary analysis of the
energy efficiency of such materials and we find that for every electron-hole produced by the incoming
radiation, there is roughly one photon produced. Further, note that the emitted light can essentially
pass right through the bulk of the crystal. This is because remember that the energy difference in the
pure crystal bulk between the conduction and valence bands was such that a de-excited electron from
the conduction gap produced short wavelength or high energy photons. The energy difference between
the activator excited states and its ground state is much less and so the light produced is of a longer
wavelength or smaller energy. There is thus usually no absorption of this light by the bulk of the crystal material since its emission(and hence absorption spectrum ) is peaked at a much shorter wavelength.
Organic Scintillators
The scintillation mechanism in organic materials is quite different from the mechanism in inorganic
crystals that we studied in the previous section. In inorganic scintillators, we saw that the scintillation
arises because of the structure of the crystal lattice and the impurities which we introduce. The
fluorescence mechanism in organic materials arises from transitions in the energy levels of a single
molecule and therefore the fluorescence can be observed independently of the physical state. Practical
organic scintillators are organic molecules which have symmetry properties associated with the electron
structure.
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The molecular energy levels of organic molecules which exhibit scintillation are separated by a few
electron volts and they get closer to each other as we go up. The singlet energy levels are subdivided
into a series of levels with a much finer structure because of the vibrational modes of the molecule.
The typical spacing of these is around a tenth of an electron volt.
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Now let us consider the case when radiation interacts with an organic scintillator. The average energy
at room temperature is around 0.025 eV (recall that an energy of 1 eV is roughly equal to the thermal
energy at 104 K, thus a room temperature of ∼ 300 K is about 0.03 eV) and so all the molecules are in
the ground state (called the S00 state where the first subscript indicates the singlet spin state and the
second the fine structure state). When radiation deposits its energy into the material, the electrons
are excited to the upper levels. The higher states like S2 , S3 etc. de-excite in a matter of picoseconds
to the S1 state via transitions which do not produce any radiation. The S1 states like S11 , S12 etc
with higher vibrational energy also lose energy and soon we have all the excited molecules in the S10
state. When these electrons in the S10 state de-excite to the S0 state, we get luminescence. Again,
essentially all the emission light is of a lower energy than that required for absorption and therefore
the organic material is transparent to the luminescence produced like in the case of inorganic scintillator.
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In addition to the transitions in the singlet states, there are also transitions from the triplet states to
the ground state. The triplet states are typically longer lived and therefore the typical time scale of
this transition is longer leading to a phosphorescence. A schematic illustration of the states is shown
in Figure 10.3.
Figure 10.3: Energy States in an organic scintillator§
§(Source:
”Pistates” by Napy1kenobi - Own work.
Licensed under CC BY-SA 3.0 via Wikimedia Commons -
https://commons.wikimedia.org/wiki/File:Pistates.svg#/media/File:Pistates.svg)
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Lab Manual for Nuclear Physics
Photomultiplier Tube
We have seen now how ionising radiation or a charged particle from a radionuclide when it interacts with
a scintillator would produce light because of fluorescence. However, to convert this light into something
which can be detected or its properties measured requires some kind of sensor which is sensitive to the
light. The most often used sensor is a Photomultiplier Tube (PMT). Let us now see how this works.
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The basic purpose of a photomultiplier tube is to convert the light signal into an electrical signal
that is, ultimately convert a photon into one or more electrons which can be detected and whose
properties measured. The PMT basically consists of three components- a photocathode which
will produce the initial or primary electron on interaction with the photon; an arrangement to
accelerate the electron(s) produced and an arrangement to measure the current produced by these
electrons. The whole arrangement has to be in a vacuum tube. As we will see, the efficiency of the
photocathode to produce the primary electrons is not very high. Thus, what is usually done in a
PMT is to have an arrangement which multiplies the number of electrons that is produce a number of
secondary electrons from the primary electrons. Let us see how each of these three components function.
Figure 10.4: Schematic of a Photomultiplier Tube
§(Source: Wikipedia)
The light from the scintillator is made to fall on a photocathode. This is, as the name suggests, made
of a material which by using the photoelectric effect, produces electrons when photons impinge on it.
Obviously, we need to choose the material and the design of the photocathode in such a way that the
energy of the photon is transferred efficiently to the primary electron which is produced. Firstly, recall
that when a photon transfers its energy to an electron, for the electron to emerge from the material
and be detected, it needs to overcome not only the collisions with other electrons and the lattice within
the material but also the surface potential barrier. Clearly then, there has to be a minimum energy of
the photon when this is possible. This means that any photocathode will have a long wavelength cutoff
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depending on the material and the geometry of the photocathode. For our purposes, typical light
given off in the scintillation is in the blue region and thus has an energy of around 3 eV. ( ∼ 400 − 410
nanometers). It turns out that the surface potential barrier or work function for most metals is more
than 3 eV while that for some semiconductors is only around 2 eV. Thus we see that to detect the light
from the scintillator, we need a photocathode made from a suitably prepared semiconductor.
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But this is not enough- as we saw above, the electron on absorbing the energy from the photon,
needs to travel to the surface of the material to be ejected out. In its motion to the surface, it loses
energy also by collisions with other electrons. Clearly, the probability of collisions with other electrons
increases with increased electron density in the material as well as the distance travelled. Thus, metals
have a high energy loss and so the electrons travel only a small distance before losing enough energy to
be unable to escape from the surface. The maximum depth from which an electron can travel to the
surface of a material and still escape is called the escape depth. In metals it is only a few nanometers.
This fact then determines the geometry of the photocathode since we thus need a very thin layer of the
photosensitive material or else most of the electrons produced will not be able to escape. The situation
in semiconductors is better where the escape depth is a few tens of nanometers. Here again, we need to
have a very thin layer in order to maximise the probability of the electrons produced to escape from the
surface. This however, leads to another issue which effects the efficiency of the photocathode- a very
thin layer of photosensitive material means that it will allow a large fraction of the incident radiation
through, thereby reducing the efficiency of the photocathode. Thus, there is a trade off which needs to
be made between these two factors while determining the thickness of the photosensitive material.
The efficiency of the photocathode is usually described by a quantity called quantum efficiency which
is defined as
number of photoelectrons emitted
number of incident photons
b
Quantum Efficiency =
La
Clearly, this is a function of the wavelength of the incident light. Most PMTs have an efficiency of
around 15 − 25%.
The photomultiplier tube, as the name suggests, does more than just produce photoelectrons- it
also has a multiplying effect which we now turn to. First the primary electrons produced in the
photocathode layer are focussed onto a narrow region. In this process, the electrons are also accelerated
in an electric field of a few hundred volts. The focussing is done by using a focussing electrode.
The accelerated primary electrons are then made to produce secondary electrons by the process of
secondary emission using a series of electrodes called dynodes.
A dynode is basically an electrode in a vacuum tube that serves as an electron multiplier through
secondary emission. When the accelerated primary electrons are focussed on a dynode, they transfer
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their energy to the electrons in the dynode material which are ejected out. Of course, just like the
photoelectric effect, the electrons in the dynode material need to have enough energy to overcome the
surface potential barrier which, as we have seen is around 3 − 4 eV. However, the primary electrons
produced by the photocathode were emerging with very low energies, (∼ 1 eV) but these are accelerated
through around 100 V and so when they strike the dynode, have an energy of 100 eV. Thus, if all this
energy was transferred to the electrons in the dynode, we could in principle get around 30 secondary
electrons for each primary electron striking the dynode. Clearly, this is the maximum number of
secondary electrons that can be produced per primary electron. The actual number is significantly less
because once again, for the electrons to emerge from the material, they need to travel to the surface
and in this process lose energy. Only those electrons which reach the surface with energy more than the
work function can escape. Typically, around 6 − 8 secondary electrons are produced for each primary
electron impinging on the dynode.
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The next stage in the PMT is to multiply this number of secondary electrons. This is done by using
different geometries of an array of dynodes. Each dynode when struck by the secondary electrons from
the previous dynode produces more secondary electrons which then impinge on the next dynode to
produce an even larger number, leading to a cascade effect. Various kinds of geometries are used to
achieve this. A fairly simply, box and grid type of arrangement is shown in Figure 10.4. Finally, the
multiplication achieved is around 107 by a PMT. The electrons which finally emerge from the dynode
arrangement are collected and analysed using the associated electronics to which the PMT is connected.
A simplified model will give us an idea of the amount of charge produced by the PMT. Let us define:
La
b
p: The number of light photons produced in the scintillating crystal.
k : Optical efficiency of the scintillating crystal, that is the fraction of transmitted photons.
l : Quantum efficiency of the photoelectrode, that is the number of electrons produced per photon
striking it.
n: The number of dynodes in the setup.
R: The dynode multiplication factor that is the number of secondary electrons produced in a dynode
per primary electron absorbed.
e: The charge of an electron.
Then, we can simply write the amount of charge which comes of the photomultiplier when a gamma
ray photon hits the scintillator crystal. This is simply
Q = pklRn e
To get an estimate of this, assume that the number of light photons produced in the crystal is 1000 and
its optical efficiency is 50%. Further assume that the quantum efficiency of the photocathode is about
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10%. Let the number of dynodes by 10 in the setup and each dynode have a multiplication factor of 5.
Then
Q = 103 × 0.5 × 0.1 × 510 × 1.9 × 10−19 C ≈ 92 picoC
which is an extremely small amount of charge.
10.2.4
Gamma Ray Spectrum
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The PMT output as we have seen is a very small pulse since the amount of charge, even with the
multiplication by the dynode stages is very small. A very sensitive amplifier is connected to the PMT
output to amplify the signal. This is called a pre-amplifier. The amplifier makes the pulse narrower
and the Pulse Height Analyzer then creates an output pulse for input pulses of acceptable height. A
Single Channel Analyzer (SCA) or a Multi Channel Analyzer (MHA) are two kinds of Pulse Height
Analyzers.
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The basic purpose of the experiment, apart from studying the working of a Scintillation counter, is
to detect and study the gamma ray emission from various nuclei. To detect and measure the energy
of the gamma rays, we first need to see how an incoming gamma ray photon would interact with the
scintillation detector. In our case, the detector is a Thallium activated Sodium Iodide (NaI)
scintillating crystal.
La
b
We have already studied the interaction with matter of gamma rays in Section 3.2.3. Recall that there
are basically three different ways for a gamma ray photon to interact with matter- photoelectric effect,
Compton scattering and pair production. Which process dominates the overall cross section depends
on the energy of the incoming gamma rays. Note that in all three processes, a high energy electron is
produced by the incoming gamma ray. (In Compton effect, the free electron is provided with energy
by the gamma ray in elastic scattering.) The production of this high energy electron is crucial since
gamma ray photons being charge neutral, will otherwise not be detected by the scintillator. Since the
spectral distribution of the light produced by the scintillator will depend on the interaction of the
gamma rays with the crystal and we now discuss this.
When a gamma ray photon strikes an atom of the scintillator material, it is absorbed completely and
all of its energy is transferred to one of the bound electrons. Since the energy of the gamma rays is
typically of the order of MeV, it is much larger than the binding energies of the bound electrons. The
electron is therefore released from the atom and moves rapidly through the crystal since it is carrying
the balance energy of the gamma ray photon (the original photon energy minus the binding energy).
This fast moving electron produces the scintillation as we have seen above. However, another process is
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also operative- when the bound electron is freed by the gamma rays, another bound electron in the atom
will fall into the vacancy left by the bound electron which has been freed. Typically, a K-shell electron
is produced by the gamma rays and another bound electron falls into the vacancy in the K-shell. This
process produces x-rays which in turn will free more loosely bound electrons and these too shall produce
the scintillation light pulses. The whole process, that is the original flash produced by the gamma ray
initially and the subsequent flashes will happen within the resolution time of the counter and so usually
cannot be distinguished by the counter. In the end, if the photoelectron stops in the crystal and no
light escapes the crystal, then the total incoming gamma ray energy would have been converted into
the photomultiplier output pulse. This pulse is called the photopeak which has an energy equal to
the incoming gamma ray photon.
However, we know that photoelectric effect is not the only interaction that the gamma rays have with
matter. Compton effect is also operative. We know that when a gamma ray photon scatters off a free
electron, the scattered electron takes away some of the energy of the incoming gamma ray. This electron
then interacts with the scintillator crystal and is detected. What happens to the scattered gamma ray?
The scattered gamma ray escapes from the crystal without any further interaction. This is because
the probability of Compton scattering of a typical gamma ray in a typical scintillator crystal is about
1 − 10%. Thus it is extremely unlikely that the scattered gamma ray will have a second scattering
event. We know that the energy of the scattered electron is given by Eq 3.41
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R
 
(1 − cos θ)

Ee = hνi − hνf = hνi 
hνi
1 + me c2 (1 − cos θ)
hνi
me c2
(10.1)
La
R
b
or the energy of the scattered photon is given by
Eγ0 =
1+
Eγ
Eγ
(1 −
m e c2
cos θ)
(10.2)
Here Eγ is the energy of the incoming gamma ray and Eγ0 is the energy of the scattered gamma photon.
The energy of the scattered electron (which remember is what the scintillator is detecting) is a function
of the angle by which the gamma ray is scattered, θ. The energy of the electron goes from 0 when
2Eγ2
θ = 0 that is the gamma ray is not scattered, to me c2 +2E
when θ = 180 that is the gamma ray photon
γ
is backscattered. This maximum energy of the scattered electron is called the Compton Edge. Also
note that the angular dependence of the scattered electron energy is a slowly varying w.r.t θ. Hence
the energy of the of the electron is essentially constant till it falls off sharply at the Compton Edge.
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In addition to the Compton edge and the Photopeak, there is another effect which can take place.
Suppose a gamma ray encounters Compton scattering outside the material of the scintillator. This can
be in the shielding of the detector for instance. In this case, the scattered gamma ray enters the crystal
and produces a peak by photoelectric effect. However, since the geometry of the detector is such, only
those scattered gamma rays with large values of θ can enter the detector and hence these gamma rays
will have energies close to the maximum photon energy at θ = 180, that is
me c2 Eγ
me c2 + 2Eγ
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Eγ0 =
We can easily see that this is lower than the Compton edge. So typically, one would also see this peak
in the detector at a lower energy. This peak is called the backscatter peak.
La
b
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Finally, we know that for gamma ray energies greater than 2me c2 , pair production is possible. Thus
for high energy gamma rays, we will get a pair of electron and positron. If both of these lose all their
energy within the scintillator, then we will see a peak at the energy of the gamma ray minus the rest
mass energy of the electron and positron. However, it might happen that the positron, before it loses all
its energy, annihilates with an electron and produces two gamma ray photons. Each of these photons
will have an energy of atleast me c2 . Usually the gamma ray photons will have a higher energy since
the positron will have some kinetic energy before it annihilates.
Figure 10.5: Typical Cs-137 spectrum with a NaI(Tl) scintillator
We also know from Section 3.2.3 that the cross section for these three processes of the interaction of
gamma ray photons with matter depend on the energy of the gamma rays. For low energies, we have
seen that photoelectric effect dominates the cross section. And since the photopeak gives us the total
energy of the incoming gamma ray, most scintillators are optimised for this energy range. For the
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NaI:T1 scintillator that we use, a small amount of Thallium is added to the sodium iodide. Thallium
being a heavy metal, has a lot of electrons and the photoelectric cross section depends strongly on the
number of electrons in the atom. Of course, as the energy of the gamma ray increases, the photoelectric
cross section decreases fast but the Compton scattering cross section decreases more slowly and above
a few hundred keV, it is the Compton scattering which dominates. Finally, above 1.02 MeV, pair
production becomes operative.
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The idea behind the experiment is to use a scintillator counter to determine the energy spectrum
of gamma rays from different emitters. However, the set up of the scintillator counter will give us
information on the output voltage or the pulse height obtained from the counter. This needs to
be correlated to the energy of the gamma ray entering the counter. For this purpose calibration
needs to be done. As we have seen above, gamma ray sources typically give a photopeak when
they interact with the material of the scintillator. These are at distinct energies which depends on
the energy of the gamma ray produced by the source. Thus, for instance, 137Cs gives a gamma ray
photon with energy Eγ ≈ 0.662 MeV. So we use various sources to obtain the photopeak voltages
and to calibrate the counter. The relationship between the energy of the incoming gamma ray and
the pulse height (or voltage in our case) is a linear one. By fitting the experimentally obtained
points, we can find the coefficients (the slope and the y-intercept) and therefore get the relationship. This then will allow us to find the energy spectrum for any source by measuring the output voltage.
La
b
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An analysis of the spectrum, specitically finding the peaks requires a Pulse Height Analyser (PHA).
This is usually done with two kinds of circuits, a Lower Level Discriminator (LLD), which as
the name suggests, allows only voltages which are higher than the setting and an Upper Level
Discriminator (ULD) which allows only voltages which are lower than the setting. It is easy to see
that if we choose the two settings of LLD and ULD appropriately, we will be able to get a window in
which we can determine the peak at any part of the spectrum. We can, for instance, start with the
lower end of the spectrum and then keep adjusting the LLD and ULD so as to scan the whole spectrum
at various energies.
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Shobhit Mahajan
Figure 10.6: Use of LLD and ULD to obtain a voltage/energy window§
§(Source: Wikicommons)
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Once we have a window with a Pulse Height Analyser, then the output is fed into a Single Channel
Analyser (SCA). A SCA basically consists of a Ratemeter which measures the number of pulses
produced in unit time in a particular channel or window. A more sophisticated instrument is an or a
Multi Channel Analyser (MCA). This replaces the PHA and the ratemeter with a single instrument
and has an electronic circuit which allows us to look at many channels simultaneously and therefore
get the complete spectrum in one go. Typically, an MCA has 1024 channels.
La
b
A simplified diagram of a Scintillation detector setup with a Single Channel Analyser is given in Figure
10.7.
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Shobhit Mahajan
Figure 10.7: Block Diagram of a scintillation counter with a Single Channel Analyser§
§(Source: Wikicommons)
La
b
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For the Multi Channel Analyser, the set up is shown in Figure 10.8.
Figure 10.8: Block Diagram of a scintillation counter with a Multi Channel Analyser§
§(Source: Wikicommons)
We can now summarise the whole operation then as follows:
Gamma ray photons from a source placed near the scintillator interact with the crystal and produce
photons. These pass through the Photo Multiplier Tube and produce electrons which are fed into a
Pre Amplifier. The Pre Amplifier output is fed into an amplifier which amplifies it. The output of
the Amplifier is then passed through a Pulse Height Analyser where the LLD and ULD settings are
chosen appropriately. Finally, the PHA output goes to a SCA to get the counts in a particular channel
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which can then be correlated to a corresponding voltage. Alternatively, the amplifier output can
go directly to a MCA which allows us to look at many channels simultaneously and obtain the spectrum.
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The decay schemes for some common gamma emitters are given below.
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Figure 10.9: Decay Scheme for
60
Co- Gamma decay
Figure 10.10: Decay Scheme for
22
Figure 10.11: Decay Scheme for
137
La
b
Na- Gamma decay
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Figure 10.12: Decay Scheme for
57
Co- Gamma decay
Thus, we can tabulate the gamma ray energies for 4 known sources which can be used to calibrate the
instrument.
Gamma Energy (MeV)
0.122
1.2746
0.6616
1.1732
1.3325
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Parent Nucleus Lifetime(years) Daughter Nucleus
57
57
Co
271.8 (days)
Fe
22
22
Na
2.605
Ne
137
137
Cs
30.17
Ba
60
60
Co
5.272
Ni
Table 10.1: Nuclear Decay Data for Gamma Ray Sources
La
b
In the case of 22Na, we see from the decay scheme above, that a positron is also emitted. We have
already seen that a positron will annihilate with an electron to produce two gamma ray photons. The
typical energy of these gamma ray photons is 511 keV each (corresponding to the rest mass energy of
the electron/positron). When we measure the gamma ray spectrum of using nuclides which produce
positrons, then we typically see this annihilation peak at 511 keV.
Another important characteristic of the counter is energy resolution. This as the name suggests, is a
measure of the ability of the detector to resolve adjacent peaks in the gamma spectrum. We can define
it is
δE
× 100%
(10.3)
E
Here δE is the Full Width at Half Maximum of the photopeak and E is the energy of the photopeak.
The resolution is basically controlled by the statistical fluctuations of the number of photoelectrons
produced at the photocathode of the photomultiplier tube in the counter. In addition, some of the
other factors controlling the resolution are:
R=
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• The number of photons per scintillation event.
• The number of photons that strike the photocathode.
• The number of photoelectrons released from the photocathode per photon hitting it.
• The number of photoelectrons striking the first dynode.
• The multiplication factor of the photomultiplier tube.
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Typically, the resolution of Na(Tl) detectors is seen to be
k
R ≈ √ × 100%
E
where k is a factor which is characteristic of the particular detector.
10.3
Procedure
Before starting the experiment, it is important to set up the electronics that are used. For this purpose,
the following steps need to be followed:
1. Ensure that the voltage is at a minimum, that is the power supply knob is at its lowest position.
3. NIM-BIM is off.
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2. High Voltage (HV) power supply is switched off.
4. Switch on the mains, that is the AC input.
5. Switch on the NIM-BIM.
b
6. Switch on the power supply module.
La
The operating voltage of NAI(Tl) scintillator used along with the PMT is 500 Volts. Before applying
the operating voltage to the counter, connect the output of the detector to DSO.
Place the source at the top of the detector mount. Slowly increase the voltage in the High Voltage
power supply from 0 to 500 Volts and simultaneously observe the shifting base line in DSO. This
ensures that the connections are done properly as bias voltage is reaching the detector.
Connect the output of the detector to the linear amplifier.
Settings for the amplifier:
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• Fine Gain: 0
• Coarse Gain: 1
• Shaping µs: 2
• Atn: ×1
• Polarity: +
DSO setting:
Voltage Scale: 1 V/div
Timing in µs
Use trigger to observe the output properly.
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Set the amplified voltage to around 6 Volts.
Connect the output of shaping amplifier to the input of the Single Channel Analyzer (SCA).
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SCA Settings:
Mode: WIN
LLD: 0.1 V (Range: 0-10 V)
∆V : 0.1 V (Range: 0-1 V)
Connect output of SCA to the input of the counter. Polariy : + ve. Preset time: 50 seconds.
After setting up the electronics.
137
Cs is placed on the PMT setup. The voltage is set at 500 V.
b
1. Source
La
2. A preset time of 50 seconds is set, with the polarity of the counter timer set at + ve.
3. ULD window is fixed at 0.1 V and the LLD is varied in steps of 0.1 V and the corresponding counts
are noted for 50 seconds each.
4. The graph of Counts versus Voltage of pulses is plotted
5. The position of the photopeak is obtained from the graph.
6. This procedure is repeated for all the sources.
7. Using the photopeaks from these sources, an energy calibration graph is plotted and fit to a straight
line.
8. Using the energy calibration equation, graphs of the energy spectrum are plotted for each source.
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9. From the energy spectrum graphs, the energy resolution of the counter for different energies (of
the photopeaks of the different sources) is obtained.
10. Using the energy spectrum graphs, the back scatter peaks are obtained and compared with the
theoretical values for the sources.
Sample Data
Source: Cs-137 (Gamma Source)
Half-Life: 30.08 years
Activity: 163 kBq
Date of Manufacturing: 01-08-13
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10.4
La
b
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LLD (Volts) Counts (Ni )
0.1
8235
0.2
12699
0.3
25814
0.4
9132
0.5
3874
0.6
3862
0.7
3711
0.8
3623
0.9
3506
1.0
3367
1.1
3510
1.2
3501
1.3
3416
1.4
3500
1.5
3380
1.6
3563
1.7
3873
1.8
4216
1.9
4225
2.0
4095
2.1
3763
2.2
3424
2.3
3170
2.4
2904
237
LLD (Volts) Counts(Ni )
3.6
2065
3.7
2061
3.8
2067
3.9
1934
4.0
1834
4.1
1730
4.2
1391
4.3
1041
4.4
757
4.5
552
4.6
405
4.7
311
4.8
276
4.9
200
5.0
238
5.1
242
5.2
270
5.3
344
5.4
438
5.5
694
5.6
1008
5.7
1747
5.8
2618
5.9
3446
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2662
2512
2445
2276
2224
2240
2137
2057
2035
2073
2070
6.0
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
3985
3727
2752
1724
774
316
126
57
46
37
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2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
Table 10.2: LLD Voltage and Counts for
137
Cs
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We can plot the counts versus voltage. The plot is shown in Fig 10.13.
Gamma Spectrum for Cs-137
4500
"cal1.txt" u 1:2
"cal1.txt" u 1:2
4000
3500
2500
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Counts
3000
2000
1500
1000
500
0
0
1
2
3
4
5
6
Voltage (V)
Figure 10.13: Voltage vs Counts for
Source: Na-22 (Gamma Source)
238
137
Cs
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Half-Life: 2.602 years
Activity: 174 kBq
Date of Manufacturing: 01-08-13
Note: The gain of the amplifier was changed to accommodate the higher energy pulses.Gain: x2.3674
LLD (Volts) Counts(Ni )
3.2
1702
3.3
1741
3.4
1591
3.5
1633
3.6
1592
3.7
1573
3.8
1672
3.9
1702
4.0
1688
4.1
1679
4.2
1528
4.3
1227
4.4
881
4.5
535
4.6
441
4.7
39
4.8
409
4.9
477
5
705
5.1
1168
5.2
1639
5.3
1820
5.4
1367
5.5
621
5.6
236
5.7
125
5.8
109
5.9
147
6.0
107
6.1
122
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LLD (Volts) Counts (Ni )
0.0
15632
0.1
15624
0.2
15660
0.3
15557
0.4
15456
0.5
16332
0.6
16926
0.7
18841
0.8
19093
0.9
17334
1.0
14721
1.1
13490
1.2
12246
1.3
10341
1.4
7072
1.5
4832
1.6
4414
1.7
4994
1.8
6514
1.9
10700
2.0
19735
2.1
30084
2.2
21687
2.3
6027
2.4
2324
2.5
2114
2.6
2070
2.7
1896
2.8
1839
2.9
1795
3.0
1788
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3.1
1768
Table 10.3: LLD Voltage and Count rate for
22
Na
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We can plot the counts versus voltage. The plot is shown in Fig 10.14.
Gamma Spectrum for Na-22
35000
"cal3.txt" u 1:2
"cal3.txt" u 1:2
30000
25000
Counts
20000
15000
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10000
5000
0
1
2
3
4
5
6
7
Voltage (V)
Figure 10.14: Voltage vs Counts for
22
Na
La
b
0
Source: Co-60 (Gamma Source)
Half-Life: 5.2712 years
Activity: 134 kBq
Date of Manufacturing: 01-08-13
Note: The gain of the amplifier was changed to accommodate the higher energy pulses.Gain: x2.3674
LLD (Volts) Counts (Ni )
0
1077
0.1
1128
0.2
1201
LLD (Volts) Counts(Ni )
2.1
579
2.2
543
2.3
554 4.3
240
LLD (Volts) Counts(Ni )
4.1
349
4.2
333
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1311
1321
1273
1187
1136
1054
1222
1231
1142
1000
911
795
702
655
599
617
567
572
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
-
553
525
532
509
520
502
516
523
550
578
523
525
587
576
603
455
440
-
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
-
299
273
223
263
417
546
477
212
75
69
155
257
356
238
96
30
-
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0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
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Table 10.4: LLD Voltage and Count rate for
60
La
b
We can plot the counts versus voltage. The plot is shown in Fig 10.15.
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Gamma Spectrum for Co-60
1400
"cal4.txt" u 1:2
"cal4.txt" u 1:2
1200
1000
600
400
200
0
0
1
Nu
cle
ar
Ph
ys
ics
Counts
800
2
3
Voltage (V)
4
Figure 10.15: Voltage vs Counts for
La
b
M
an
ua
l
Source: Co-57 (Gamma Source)
Half-Life: 271.8 days
Activity: 149 kBq
Date of Manufacturing: 17-08-15
LLD (Volts) Counts (Ni )
0.0
5594
0.1
5768
0.2
2700
0.3
2109
0.4
2012
0.5
2073
0.6
3350
0.7
6197
0.8
12020
0.9
19935
1.0
29708
1.1
52454
1.2
94828
242
5
60
Co
6
Shobhit Mahajan
Lab Manual for Nuclear Physics
87560
31426
7605
2127
1403
1363
1267
1385
1329
1479
1364
Nu
cle
ar
Ph
ys
ics
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
Table 10.5: LLD Voltage and Count rate for
57
Co
M
an
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We can plot the counts versus voltage. The plot is shown in Fig 10.16.
Gamma Spectrum for Co-57
110000
"cal2.txt" u 1:2
"cal2.txt" u 1:2
100000
90000
80000
60000
La
b
Counts
70000
50000
40000
30000
20000
10000
0
0
0.5
1
1.5
2
Voltage (V)
Figure 10.16: Voltage vs Counts for
Energy Calibration
243
57
Co
2.5
Shobhit Mahajan
Lab Manual for Nuclear Physics
Energy (keV)
122
511
662
1275
1173
1333
Nu
cle
ar
Ph
ys
ics
Voltage (Volts)
1.24
5.05
6.00
12.535
11.644
13.402
Table 10.6: Voltage vs Energy of Photopeaks (and annihilation peak for
22
Na) for different gamma sources
One can see that the relationship is roughly linear. On doing the least square fit to a straight line, one
gets
m = 99.454 c = 19.353
and therefore the energy-voltage relationship is given by
M
an
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E = 99.454V + 19.353
La
b
where V is in Volts and E is in keV. The points and the best fit line are plotted in Figure 10.17.
244
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Energy Calibration
1400
"res4.txt" u 1:2
99.454*x+19.353
1200
800
600
400
200
0
0
2
4
Nu
cle
ar
Ph
ys
ics
Energy(keV)
1000
6
8
10
12
14
Voltage
Figure 10.17: Energy Calibration
M
an
ua
l
One can also use Eq(1.79) and Eq(1.80) to find the errors in the slope and the intercept. This gives us
σm = 2.143 σc = 20.247
La
b
The error as we see, in the slope is not high (∼ 2%) but that in the intercept is high, almost 100%.
However, remember that the scale on the y axis is greater and therefore an error of 20 in the y-intercept
is not significant.
To test how good our fit is, we can use the χ2 test. Recall from Eq(1.68) that is χ2 is defined as
2
χ =
N
X
(yi − M xi − C)2
σy2
i=1
where
r
σy =
1 X
(yi − M xi − C)2
N −2
The quantity of interest for us is the reduced χ2 , that is χ̃2 , which is defined as
χ̃2 =
245
χ2
ν
Shobhit Mahajan
Lab Manual for Nuclear Physics
where
ν = N − Nc
where N is the number of sample frequencies and Nc is the number of constraints. In our case, N is
the number of data points that is 6 and the number of constraints is 1. A fit is considered good if the
value of the reduced χ2 , that is χ̃2 is close to one.
Nu
cle
ar
Ph
ys
ics
In our case, we can calculate the reduced χ2 and we get a value of χ̃2 = 0.8 which indicates that the
fit of our energy calibration equation is a very good one.
With the calibration done, we can now plot the counts versus energy to get the actual gamma ray
spectrum.
Source: Cs-137 (Gamma Source)
Half-Life: 30.08 years
Activity: 163 kBq
Date of Manufacturing: 01-08-13
La
b
M
an
ua
l
LLD (Volts) Counts (Ni ) Energy (keV)
0.5
3874
69.080002
0.6
3862
79.02
0.7
3711
88.97
0.8
3623
98.91
0.9
3506
108.86
1.0
3367
118.80
1.1
3510
128.75
1.2
3501
138.69
1.3
3416
148.64
1.4
3500
158.58
1.5
3380
168.53
1.6
3563
178.47
1.7
3873
188.42
1.8
4216
198.37
1.9
4225
208.31
2.0
4095
218.26
2.1
3763
228.20
2.2
3424
238.15
246
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Lab Manual for Nuclear Physics
La
b
3170
2904
2662
2512
2445
2276
2224
2240
2137
2057
2035
2073
2070
2065
2061
2067
1934
1834
1730
1391
1041
757
552
405
311
276
200
238
242
270
344
438
694
1008
1747
2618
3446
3985
248.09
258.04
267.98
277.93
287.87
297.82
307.76
317.71
327.66
337.60
347.55
357.49
367.44
377.38
387.332
397.27
407.223
417.16
427.11
437.05
447.00
456.95
466.89
476.84
486.78
496.73
506.67
516.62
526.56
536.51
546.45
556.40
566.35
576.29
586.24
596.18
606.13
616.07
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cle
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Ph
ys
ics
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2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
247
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Lab Manual for Nuclear Physics
3727
2752
1724
774
316
126
57
46
37
626.02
635.96
645.91
655.85
665.80
675.74
685.69
695.64
705.58
Nu
cle
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Ph
ys
ics
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
Table 10.7: LLD Voltage, Count rate & Energy for
137
Cs
Energy Spectrum for Cs-137
4500
"cal11.txt" u 3:2
"cal11.txt" u 3:2
4000
3500
M
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Counts
3000
2500
2000
1500
b
1000
La
500
0
0
100
200
300
400
Energy (keV)
500
Figure 10.18: Energy Spectrum for
600
700
800
137
Cs
We can use this graph to find the energy resolution. For this we need the Full Width at Half Maximum.
From the graph, the peak is at E = 618 keV. The Full Width at Half Maximum, δE is 53.8 keV. Then,
by using Eq(10.3), we get
R=
δE
× 100% = 8.71%
E
248
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Lab Manual for Nuclear Physics
Source: Na-22 (Gamma Source)
Half-Life: 2.602 years
Activity: 174 kBq
Date of Manufacturing: 01-08-13
Note: The gain of the amplifier was changed to accommodate the higher energy pulses.Gain: x2.3674
La
b
M
an
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l
Nu
cle
ar
Ph
ys
ics
LLD (Volts) Counts (Ni ) Energy (keV)
0.0
15632
19.35
0.1
15624
29.29
0.2
15660
39.24
0.3
15557
49.18
0.4
15456
59.13
0.5
16332
69.08
0.6
16926
79.02
0.7
18841
88.97
0.8
19093
98.91
0.9
17334
108.86
1.0
14721
118.80
1.1
13490
128.75
1.2
12246
138.69
1.3
10341
148.64
1.4
7072
158.58
1.5
4832
168.53
1.6
4414
178.47
1.7
4994
188.42
1.8
6514
198.37
1.9
10700
208.31
2.0
19735
218.26
2.1
30084
228.20
2.2
21687
238.15
2.3
6027
248.09
2.4
2324
258.04
2.5
2114
267.98
2.6
2070
277.93
2.7
1896
287.87
2.8
1839
297.82
2.9
1795
307.76
249
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Lab Manual for Nuclear Physics
La
b
1788
1768
1702
1741
1591
1633
1592
1573
1672
1702
1688
1679
1528
1227
881
535
441
39
409
477
705
1168
1639
1820
1367
621
236
125
109
147
107
122
317.71
327.66
337.60
347.55
357.49
367.44
377.38
387.33
397.27
407.22
417.16
427.11
437.05
447.00
456.95
466.89
476.84
486.78
496.73
506.67
516.62
526.56
536.51
546.45
556.40
566.35
576.29
586.24
596.18
606.13
616.07
626.02
Nu
cle
ar
Ph
ys
ics
M
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3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
6.1
Table 10.8: LLD Voltage, Count rate & Energy for
22
Na
While plotting the energy spectrum, note that we need to factor in the gain of 2.3674. Consequently,
the voltage values need to be multiplied by this factor before one finds the energy using the calibration
250
Shobhit Mahajan
Lab Manual for Nuclear Physics
equation.
Energy Spectrum for Na-22
35000
"cal31.txt" u 3:2
"cal31.txt" u 3:2
30000
Counts
20000
15000
10000
5000
0
0
200
400
Nu
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Ph
ys
ics
25000
600
800
Energy (keV)
1000
22
1400
1600
Na
M
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Figure 10.19: Energy Spectrum for
1200
We can use this graph to find the energy resolution. For this we need the Full Width at Half Maximum.
From the graph, the peak is at E = 520 keV. The Full Width at Half Maximum, δE is 67.4 keV. Then,
by using Eq(10.3), we get
δE
× 100% = 12.97%
E
b
R=
La
Source: Co-60 (Gamma Source)
Half-Life: 5.2712 years
Activity: 134 kBq
Date of Manufacturing: 01-08-13
Note: The gain of the amplifier was changed to accommodate the higher energy pulses.Gain: x2.3674
LLD (Volts) Counts (Ni ) Energy (keV)
0.0
1077
19.35
0.1
1128
29.29
0.2
1201
39.24
0.3
1311
49.18
251
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Lab Manual for Nuclear Physics
La
b
1321
1273
1187
1136
1054
1222
1231
1142
1000
911
795
702
655
599
617
567
572
579
543
554
553
525
532
509
520
502
516
523
550
578
523
525
587
576
603
455
440
349
59.13
69.08
79.02
88.97
98.91
108.86
118.80
128.75
138.69
148.64
158.58
168.53
178.47
188.42
198.37
208.31
218.26
228.20
238.15
248.09
258.04
267.98
277.93
287.87
297.89
307.76
317.71
327.66
337.60
347.55
357.49
367.44
377.38
387.33
397.27
407.22
417.16
427.11
Nu
cle
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Ph
ys
ics
M
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0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
252
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Lab Manual for Nuclear Physics
333
306
299
273
223
263
417
546
477
212
75
69
155
257
356
238
96
30
437.05
447.00
456.95
466.89
476.84
486.78
496.73
506.67
516.62
526.56
536.51
546.45
556.40
566.35
576.29
586.24
596.18
606.13
Nu
cle
ar
Ph
ys
ics
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
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Table 10.9: LLD Voltage, Count rate & Energy for
60
Co
La
b
While plotting the energy spectrum, note that we need to factor in the gain of 2.3674. Consequently,
the voltage values need to be multiplied by this factor before one finds the energy using the calibration
equation.
253
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Energy Spectrum for Co-60
1400
"cal41.txt" u 3:2
"cal41.txt" u 3:2
1200
1000
600
400
200
0
0
200
400
Nu
cle
ar
Ph
ys
ics
Counts
800
600
800
Energy (keV)
1000
Figure 10.20: Energy Spectrum for
1200
60
1400
1600
Co
M
an
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l
We can use this graph to find the energy resolution. For this we need the Full Width at Half Maximum.
From the graph, the peak is at E = 1336 keV. The Full Width at Half Maximum, δE is 69 keV. Then,
by using Eq(10.3), we get
R=
δE
× 100% = 5.10%
E
La
b
Source: Co-57 (Gamma Source)
Half-Life: 271.8 days
Activity: 149 kBq
Date of Manufacturing: 17-08-15
LLD (Volts) Counts (Ni ) Energy (keV)
0.0
5594
19.35
0.1
5768
29.29
0.2
2700
39.24
0.3
2109
49.18
0.4
2012
59.13
0.5
2073
69.08
0.6
3350
79.02
0.7
6197
88.97
254
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Lab Manual for Nuclear Physics
12020
19935
29708
52454
94828
87560
31426
7605
2127
1403
1363
1267
1385
1329
1479
1364
98.91
108.86
118.80
128.75
138.69
148.64
158.58
168.53
178.47
188.42
198.37
208.31
218.26
228.20
238.15
248.09
Nu
cle
ar
Ph
ys
ics
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
La
b
M
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Table 10.10: LLD Voltage, Count rate & Energy for
255
57
Co
Shobhit Mahajan
Lab Manual for Nuclear Physics
Energy Spectrum for Co-57
110000
"cal21.txt" u 3:2
"cal21.txt" u 3:2
100000
90000
80000
60000
50000
40000
30000
20000
10000
0
0
50
Nu
cle
ar
Ph
ys
ics
Counts
70000
100
150
200
250
Energy (keV)
Figure 10.21: Energy Spectrum for
57
Co
M
an
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l
We can use this graph to find the energy resolution. For this we need the Full Width at Half Maximum.
From the graph, the peak is at E = 143.6 keV. The Full Width at Half Maximum, δE is 26.5 keV.
Then, by using Eq(10.3), we get
δE
× 100% = 18.5%
E
Once we have the energy calibration done and have the energy spectrum for the sources that we have
used, we can now find the back scatter peaks in the spectrum and compare with the theoretical values
from the tables.
La
b
R=
Backscatter peaks- Experimental and Theoretical values
Source Theoretical Value (keV) Experimental Value (keV) % Error
137
Cs
184.4
204.9
11.1 %
22
Na
199.8
198.8
0.5 %
60
Co
210 & 214
248.7
17.3 %
Table 10.11: Backscatter peaks- Theoretical & Experimental values for various Sources
256
Shobhit Mahajan
10.5
Lab Manual for Nuclear Physics
Questions
1. What is the difference between Fluorescence & Phosphorescence
2. How are bands formed in a solid?
3. What is band gap? How is the band gap different between an insulator, a conductor
and a semiconductor?
Nu
cle
ar
Ph
ys
ics
4. What are activators? What is their function in a crystal?
5. Describe the functioning of a photomultiplier tube.
6. What is a dynode and why is it needed in a PMT?
7. Describe the various interactions of a gamma ray photon with matter.
8. What is the origin of the Compton Edge in a gamma ray spectrum?
9. How are backscatter peaks formed in a gamma ray spectrum?
La
b
M
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10. What is a photopeak and how is it formed?
257
Shobhit Mahajan
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La
b
M
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p-Value Tables
Nu
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Appendix A
258
La
b
M
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Nu
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Ph
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La
b
M
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l
Nu
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Ph
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La
b
M
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Nu
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Ph
ys
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Shobhit Mahajan
Lab Manual for Nuclear Physics
Using MS Excel
Nu
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Ph
ys
ics
Appendix B
The experimental data that we obtain in our experiments needs to be analysed. Frequently, this
means computing the errors, plotting the data with the error bars, finding the best fit curve through
the data etc. Sometimes, we also need to compare the data with some expected theoretical values,
as for instance in the case of Counting Statistics for G.M Counter (Chapter 5). All this analysis,
including plotting of data can be easily done using the Microsoft Excel package. This appendix will give
some basic tips on how to use MS Excel for the purposes of analysis of data obtained in the experiments.
Microsoft (MS) Excel is a simple spreadsheet program.
R
M
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l
SOME HANDY TIPS ON MS EXCEL
La
b
You can highlight a group of cells by clicking on one cell, holding the mouse button
down, and dragging the mouse over the spreadsheet. The cells that are highlighted
will appear black with a black cell border, except for the first cell highlighted, which
will remain white.
To move the contents of a cell (or many cells) from one place to another, highlight the
cell or group of cells, place your cursor on the sides of the cell (on the black outline
of the cell) – the mouse cursor will change to an arrow from a fat cross, click and
hold the mouse button on the border, and drag the cell to its final destination. The
destination cell will be overwritten!
To copy the cell or group of cells, highlight the cell(s), click Edit à Copy. Then
highlight the destination cell(s) and click Edit à Paste. If you want to paste the
formulas in the cells or just the values of the cells, you can select Edit à Paste Special.
262
Shobhit Mahajan
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You can’t solve symbolic equations in Excel. This means that whatever complicated
mathematical expression you type in as an equation must return a number. You can
have cell references in an equation, but the cell that is referenced must contain a
number. If you reference a blank cell, the number 0 is automatically inserted. The
standard operations are +, −, ∗, /, ab
Nu
cle
ar
Ph
ys
ics
When you want to type in an equation into a cell, the first character you type must be
an ”=” (equals sign). This tells Excel to evaluate whatever comes after it, otherwise
Excel will just treat it like a string (a bunch of letters) and not evaluate the equation.
Simple calculations
La
B.1
b
M
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l
You can also enter in a cell reference into an equation. While typing the equation,
you can either manually type in the appropriate cell reference or click on the cell
with your mouse. You can drag and fill equations to make a series of equations as
well as dragging and filling in numbers . For example, let’s say you had a series of x
values from 0 to 10, incremented by 1. You want to multiply each of those cells by
2 and put those new values in a separate column. The A column has the first series
of numbers. The B column is where you type in the equation that multiplies the A
column by 2. You can manually type in the equation into B1, referencing A1 in the
equation. Now, you hit enter to enter in the equation into the cell. B1 shows the
value that the equation returns (0 ∗ 2) = 0. Now select the lower right-hand corner of
the highlighted B1, hold the mouse button down, and drag down to B11.There are
now similar equations in all of the B column cells.
The first step for any analysis is to enter the data. As an example to illustrate some of the things one
can do with MS Excel, we will choose the data for counting statistics from Chapter 5.
The data for the preset time of 5 seconds is as follows:
263
Shobhit Mahajan
Lab Manual for Nuclear Physics
fi
27
55
60
31
16
06
05
Nu
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xi
0
1
2
3
4
5
6
Table B.1: Raw Data
M
an
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Figure B.1: Raw Data
fi
pei = P
fi
La
b
The data in Table B.1 gives us the frequency fi for the number of counts x, with a total number of
P
readings that we have taken, which in this case is
fi = 200. We need to determine the probability
e
pi , of getting the various number of counts. Clearly, this is simply
We can now start using Excel. When you start Excel, you start with a blank ”sheet” with rows and
columns identified as numbers for rows and Capital letters for columns. Enter the data above in a
blank worksheet. We now have a table where each entry can be identified by a number and a letter.
Thus, for instance, A2 will be 0, B5 will be 31 etc as shown in Figure B.1.
P
Next we need to determine the probability of each of those counts. For this we need
fi . It is easy
to do this in Excel. Simply select all the data in column B (which contains the frequencies fi ) and use
the AUTOSUM function which you can see on the top right hand side of the screen. This will give
P
you
fi for the data in column B which will be displayed in cell B9.
To find the probabilities for each count, we need to use the formula above. Excel provides a way to do
this calculation easily. In the cell C2, write
264
Shobhit Mahajan
Lab Manual for Nuclear Physics
= (B2/200)
and press Enter.
What this will do is to take the value of the cell B2 and divide the value by 200 and display the result in cell C2. However, we need to do this for all the data in column B. There is an easy way to do this.
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First one needs to enable a functionality of Excel called AUTOFILL. To do this,
1. Click the File tab, and then click Options.
2. Click Advanced, and then under Editing options, select or clear the Enable fill handle
and cell drag-and-drop check box to show or hide the fill handle.
3. To avoid replacing existing data when you drag the fill handle, make sure that the
Alert before overwriting cells check box is selected. If you don’t want to see a message
about overwriting nonblank cells, you can clear this check box.
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Now, you need to simply select the cell C2, and take the mouse to the LOWER right corner of the
cell. Then while HOLDING the LEFT button of the mouse, simply move down to C8 (since the
data is from B2 to B8). This will automatically perform the SAME operation for the different values
of the column B , that is in C3 it will give B3/200, in C4, it will give B4/200 etc. Thus we have all
the probabilities now with the count rates. To check, we can use AUTOSUM to find the sum of the
probabilities in column C and display it in C9. The table will now look like Table B.2
xi
fi
0
1
2
3
4
5
6
27
55
60
31
16
06
05
200
pe =
Pfi
fi
0.135
0.275
0.300
0.155
0.080
0.030
0.025
1
Table B.2: Raw & Probability Data
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Figure B.2: Raw & Probability Data
However, we need the sample mean and sample variance for this data to analyse and compare the
experimental data with the theoretically expected result. Though Excel has inbuilt statistical functions
for mean and variance AVERAGE & STDEV, one cannot use them with a frequency table. However,
this is easy to do using the defining formulae for the sample mean and the sample variance. The sample
mean is given by
P
xi f i
x̄ = P
fi
and the sample variance is given by (Equation 1.9)
"
N
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1 X
σE2 =
(xi − x̄)2
N − 1 i=1
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To calculate the sample mean, we use the inbuilt formulae in Excel called SUMPRODUCT and
SUM. SUMPRODUCT(array 1, array 2) has arguments array 1, array 2, array 3, etc. In our
case, we want to multiply the corresponding data of the column A with the data of column B and then
sum it up. So our array 1 is A2:A8 and array 2 is B2:B8. The function SUM(number 1, number 2,
...) simply sums up numbers number 1, number 2, etc. Therefore to find the sample mean, we do the
following. We click on any empty cell, say F 10 and enter the formula
=SUMPRODUCT(A2:A8,B2:B8)/SUM(B2:B8)
Here remember that we dont need to write A2:A8 by hand. All we need to do is to select the cells from
A2 to A8 and this gets entered automatically. Similarly for B2:B8. To calculate the sample variance,
we use these formulae in another empty cell , say F 11, by typing
=(SUMPRODUCT(A2:A8ˆ 2,B2:B8)-SUM(B2:B8)*F10ˆ 2)/(SUM(B2:B8)-1)
where recall that F10 has the sample mean x̄ = 1.94.
We thus have the sample mean x̄ and the variance σE2 and also σE =
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The theoretical expectation is that the distribution follows a Poisson distribution whose probability
distribution function P (xi , µ), is given by
P (xi , µ) =
µxi −µ
e
xi !
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We take the mean µ of the underlying distribution to be the sample mean x̄ and then try to see how
closely the sample distribution follows the expected Poisson distribution with the SAME mean. To do
this, we need to calculate the Poisson distribution function P (xi , x̄) for various xi . This is easy to do
in Excel. We select an empty cell, say D2 and enter
=(D10ˆ (A2)*exp(-D10))/fact(A2)
This will give us
1.940 ∗ exp(−1.94)/(0!) = 0.1437
Now we need to calculate this for all the values of xi . For this, we again point the cursor to the
BOTTOM RIGHT corner of the cell D2 and pull it down using the LEFT button on the mouse.
Excel then calculates this formula for each value of A2:A8 and displays it in the cells D2:D8. We can
also use AUTOSUM to find the sum of the Poisson probabilities in D2:D8 and display it in D9 The
table will now look something like this
fi
pe =
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xi
0
1
2
3
4
5
6
27
55
60
31
16
06
05
200
Pfi
fi
0.135
0.275
0.300
0.155
0.080
0.030
0.025
1
Poisson
0.144
0.279
0.271
0.175
0.084
0.030
0.011
0.9961
Table B.3: Data & Poisson Distribution
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Figure B.3: Raw & Probability Data
B.2
Plotting Data
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Now we are ready to plot this data to get the shape of the probability distribution. Excel also allows us
to do this easily. Go to the Tool Bar and click on INSERT. In the INSERT Menu, click on Scatter in
the CHARTS section. Within the SCATTER section of the CHARTS, click on SMOOTH LINES
WITH MARKERS. An empty window will open on the screen. You can move and resize the window
if you want. Next, click SELECT DATA. This will open a dialog box. Now, remember we want to
plot two curves. One, the plot of the experimental probabilities versus the counts, that is the values in
column A and column C. We also want to plot the theoretically expected Poisson probabilities versus
counts that is the values in column A and column D. So after clicking on SELECT DATA, take the
cursor to A1 and press the LEFT button on the mouse. Press the CTRL key while KEEPING THE
MOUSE BUTTON PRESSED. Now scroll the mouse down to A8 and leave the Mouse Key BUT
KEEP THE CTRL key pressed. Take the cursor to C1 and press the left button on the mouse and
scroll down to C8 and then while keeping the CTRL key pressed once again go to D1 and press the
mouse button and scroll down to D8. Now we have selected three columns, A, C and D. Pressing OK
in the dialog box will give you the two graphs in the empty window. That is it! It is as simple as that.
Figure B.4: Experimental & Poisson Graph & Data
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The graph will look like Figure B.5.
Figure B.5: Experimental & Poisson Graph
B.2.1
Error Bars
What about the error bars? The error bars are the error on pe . However, we know that pe is actually
a derived quantity, since
fi
pe = P
fi
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√
P
Now
fi = 200 and obviously there is no error in this. We take the error in fi as fi and get the
√
fi
error bars by taking the error in pe to be 200
.
So we need to calculate this quantity for each value of pe . Once again, this is easy to do. In the cell
E2, we can write the formula
=(SQRT(B2)/200)
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This will take the value in the cell B2, which in our case is 27, take its square root and divide by 200.
This will give us the error in the corresponding pe . Once again, we need to repeat this for every value
of fi and so we select the cell E2, take the pointer to its lower right corner and while pressing the left
button on the mouse, scroll down to E8. This will take the corresponding values in the cells B3,B4,
etc. and calculate the error by the formula used in E2. Now we have the errors in our experimental
distribution.
The screen will look like Figure B.6.
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Figure B.6: Error Bar Calculation
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Next, we need to plot these on the graph we have obtained earlier in Figure B.5. This is easy to do in
MS Excel. Go to the Layout tab on the top. There you will see a tab for Error Bars. When you press
it, it gives you several options like None, Error Bars with Standard Error etc.. Press the button
for More Error Bar Options. This will open a dialog box with Add Error Bars and in our case
give you two options- Adding error bars to the series pE and to Poisson distribution graph. Choose the
series pE . This will open another dialog box with Format Error Bars. Choose the option Vertical
Error Bars and in Error Amount, choose Custom and press the button Specify Value. This will
open another dialog box with specify positive and negative values of the errors. In BOTH choose the
column E2:E8. To do this, take the cursor to E2 and with the mouse button pressed, move down to
E8. This will automatically put the values in these cells in the positive values of errors. DO the same
for the negative values of the errors and press OK. This will draw the error bars on the experimental
data points. The graph will look look Figure B.7.
Figure B.7: Error Bars on Experimental Curve
You can also change the shape, thickness, color etc of the error bars using the format error bars option.
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Lab Manual for Nuclear Physics
Formatting Graphs & Plots
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Now that we have the basic graph done,we can play around with Excel and try to make it better. We
can resize it by just going to one of the corners and clicking on the mouse and dragging the mouse to
make the graph bigger or smaller. Or we can click on the graph and then on the TOOLBAR open
the LAYOUT tab. This allows us various options of choosing the Chart Title (the size, the placement
etc.), Titles for the axis ( the size, the positions, the orientations etc.), Gridlines both in the horizontal
and vertical directions (No lines, major lines in one or both directions, minor lines in one or both
directions, both major and minor lines in one or both directions), changing the colors of the graphs etc.
You are encouraged to try out various options and see how the graph changes. When you are finished,
you can right click on the graph and copy the graph and paste it in one of the imaging programs like
MS Paint and then save it in the format you want (PNG, JPEG, BMP etc.). The final output could,
for instance, be like in Figure B.8.
B.3
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Figure B.8: Experimental & Poisson Graph with Error Bars
Fitting Data
We have seen how to plot data graphs in Excel from the given data. Most of the times in our
experiments, we need to fit a straight line graph to the data and determine the slope and the intercept
of the data. This too is very simple in Excel. Consider the data that we obtain from the experiment
on verification of inverse square law for gamma rays, Chapter 8. We see from Table 8.3, that the data
for the count rate and the distance is certainly not linear. However, we expect that the plot between
count rate and d12 will be a straight line. So let us see how to obtain the graph and also the least square fit.
Our data is basically as follows :
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NB = 18.8/ minute
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d(cm) N(counts/minute)
2
8728
2.5
4858
3.0
2976
3.5
1969
4.0
1418
4.5
1061
5.0
867
5.5
711
6.0
572
6.5
485
7.0
407
Table B.4: Count rates at various distances from
137
Cs
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We need to first find out the count rate net of background per second and then also value of d12 ( in
units of m−2 ) for various d in the data. This is easily done in Excel. Open a new sheet in Excel and
in column B, enter the values of d and in column C, enter the values of N . Next in column A, in the
cell A2 (A1 we will have the labels), enter the formula
= (10000*1/(B2*B2))
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This as we know will take the value in the cell B2 and do the operation as specified above. But we want
this to be repeated for all the values of column B. So again we do the same AUTOFILL technique
described above by clicking on the right bottom corner of A2 and pulling the mouse down all the time
B
. Now
keeping the left mouse button pressed. Now we have the values of d12 . Next we need to find N −N
60
in the cell D2, enter the formula
= (C2-18.8)/60
This will take the value in the cell C2 and perform the operation described. Again we want to repeat
it for all the values in column C. So we use AUTOFILL again to get the values. Now we have all the
data we need. The sheet will look like Figure B.9
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Figure B.9: Data for Inverse Square law
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An important thing to note is that the variable you want to plot on the x-axis should come in the first
column as we have done above. Now with the data at hand, we are ready to plot the curve and also the
least square fit straight line. First, as usual, we go to INSERT Menu on the Toolbar. We then click
the SCATTER chart button and get a choice of charts/graphs we can plot. Choose the SCATTER
WITH SMOOTH LINES AND MARKERS. This will open an empty chart box on the sheet.
Once again, you can move the position and resize the cart area as you wish. Next press the SELECT
DATA button on the Toolbar. This will open a dialog box with Select Data Source. We want to plot
a graph of Rate vs d12 . So we select the cells (by pressing the CTRL key and using the mouse) from
A1 to A12 and from D1 to D12. You will see that the chart area gets filled up with the points which
are joined with curve. We next need to label the graph and the axes. To do this, take the mouse to
any point in the Chart Area and press the left mouse key. This will open the CHART TOOLS in
the toolbar at the top. In this toolbar, press LAYOUT. Now double click on the Title in the Chart
Area and enter any text. In our case, we enter the title of the graph, namely ”R vs d12 ”. Next we want
to label the axes. In the LAYOUT toolbar, press the AXIS TITLES button. It will ask you about
the placement and the orientation of the titles. Enter the titles that you want on the x and y axis. In
our case it is Rate and d12 . Lastly, we see that the graph is approximately a straight line. So we want
to fit a straight line to the data using the Method of Least Squares. This is easy to do with Excel.
In the LAYOUT toolbar, there is a button TRENDLINES. When you press this, you get many
options of choosing the kind of curve you want to fit. Choose a MORE OPTIONS and in this choose
LINEAR. Also check the box at the bottom of the dialog box which says DISPLAY EQUATION
ON CHART. This will plot the best fit straight line and also give you the equation of the line. In our
case it is
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y = 0.0593x − 10.237
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The sheet would now look something like Figure B.10.
Figure B.10: Graph of Rate vs
1
d2
and the Best fit line
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You can copy the chart area with the graphs and then use any image processing program like MS Paint
or Photoshop etc to resize and do other things to make the graph look better. The graph will finally
look like Figure B.11
Figure B.11: Graph of Rate vs
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1
d2
and the Best fit line
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B.4
Lab Manual for Nuclear Physics
Statistical Analysis
MS Excel is not just good for plotting and doing simple calculations. It is also extremely useful for
statistical analysis. There are many in-built functions which one can use to determine things like
cumulative distribution functions, p values, probability distribution functions etc.
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Before one starts to use Excel for statistical analysis, one needs to make sure that the Analysis Toolpak
is installed. To check this, one can go to the Data Menu and see if the Data Analysis dropdown menu
is functional or not. In case it is not, one would need to install this Add-in from the Excel software
DVD or any other source.
Once this is done, we can use a host of functionalities of this tool. To illustrate the use of some of
the functionalities, let us consider the following data for the counting statistics of a GM counter. We
set the preset time of the GM counter to 20 seconds. We know that the background counts, which is
what the counter is measuring in this experiment, should follow a Poisson distribution. Now we also
know that for a large value of µ, the mean of the Poisson distribution, the distribution goes to a normal
distribution. We want to check this with our data. The data is given below:
2
3
4
5
6
7
8
9
10
11
12
13
14
5
5
15
12
23
25
21
30
26
15
09
05
02
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fi
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Preset Time: 20 seconds
Operating Voltage: 460 Volts
xi f i
(xi − x)
(xi − x)2
10
15
60
60
138
175
168
270
260
165
108
65
28
-6.155
-5.155
-4.155
-3.155
-2.155
-1.155
-0.155
0.845
1.845
2.845
3.845
4.845
5.845
37.88
26.57
17.26
9.95
4.64
1.33
0.024
0.714
3.4
8.09
14.78
23.47
34.16
275
pe =
Pfi
fi
0.025
0.025
0.075
0.060
0.115
0.125
0.105
0.150
0.130
0.075
0.045
0.025
0.010
(xi − x̄)2 p
0.947
0.664
1.294
0.597
0.533
0.166
0.002
0.107
0.442
606
0.665
0.586
0.341
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04
60
02
32
01
17
P
P
fi = 200
xi fi = 1631
6.845
7.845
8.845
46.55
61.54
78.23
0.020
0.937
0.010
0.615
0.005
0.391
P
P
pe = 1 σE2 = (xi − x̄)2 p = 8.893
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Table B.5: Counting Statistics for Preset time = 20 seconds
We can easily calculate the sample mean as
P
x i fi
1631
= 8.155
x̄ = P
=
fi
200
(B.1)
The corresponding sample variance can be calculated from the expectation value of the square of the
deviations from the sample mean
σE2 =
X
(xi − x̄)2 p = 8.893
(B.2)
σE = 2.98
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We are now ready to plot this distribution that is the value of pe for the experimental data. To do this,
we simply enter the values of x (the number of counts) and f (the frequency of the counts or x) and
then sum the frequency. We are taking 200 readings. The excel worksheet will look like Fig B.12.
Figure B.12: Screenshot for data
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Now we need to find the probabilities corresponding to each value of x. This is easy to do since we
know that
fi
pe = P
fi
So in cell C3, we put in the formula for probability. We want to calculate the probability for x = 2
that is the value in cell A3. So we write
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= A3/200
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and then use the AUTOFILL command discussed above to fill in all the remaining values of Column
C as shown in Fig B.13.
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Figure B.13: Screenshot for data
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Now we are ready to plot this experimental data. We want to plot p versus x. We choose the column
for the probability and then insert CHART. Depending on the kind of chart you want you can get a
scatter plot or a bar graph etc. Then by double clicking the chart one can format the axis, the title etc.
When we are finished, we will get something like Fig B.14
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Figure B.14: Plot for Experimental Data
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Now that we have the experimental data plotted, we need to compare it with a Poisson distribution.
Now to get a Poisson distribution, we need one parameter, that is the mean. However, we don’t know
the mean of the underlying distribution. But we do know the sample mean Eq (B.1). We use it to find
the Poisson probabilities. Excel allows us to do this easily. The function is
=POISSON.DIST(x,µ, TRUE/FALSE)
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The first argument is the value x at which we need the probability. In our case, it is the the values in
Column A. Next is the mean, µ, which we use as the sample mean. Finally, the last logical argument
is TRUE/FALSE. TRUE will give us the cumulative Poisson distribution while FALSE will give us
the Poisson distribution function. Once we do this for the first value of x, all we need to do is to use
AUTOFILL to get all the other values. When we do this, our spreadsheet looks like Fig B.15.
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Figure B.15: Poisson probabilities
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We then repeat the same procedure to get a graph of Poisson probabilities. If we do this, we get Fig
??.
Figure B.16: Plot for Poisson distribution
Finally, we want to check if the distribution is close to a Normal distribution. Here we need the mean
and the variance. We again use the sample mean and the sample variance (Eq(B.2)). The function to
get the Normal distribution is
=NORM.DIST(x,µ, σ,TRUE/FALSE)
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Again we use the AUTOFILL function to get all the values for the variable x. The spreadsheet will
now look like this.
Figure B.17: Gaussian probabilities
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We can now plot this in exactly the same was as above to get Fig B.18.
Figure B.18: Plot for Gaussian distribution
There are many other functions in Excel which we can use to carry out statistical analysis. For instance,
we can use the statistical function
NORM.S.DIST(x,TRUE/FALSE)
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to find the standard normal distribution with mean 0 and variance 1. The TRUE/FALSE allows us to
get the cumulative distribution or not. Similarly,
NORM.S.INV(p)
to return the value of x for which the inverse of the standard normal distribution has the value p. Ee
can use the function
CONFIDENCE.NORM( α, σ, size )
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to get the confidence limits assuming that the sample distribution is normal. Here α is the significance
level and σ is the sample standard deviation and size is the size of the sample.
For our purposes, p-values for the χ2 distribution can also be easily found with Excel. The function is
CHI.SQ.DIST(x,degrees of freedom,TRUE/FALSE)
where x is the value at which we want to find the probability of the χ2 distribution, and TRUE/FALSE
indicates whether one wants a cumulative probability or not. This allows us easily to find the p-value
corresponding to the value of χ2 that we obtained from our experimental data. We basically need
1- CHI.SQ.DIST(x,degrees of freedom,TRUE)
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and this will return us the p-value.
Some of the other useful functions for our purposes are:
BINOM.DIST( number s, trials, probability s, cumulative )
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where numbers is the number of successes,trials is the number of trials, probabilitys is the the
probability of success in one trial and cumulative is once again TRUE/FALSE.
EXPON.DIST( x, λ, cumulative )
is the function which returns the exponential distribution.
GAMMA.DIST( x, α, β, cumulative )
returns the gamma distribution with parameters α and β.
T.DIST( x, degrees freedom, cumulative )
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returns the Student’s t-distribution.
VAR.S( number1, [number2], · · · )
where number1 etc are the samples. One can either add a range of cells instead or acutal numbers.
COVARIANCE.P( array1, array2 )
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returns the population Covariance between two arrays.
COVARIANCE.S( array1, array2 )
returns the sample covariance between two arrays.
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There are a host of other statistical function which can be very useful. You can find more information
at http://www.excelfunctions.net/Excel-Statistical-Functions.html.
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Using Gnuplot
C.1
Introduction
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Appendix C
Most of you are familiar with MS-Excel and we have seen how to use it to analyse data and plot graphs
in the previous chapter. However, if one is using a Linux based machine, then the chances are that one
does not have access to MS Excel. However, Linux has a package called Gnuplot which comes with
Linux. It is a very powerful package which allows on to generate two- and three-dimensional plots of
functions and data. The program runs on all major computers and operating systems (Linux, UNIX,
Windows, Mac OSX...).
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The software is copyrighted but freely distributed (i.e., you don’t have to pay for it). It was originally
intended to function as a software for plotting mathematical functions and data but has outgrown
its credentials. It now supports many non-interactive uses, including web scripting and integration
as a plotting engine for third-party applications like Octave. Gnuplot has been supported and under
development since 1986.
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Gnuplot supports many types of plots in either 2D and 3D. It can draw using lines, points, boxes, contours, vector fields, surfaces, and various associated text. It also supports various specialized plot types.
Gnuplot supports many different types of output: interactive screen terminals (with mouse and hotkey
functionality), direct output to pen plotters or modern printers (including postscript and many color
devices), and output to many types of graphic file formats (eps, fig, jpeg, LaTeX, metafont, pbm, pdf,
png, postscript, svg, ...). Gnuplot is easily configurable to include new devices.
R
THIS CHAPTER ASSUMES YOU ARE USING A LINUX MACHINE OR A
LINUX EMULATOR ON A WINDOWS MACHINE
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Plotting with inbuilt functions of GNUPLOT
We will first demonstrate gnuplot using built-in functions.
C.2.1
Interactive plotting
In the following example, we plot cos(x) between −2π < x < 2π with labels on x and y axis.
$gnuplot
Figure C.1: Screen Shot of opening GNUPLOT
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You will see a screen like this
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Open a terminal and type at the prompt
On the gnuplot prompt type the following lines one by one (self explanatory)
gnuplot>
gnuplot>
gnuplot>
gnuplot>
gnuplot>
set xlabel ‘x’
set ylabel ‘Cos(x)’
set grid
set title ‘Cos Function’
plot [-2*pi : 2*pi] cos(x) w lp
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You will get a plot like this
Figure C.2: Screen Shot of Cos(x)
To turn off the grid, you can “unset grid”, to turn off the xlabel, you can type “set xlabel ’ ’ ”. Type
set at the gnuplot prompt to see all of the options you can turn on and off. To turn on auto-scaling
(without any ‘x’ or ‘y’ labels: default), type “set auto” at gnuplot> prompt.
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Note that many of the gnuplot keywords including: using, title, and with can be
abbreviated with a single alphabet as: u, t and w but should be avoided by beginners.
Also, each line and point style has an associated number.
In order to draw two plots on top of each other, you can replace the last line by
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gnuplot>plot [-2*pi : 2*pi] sin(x) t ‘Sine Wave’ with linespoints, cos(x) t ‘Cosine Wave’
with linespoints
Figure C.3: Screen Shot of Sin(x) & Cosine(x)
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You can exit from gnuplot by typing “exit” (or “quit”) on the gnuplot prompt.
gnuplot>exit
C.3
Saving Plots
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To save the above image as an enhanced postscript file with “.eps” or as a postscript file with “.ps”
extension, instead of displaying it to the screen, enter the following commands:
gnuplot> set terminal postscript
gnuplot>set output ‘plot.ps’
gnuplot>set xlabel ‘x’
gnuplot>set ylabel ‘y’
gnuplot>set title ‘Sine Wave’
gnuplot> plot [-2*pi : 2*pi] sin(x) w lp
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This will create a postscript image file called “plot.ps” of the previous plot. It will be placed in the
same folder in which you are working. You can then use any postscript viewer program like “gv” (or
“evince”) to open your saved graphics file. Remember, the plot will not appear on the screen
when you redirect the terminal type to postscript (first line of the example above), so it
may appear as if nothing has happened. Exit from gnuplot prompt, and then type on the terminal
$gv plot.ps &
C.3.1
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You can also use any converter available on the Internet (like ps2pdf.com or online2pdf.com) to convert
the postscript (.ps) files to PDF files which can then be used.
Customization
Customization of the axis ranges, axis labels, and plot title, as well as many other features, are specified
using the set command. Specific examples of the set command follow. (The numerical values used in
these examples are arbitrary.) To view your changes type: replot at the gnuplot> prompt at any time.
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Command
Create a title:
> set title ”Beta Particle Spectrum”
Put a label on the x-axis:
> set xlabel ”Energy (keV)”
Put a label on the y-axis:
> set ylabel ”Intensity”
Change the x-axis range:
>set xrange [0.001:0.005]
Change the y-axis range:
> set yrange [20:500]
Have Gnuplot determine ranges:
> set autoscale
Put a label on the plot:
> set label ”Q-Value” at 0.003, 260
Remove all labels:
> unset label
Plot using log-axes:
> set logscale
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Action
Plot using log-axes on y-axis:
> unset logscale; set logscale y
Change the tic-marks:
> set xtics (0.002,0.004,0.006,0.008)
Return to the default tics:
>unset xtics; set xtics auto
Other features which may be customized using the set command are: arrow, border, clip, contour,
grid, mapping, polar, surface, time, view, and many more. The best way to learn is by reading the
on-line help information, trying the command, and reading the Gnuplot manual.
FUNCTION
abs(x)
acos(x)
absolute value of x, |x|
arc-cosine of x
arc-sine of x
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asin(x)
RETURNS
atan(x)
arc-tangent of x
cos(x)
cosine of x, x is in radians.
cosh(x)
hyperbolic cosine of x, x is in radians
erf(x)
error function of x
exp(x)
exponential function of x, base e
inverf(x)
inverse error function of x
invnorm(x)
inverse normal distribution of x
log(x)
log of x, base e
log10(x)
log of x, base 10
norm(x)
normal Gaussian distribution function
rand(x)
pseudo-random number generator
sgn(x)
1 if x > 0, −1 if x < 0, 0 if x = 0
sin(x)
sine of x, x is in radians
sinh(x)
hyperbolic sine of x, x is in radians
sqrt(x)
the square root of x
tan(x)
tangent of x, x is in radians
tanh(x)
hyperbolic tangent of x, x is in radians
Table C.1: Some Inbuilt Functions in Gnuplot
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Bessel, Beta and Gamma functions are also supported. Many functions can take complex arguments.
Binary and unary operators are also supported. The supported operators in Gnuplot are the same
as the corresponding operators in the C programming language, except that most operators accept
integer, real, and complex arguments. The ** operator (exponentiation) is supported as in FORTRAN.
Parentheses may be used to change the order of evaluation. The variable names x, y, and z are used as
the default independent variables.
Plotting using data from a file
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C.4
This is the section which will be most important for us. We normally will have our data and we would
like to plot it. First we need to create a file with the data in it. Let us suppose that the file is called
data1.txt and it looks something like Table C.2.
n2
n3
1
1
1
2
4
8
3
9
27
4
16
64
5
25
125
6
36
216
7
49
343
8
64
512
9
81
729
10
100
1000
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#n
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Table C.2: Experimental Data
La
Again start gnuplot in a terminal by writing “gnuplot” on the terminal prompt and type the following
line
gnuplot> plot “data1.txt” u 1:2 w lp,“data1.txt” u 1:3 w lp
gnuplot ignores lines starting with # (comment lines.) Also, you can combine any number of plots in
one figure. Thus, in this example, we have plotted n vs n2 and n vs n3 by the command u 1:2 and
then again u 1:3. It should be clear that the 1 refers to the first column of the data file and the 2 and
3 refer to the second and third column. Similarly, one can plot data from different data files in this
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manner. The above command will create a plot like that in Fig C.4.
Figure C.4: Screen Shot of Plot using data file
You can modify these plots again using the x and y-labeling.
You can also save the above plot in a .ps file using the following commands:
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gnuplot>set terminal postscript
gnuplot>set output ‘plots2.ps’
gnuplot>plot “data1.txt” u 1:2 w lp, “data1.txt” u 1:3 w lp
C.5
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Exit the gnuplot and then type gv plots2.ps on the terminal command prompt to see the output.
Plotting using data from file and fitting to a smooth curve
Frequently, in our experiments we will get data which visually looks scattered. Now if we join the
points on a plot, we get a set of connected straight lines. This obviously is not how nature would
behave. We all know that in nature there would be no discontinuities and so we would like to join
the points with a smooth curve rather than a collection of straight lines. Of course, some plots are
straight line plots and these can be obtained easily by using the Method of Least Squares as we saw in
Section 1.7 in Chapter 1. For other kinds of plots, we basically need to interpolate between points to
get a smooth curve. Gnuplot allows one to do this easily. To demonstrate this, consider the example
of Counting Statistics from Section 5.3.2 in Chapter 5. For this example we will take the data for a
preset time of 5 seconds as given in Table 5.3.
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0
1
2
3
4
5
6
fi
xi f i
27
0
55
55
60
120
31
93
16
64
06
30
05
30
P
P
fi = 200
xi fi = 387
(xi − x) (xi − x)2
-1.94
-0.94
0.06
1.06
2.06
3.06
4.06
3.76
0.880
0.086
1.12
4.24
9.86
16.48
pe =
Pfi
fi
(xi − x̄)2 p
0.135
0.507
0.275
0.242
0.300
0.010
0.155
0.173
0.080
0.339
0.030
0.280
0.025
0.412
P
P
2
pe = 1 σE = (xi − x̄)2 p = 1.96
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Table C.3: Counting Statistics for Preset time = 5 seconds
From this we can derive the probabilities since we know the frequencies and the total number of readings
as explained in Eq(5.3).
fi
pe = P
(C.1)
fi
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The error in the experimental data can also be easily calculated since we know that pe is actually
a derived quantity and therefore the error has to be evaluated using the error propagation equaP
tion. However, we know that the denominator, that is
fi has no error since it is fixed at 200 in
√
√
fi
our experiment. We take the error in fi as fi and get the error bars by taking the error in pe to be 200
.
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Further, we expect that the counting statistics follows a Poisson distribution with a mean given by the
sample mean which can be calculated. This table is given below.
xi
0
1
2
3
4
5
6
PPoisson
0.144
0.279
0.271
0.175
0.084
0.030
0.011
pe
0.135
0.275
0.300
0.155
0.080
0.030
0.025
Table C.4: Experimental and Poisson probabilities: Preset Time 5 seconds
Now let us see how to plot these. First let us use Gnuplot to plot the experimental data which is stored
in a file called “nuc1.dat” for instance which you have created. We will also need the error bars on the
experimental data. The data with the error bars thus becomes
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PPoisson
0.144
0.279
0.271
0.175
0.084
0.030
0.011
pe
error in pe
0.135
0.026
0.275
0.037
0.300
0.038
0.155
0.028
0.080
0.020
0.030
0.012
0.025
0.011
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xi
0
1
2
3
4
5
6
Table C.5: Experimental and Poisson probabilities with errors: Preset Time 5 seconds
Now we are ready to plot this data using Gnuplot. First let us plot the raw data for a plot of pe vs xi .
gnuplot> plot “nuc1.dat” u 1:3 w lp
since the experimental data is in the first and third column in the Table C.5. This will generate a graph
as in Figure C.5.
0.3
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"nuc1.dat" u 1:3
0.25
0.15
b
p
0.2
La
0.1
0.05
0
0
1
2
3
x
4
5
6
Figure C.5: Counting Statistics : Preset 5 seconds: Experimental Data
Next we need to get the graph with the errorbars. For this, we need to just do the following
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gnuplot> plot “nuc1.dat” u 1:3 w lp, “nuc1.dat” u 1:3:4 w yerrorbars
The graph will look like that in Figure C.6
0.35
Exp. data
Exp. data with errorbars
0.25
y
0.2
0.15
0.1
0.05
0
1
2
3
x
4
5
6
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0
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0.3
Figure C.6: Counting Statistics : Preset 5 seconds: Data with errorbars
But this is not a smooth curve. So we need to use some interpolation to generate a smooth curve.
Curve Fitting & Interpolation
b
C.5.1
La
Typically in any experiment, we will get a set of data points (x, y). It is a good assumption that there
is some function f (x) which will generate this set of data. The function f (x) does not only generate the
set of data points, but should represent all the non-data points that could be generated by the particular process that the experiment is using. The problem is that we don’t know what the function f (x) is.
In Interpolation , we try to find a function, say g(x) which approximates the function f (x) as best as
we can. This function should pass through all the data points that we have got from our experiment.
In addition, we believe that the interpolating function also will give us the values of the variables at
the non-data points. This is what makes interpolation useful.
In Curve Fitting we are NOT doing this. Instead, when we determine the best-fit line, say by using
the Method of Least Squares, we are basically only finding the best fit curve to the set of data points.
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In this case, the approximating function does not have to pass through all the data points. This point
is very important to remember. Curve fitting usually gives a good idea of the trend of the data. It
cannot be used to determine the values in between the data points. The Method of Least Squares
that we learnt in Section 1.7 only used a linear fit. Recall that in the case of a linear fit, we needed
to determine 2 parameters, namely, the slope m of the best fit line and c its y-intercept. We can
easily generalise the discussion in Section 1.7 to use instead of a straight line with two parameters, a
higher order polynomial. A typical Least Square fitting can, for instance, use a fifth order polynomial.
Normally, in the case of higher order polynomials, the determination of the coefficients becomes a bit
difficult algebraically. Therefore one uses a standard mathematical package like Scilab or Matlab
which does it for you.
Now let us consider Interpolation. The simplest interpolation would of course be a linear interpolation
where we just join the data points with a straight line. An example of this, for our case is Figure C.5.
Obviously we want a better curve rather than simply joining the adjacent points with a straight line.
This is usually done with a cubic polynomial instead of a polynomial of order 1 that is a linear function.
If one uses a cubic polynomial, then one will need to determine the coefficients of the polynomial to
get it. This is usually done by using the data points and solving a system of simultaneous equations.
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However, by simply using a cubic interpolation, one might have a problem of smoothness. This is
overcome by spline functions. These are basically different polynomials between the datapoints which
are piecewise continuous and have a high degree of smoothness where the polynomials meet. The most
often used spline functions are cubic splines or csplines. As the name suggests, these use cubic
polynomials between data points.
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A natural smoothing spline approximation can be done by using csplines between adjacent points but
weighting the coefficients with not just the adjacent datapoints but the ones which are further off also.
Obviously, the weight of the nearest data points in the cubic polynomial will be the most and that of
the farthest will be the least. Once again, spline functions don’t have to be cubic though these are the
most often used. The spline function can be a polynomial of any order.
Gnuplot has several different techniques of interpolation. These are for instance, acsplines, csplines,
bezier etc. Acsplines uses a cubic polynomial piecewise where the coefficients are weighted with
weights to generate a smooth curve. A bezier smoothing uses a bezier curve to smooth the data while
csplines uses a natural cubic polynomial. A bezier curve is a curve based on Bernstein polynomials
and is used to generate smooth curves frequently in computer graphics applications. To generate a
smooth curve, we do the following
gnuplot> plot “nuc1.dat” u 1:3:4 w yerrorbars, “nuc1.dat” u 1:3 smooth csplines,
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This will generate a graph as in Figure C.7
0.35
"nuc1.dat" u 1:3:4
Exp.data
0.3
p
0.2
0.15
0.1
0.05
0
0
1
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0.25
2
3
x
4
5
6
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Figure C.7: Counting Statistics : Preset 5 seconds: Data with errorbars & smoothed
Finally we need to see how well this experimental data fits the expected Poisson distribution. For
this,we plot the probabilities of the Poisson distribution with the sample mean as in Column 2 of Table
above and draw a smooth line between them. For this, we do the following
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gnuplot> plot “nuc1.dat” u 1:3:4 w yerrorbars, “nuc1.dat” u 1:3 smooth csplines,
“nuc1.dat” u 1:2, “nuc1.dat” u 1:2 smooth csplines
This will generate a graph as in Figure C.7
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0.35
"nuc1.dat" u 1:3:4
Exp.data
"nuc1.dat" u 1:2
Poisson
0.3
0.25
0.15
0.1
0.05
0
0
1
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p
0.2
2
3
x
4
5
6
Figure C.8: Counting Statistics : Preset 5 seconds: Data with errorbars & smoothed & Poisson distribution
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One can do many other things with Gnuplot. You can explore the other features of the software and
learn more about it from any of the references on the Internet.
Some of the common errors & good practices while using gnuplot
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1. In the plot statement, the name of the datafile to be plotted should always be in
“ ”.
2. Before you plot any data from a file, make sure you check the file and confirm
that it has the data that you expect.
3. Always first see the plot without any extra labels, titles etc. on your screen. If
it is of the form that you expect, then only go and add all the extra things like
labels, titles, legends etc and save as .ps file.
4. Do not use very long and complicated names for your datafiles or the graphic
files (.ps) since then there is a chance you will make a mistake when trying to
enter the names. Short, descriptive names are best.
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Appendix D
D.1
Introduction
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Radioactive Decay Equilibrium
We saw in Section 2.1.2 that typically a radioactive nuclide is part of a decay chain consisting of parent
and daughter nuclides etc. This process can be represented by
A → B → C → ···
D.2
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The question we need to ask is, given the half lives of the different nuclides in the decay chain, what can
we say about the behaviour with time? In particular, we will talk about different kinds of equilibrium
configurations that can be analysed depending on the half lives.
Bateman Equation
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Let us consider a decay chain as shown above. Let the nuclides A, B, C, · · · have decay constants
λ1 , λ2 , λ2 · · · . Let the number of nuclides of these types at time t be N1 , N2 , N3 , · · · and their number
at time t = 0 be N10 , N20 , N30 , · · · . Then we know that the number of nuclei of type A , that is the
parent nuclide, at any time t is given by the Activity Law, that is
N1 = N10 e−λ1 t
(D.1)
and its rate of change is given by
dN1
= −λ1 N1
(D.2)
dt
Similarly, the nuclide of type B which is being produced by A in its decay can be analysed. However,
remember that while it is being produced by the decay of A it itself is also decaying into the granddaughter nuclide C simultaneously. The rate of production of B is obviously equal to the rate of decay
of A and its rate of decay is given by its own decay constant λ2 . Thus we can write
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dN2
= λ1 N1 − λ2 N2
dt
(D.3)
dN2
= λ1 N10 e−λ1 t − λ2 N2
dt
(D.4)
dN2
+ λ2 N2 = λ1 N10 e−λ1 t
dt
(D.5)
Substituting Eq D.1 into Eq D.3, we get
Multiplying both sides by eλ2 t we get
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or
dN2
+ eλ2 t λ2 N2 = λ1 N10 e(λ2 −λ1 )t
dt
The left hand side is a complete differential and we get
e λ2 t
d N2 eλ2 t = λ1 N10 e(λ2 −λ1 )t dt
This can now be integrated to give us
N2 eλ2 t =
λ1
N10 e(λ2 −λ1 )t + C
(λ2 − λ1 )
(D.6)
(D.7)
(D.8)
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where C is the integration constant. To determine C we need to use the initial condition namely,
N2 (t = 0) = N20 . Then
C = N20 −
λ1
N10
(λ2 − λ1 )
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Thus we get for the number of nuclide B at any time t to be
N2 (t) =
λ1
N10 e−λ1 t − e−λ2 t + N20 e−λ2 t
(λ2 − λ1 )
(D.9)
In terms of activity, A, which we recall is simply
A=−
dN
= λN
dt
we can write Eq D.9 as
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A2 (t) =
λ2
A10 e−λ1 t − e−λ2 t + A20 e−λ2 t
(λ2 − λ1 )
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The interpretation of this equation is obvious. The first term on the right is the number of atoms
produced from the decay of A which have not yet decayed. The second term is the number of atoms of
B which remain from the initial number N20 that is the number that has not decayed since remember
that B is decaying to C.
Now we turn to the nuclide C. This is being produced by the decay of B and is also decaying simultaneously. Thus its number at any time is given by
dN3
= λ2 N2 − λ3 N3
dt
But N2 is given by Eq D.9. Substituting, we get
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(D.11)
λ1 λ2
dN3
+ λ3 N3 =
N10 e−λ1 t − e−λ2 t + N20 e−λ2 t
dt
(λ2 − λ1 )
(D.12)
At this point,we can make a simplification and assume that there are no daughter (B) or granddaughter
nuclide (C ) at time t = 0, that is N20 = N30 = 0. In other words, we assume a pure sample of the
parent nuclide. Then Eq D.12 becomes
dN3
λ1 λ2
+ λ3 N3 =
N10 e−λ1 t − e−λ2 t
dt
(λ2 − λ1 )
(D.13)
Once again multiplying by eλ3 t on both sides, we get
λ1 λ2
N10 e(λ3 −λ1 )t − e(λ3 −λ2 )t dt
(λ2 − λ1 )
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d N3 eλ3 t =
(D.14)
Integrating this and using the initial condition that N30 = 0, we get the number of nuclides of type C
at any time t as
b
N3 (t) = N10 λ1 λ2
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e−λ1 t
e−λ2 t
e−λ3 t
+
+
(λ2 − λ1 )(λ3 − λ1 ) (λ1 − λ2 )(λ3 − λ2 ) (λ1 − λ3 )(λ2 − λ3 )
(D.15)
This equation and its solutions are called Bateman Equations . They were proposed first by H.
Bateman in 1910.
In the general case, for the situation where all the daughter nuclides are not present at time
t=0, that is N20 = N30 = · · · = Nn0 = 0, the solutions are given by
Nn = C1 e−λ1 t + C2 e−λ2 t + · · · + Cn e−λn t
where the constants C1 , C2 , · · · are given by
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λ1 λ2 · · · λn−1
N10
(λ2 − λ1 )(λ3 − λ1 ) · · · (λn − λ1 )
λ1 λ2 · · · λn−1
=
N10
(λ1 − λ2 )(λ3 − λ2 ) · · · (λn − λ2 )
C1 =
C2
..
.
(D.17)
λ1 λ2 · · · λn−1
N10
(λ1 − λn )(λ2 − λn ) · · · (λn−1 − λn )
(D.18)
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Cn =
One can immediately see how useful these equations and their solutions will be. For instance knowing
the initial activity of a pure sample of a radioactive material, we can easily find the activities of all the
daughter nuclides in the decay chain at any time.
Example D.2.0.1
Consider the decay of a sample weighing 2µ g of pure
beta decay by the radioactive chain
20
8O
→
20
9F
→
20
10Ne
radioisotope. The isotope decays by
stable
20
9F
is 11.163 seconds. Calculate the activity
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The half life of the 208O is 13.51 seconds and that of
of the sample after 1 minute.
20
8O
The first thing we need to do is to calculate the decay constants of the two isotopes. This is simply
ln 2
= 0.0513sec−1
τ1
λ2 =
ln 2
= 0.0621 sec−1
τ2
b
λ1 =
La
Now we need to calculate the activity of the initial sample.
A10 = λ1 N10
2 × 10−6
× 6.023 × 1023 = 3.09 × 1015 Bq = 8.35 × 104 Ci
= 0.0513 ×
20
Now we can use Eq D.10 to calculate the activity at time t = 60 seconds since we know that the
initial activity of nuclide 2 , A20 = 0. Then
A2 = A10
λ2 −λ1 t
e
− e−λ2 t
λ2 − λ1
Putting in the numbers we get
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A2 (t = 60) = 1.44 × 104 Ci
D.3
Different Kinds of Equilibrium
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Looking at the Activity equations, we can think of three different situations in radioactive decay chains
depending on the relative values of the half lives of the parent and daughter nuclides. These are
1. No Equilibrium
2. Transient Equilibrium
3. Secular Equilibrium
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When the half life of the parent nuclide is shorter than that of the daughter nuclide, we have a situation
of No Equilibrium. Remember that a shorter half life means a higher decay constant. In this case,
the parent decays much faster than the daughter and therefore the build up of the daughter nuclide is
much faster than its own decay. Essentially, after a few half lives, all of the parent has decayed and
all the subsequent activity is only due to the daughter nuclide. Examples of this kind of situation are
131
Te → 131I, 210Bi → 210Po, 92Sr → 92Y which are all examples of β decay towards higher stability. An
138
example for 138
54Xe → 55Cs is shown in Figure D.1.
No Equlibrium
1
Parent
Daughter
0.9
0.8
0.6
Activity
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b
0.7
0.5
0.4
0.3
0.2
0.1
0
0
10
20
30
40
50
Time(m)
60
70
80
90
100
Figure D.1: No Equilibrium
Second case is that of Transient Equilibrium. Here the half lives of the parent and the daughter
are of the same order, though the half life of the parent is of the order of 10 times the half life of the
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daughter. Now the parent controls the decay chain. This can be easily seen from Eq D.9.
Now we have
λ2 > λ1
Then as t → ∞, we have
and therefore
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e−λ2 t e−λ1 t
N20 e−λ2 t → 0
Therefore in Eq D.9
N2 ≈ N10
Thus
λ1
λ1
e−λ1 t = N1
λ2 − λ1
λ2 − λ1
N1
λ2 − λ1
=
N2
λ1
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We see that at long times, the ratio of the daughter and parent activity becomes a constant. Example
of transient equilibrium is 140Ba → 140La where the half life of the parent is 12.8 days and that of the
99m
daughter is 40 hours. An example for 99
42Mo → 43Tc is shown in Figure D.2.
Transient Equlibrium
b
1
Parent
Daughter
La
0.9
0.8
Activity
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
10
20
30
40
50
Time(h)
60
70
80
Figure D.2: Transient Equilibrium
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Finally we have a situation of Secular Equilibrium. Here the half life of the parent is many times
(∼ 104 ) that of the daughter and therefore there is no significant change in the parent population during
observation. In this case, as in the case of Transient Equilibrium, we can see that
λ2 λ1
and therefore
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N1
λ2 − λ1
λ2
=
≈
N2
λ1
λ1
or
A1 = A2
Examples of these are the naturally occurring heavy element chains like 238U → 206Pb or 235U → 207Pb
144
because of very long half lives of the parent nuclides. An example for 144
58Ce → 59Pr is shown in Figure
D.3.
Secular Equlibrium
1.2
Parent
Daughter
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1
Activity
0.8
0.6
0.4
La
b
0.2
0
0
50
100
150
Time(m)
200
250
300
Figure D.3: Secular Equilibrium
D.4
Numerical Integration of Bateman Equations
Although we have given above the exact solutions of the Bateman equations in the general case, these
may not be easy to obtain. A simpler method is to numerically solve the equations governing the
number of nuclides (like, for instance, in the case of 3 nuclides, Eq D.2, Eq D.3 and Eq D.11)using
any standard method to solve differential equations. A powerful method to do this is to use the
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Runge-Kutta (RK) method. In our case we use RK-4 method. The input in this method is the rate
equations that is the equations governing the number of nuclides, the decay constants of the various
nuclides and the initial conditions, that is the number of nuclides of different kinds present at time
t = 0. Once these are used as inputs, we can write a program to solve for any number of nuclides in a
decay chain.
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As an example consider a pure sample of a nuclide X with a half life of 14.08 minutes. This implies
that the initial sample has only nuclide X and the initial amounts of other nuclides is 0. The nuclide
X decays to another nuclide Y which has a half life of 33.41 minutes to a nuclide Z. The nuclide
Z is stable. The rate equations are obviously coupled since the solution of one enters into the other
equations. The program below uses RK-4 method to solve three coupled equations for the change of
numbers of nuclides X, Y and Z. It also evaluates the exact solutions obtained above (EqD.1, EqD.9
and EqD.15). The evolution of the numbers of the three nuclides are then plotted in both cases. These
are shown in Fig D.4 and Fig D.5 respectively. We can see that the numerical solution and the exact
solutions are identical.
/∗ S o l v i n g Bateman E q u a t i o n s N u m e r i c a l l y ∗/
/∗ s o l v i n g dx / d t = −l 1 ∗x ,
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dy / d t = l 1 ∗x−l 2 ∗y ,
and
dz / d t = l 2 ∗y−l 3 ∗ z u s i n g RK4 method ∗/
/∗ l 1 , l 2 , l 3 a r e t h e decay c o n s t a n t s o f p a r e n t , d a u g h t e r and g r a n d d a u g h t e r ∗/
/∗ t1 , t 2 a r e h a l f l i v e s ∗/
/∗ x1 i n i t i a l amount o f p a r e n t . i n i t i a l amounts o f d a u g h t e r and g r a n d d a u g h t e r t a k e n t o be 0 ∗/
/∗ x , y , z a r e t h e number o f p a r e n t , d a u g h t e r , g r a n d d a u g h t e r a t time t from
\
n u m e r i c a l i n t e g r a t i o n ∗/
/∗ x2 , y2 , z2 a r e number o f p a r e n t , d a u g h t e r , g r a n d d a u g h t e r from e x a c t s o l u t i o n ∗/
b
/∗ ax , ay , az a r e a c t i v i t i e s ∗/
La
#include<s t d i o . h>
#include<math . h>
#define f 1 ( t , x , y , z ) (− l 1 ∗ ( x ) )
#define f 2 ( t , x , y , z ) ( l 1 ∗ ( x) −(y ) ∗ l 2 )
#define f 3 ( t , x , y , z ) ( l 2 ∗ ( y)− l 3 ∗ ( z ) )
main ( )
{
f l o a t t 1 =14.08 , t 2 =33.41 , h = 0 . 0 1 ; /∗ h a l f l i v e s and s t e p s i z e ∗/
f l o a t t , x , y , z , k1 , k2 , k3 , k4 , m1, m2, m3, m4, n1 , n2 , n3 , n4 , l 1 , l 2 , l 3 =0;
f l o a t ax , ay , az ; /∗ a c t i v i t i e s ∗/
f l o a t x1 =1 , y1 =0, z1 =0; /∗ i n i t i a l v a l u e s o f t h e n u c l i d e s ∗/
f l o a t x2 , y2 , z2 ;
FILE ∗ f p=NULL;
f p=f o p e n ( ” r e s . t x t ” , ”w” ) ;
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FILE ∗ f p 1=NULL;
f p 1=f o p e n ( ” r e s 1 . t x t ” , ”w” ) ;
l 1=l o g ( 2 ) / t 1 ;
l 2=l o g ( 2 ) / t 2 ;
t = 0 . 0 ; x = x1 ; y = y1 ; z = z1 ;
do
{ k1 = h∗ f 1 ( t , x , y , z ) ;
n1 = h∗ f 3 ( t , x , y , z ) ;
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m1 = h∗ f 2 ( t , x , y , z ) ;
k2 = h∗ f 1 ( t+h / 2 , x+k1 / 2 , y+m1/ 2 , z+n1 / 2 ) ;
m2 = h∗ f 2 ( t+h / 2 , x+k1 / 2 , y+m1/ 2 , z+n1 / 2 ) ;
n2 = h∗ f 3 ( t+h / 2 , x+k1 / 2 , y+m1/ 2 , z+n1 / 2 ) ;
k3 = h∗ f 1 ( t+h / 2 , x+k2 / 2 , y+m2/ 2 , z+n2 / 2 ) ;
m3 = h∗ f 2 ( t+h / 2 , x+k2 / 2 , y+m2/ 2 , z+n2 / 2 ) ;
n3 = h∗ f 3 ( t+h / 2 , x+k2 / 2 , y+m2/ 2 , z+n2 / 2 ) ;
k4 = h∗ f 1 ( t+h , x+k3 , y+m3, z+n3 ) ;
m4 = h∗ f 2 ( t+h , x+k3 , y+m3, z+n3 ) ;
n4 = h∗ f 3 ( t+h , x+k3 , y+m3, z+n3 ) ;
x
= x+(k1 +2.0∗( k2+k3)+k4 ) / 6 . 0 ;
y
= y+(m1+2.0∗(m2+m3)+m4 ) / 6 . 0 ;
z
= z+(n1 +2.0∗( n2+n3)+n4 ) / 6 . 0 ;
t
= t+h ;
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/∗ Exact S o l u t i o n o f Bateman E q u a t i o n s ∗/
x2=x1 ∗ exp(− l 1 ∗ t ) ;
y2=( l 1 / ( l 2 −l 1 ) ) ∗ x1 ∗ ( exp(− l 1 ∗ t )−exp(− l 2 ∗ t ))+ y1 ∗ exp(− l 2 ∗ t ) ;
z2=l 1 ∗ l 2 ∗ ( ( exp(− l 1 ∗ t ) / ( ( l 2 −l 1 ) ∗ ( l 3 −l 1 ) ) ) + ( exp(− l 2 ∗ t ) / ( ( l 1 −l 2 ) ∗ ( l 3 −l 2 ) ) ) \
+(exp(− l 3 ∗ t ) / ( ( l 1 −l 3 ) ∗ ( l 2 −l 3 ) ) ) ) ;
/∗ For A c t i v i t y ∗/
ay=l 2 ∗y ;
La
az=l 3 ∗ z ;
b
ax=l 1 ∗x ;
f p r i n t f ( fp , ”%f \ t %f \ t %f \ t %f \n” , t , x , y , z ) ;
f p r i n t f ( fp1 , ”%f \ t %f \ t %f \ t %f \n” , t , x2 , y2 , z2 ) ;
}
while ( t <=100.0);
}
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Numerical integration of Bateman Equations
1
Parent
Daughter
Granddaughter
0.9
0.8
0.7
Number
0.6
0.5
0.4
0.2
0.1
0
0
10
20
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0.3
30
40
50
60
Time(m)
70
80
90
100
110
Figure D.4: Numerical Integration of Bateman Equations
Exact Solution of Bateman Equations
1
Parent
Daughter
Granddaughter
0.9
0.8
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0.7
Number
0.6
0.5
0.4
0.3
0.2
La
b
0.1
0
0
10
20
30
40
50
60
Time(m)
70
80
90
100
110
Figure D.5: Exact Solution of Bateman Equations
Although it is easy to obtain the exact solutions to the rate equations in the case of a few nuclides, it
can get very tedious in the case of a larger number of nuclides in a decay chain. Numerical solutions
provide an easy method of solving for the number evolution of nuclides in any chain.
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Appendix E
E.1
Introduction
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Theory of Alpha Decay
As we have seen in Chapter 2, Section 2.2.1, classically it is impossible to explain how an alpha particle
can escape from the radioactive nucleus. However, quantum mechanically, this is easily explained. We
give below a simplified derivation of the relationship between the decay constant λ and the energy of
the outgoing alpha particle. This is adapted from “Concepts of Modern Physics”, by Beiser,
Mahajan & Choudhury.
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The basic assumptions of the Theory of Alpha Decay are as follows:
• An alpha particle exists as an entity within the nucleus.
• The particle is in constant motion but is held inside the nucleus because of a potential barrier
• There is a small but definite probability that the particle can tunnel through the barrier every
time it collides with the barrier.
La
b
With these assumptions, notice first that the decay probability per unit time, λ can be expressed as
λ = νT
(E.1)
where ν is the number of times the alpha particle strikes the potential barrier per unit time and T is
the probability that the particle will transmit through the barrier. We further assume that at any one
moment, one and only one alpha particle exists inside the nucleus and this moves back and forth inside
the nucleus, along the diameter. Then we can easily see that the collision frequency, ν is
ν=
v
2R0
(E.2)
where v is the velocity of the alpha particle and R0 is the nuclear radius. If we put in typical values of
these quantities, v ≈ 2 × 107 m s−1 and R0 ≈ 10−14 m, we get
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ν ≈ 1021 s−1
This is remarkable since this implies that even though the alpha particle strikes the nuclear barrier
1021 times per second, it still has to wait a long time to come out (depending on the half life of the
nuclide in question).
1-d Tunnel Effect For Rectangular Barrier
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E.2
To understand alpha particle decay, we need to revisit Tunnel Effect in Quantum Mechanics. Recall
that in 1-d quantum mechanics, if we have a finite potential barrier, a particle with a lower energy
than the height of the barrier can still have a finite probability of tunnelling through the barrier. This
is the basis, for instance of the semiconductor tunnel diode.
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Consider a barrier of height U and width L as in Figure E.1. A particle with energy E < U is incident
on the barrier from the left, that is from Region I.
Figure E.1: Tunneling in 1-dimension
Now let us write down the Schrodinger equation for the particle in Region I and Region III where there
is no potential and so the particle is a free particle.
d2 ψI 2m
+ 2 EψI = 0
dx2
~
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(E.3)
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and
d2 ψIII 2m
+ 2 EψIII = 0
dx2
~
The solutions are as expected, free particle, plane wave solutions of the form
(E.4)
ψI = Aeik1 x + Be−ik1 x
(E.5)
ψIII = F eik1 x + Ge−ik1 x
(E.6)
where
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and
√
2mE
~
Clearly, the two terms in ψI represent the right moving and left moving waves, or ψI+ and ψI− , where
the right moving wave is the incident beam of particles and the left moving is the reflected part of the
beam. That is
k1 =
ψI = ψI+ + ψI− = Aeik1 x + Be−ik1 x
The incident flux is then simply
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SI = |ψI+ |2 vI+
where vI+ is the velocity of the incident particles. Similarly, since in Region III, there can only be
transmitted particles moving right, the wavefunction in that region is simply
ψIII = F eik1 x
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The transmission probability is what we are interested in. This is simply the transmitted flux divided
by the incident flux.
T =
|ψIII+ |2 vIII+
|ψI+ |2 vI+
(E.7)
T =
F F ∗ vIII+
AA∗ vI+
(E.8)
or
R
As always, the values of the constants A and F are to be determined by the boundary conditions.
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What about the Region II? Here, classically, there cannot be any particle since E < U . However,
quantum mechanically, there is a definite probability of finding the particle. The Schrodinger equation
for this region is simply
d2 ψII 2m
+ 2 (E − U )ψIII = 0
dx2
~
(E.9)
ψII = Ce−k2 x + Dek2 x
(E.10)
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which has solutions
with
p
2m(U − E)
k2 =
~
Now we are ready to apply the boundary conditions. Recall that the wavefunction and its derivative
have to be continuous everywhere, and in particular at the boundaries, x = 0 and x = L. This gives us
A+B = C +D
ik1 A − ik1 B = −k2 C + k2 D
Ce−k2 L + Dek2 L = F eik1 L
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−k2 Ce−k2 L + k2 Dek2 L = ik1 F eik1 L
(E.11)
This system of equations allows us to solve for A and F which we need for the transmission probability
in Eq E.8. Assuming that U E and the barrier is wide enough that is k2 L 1, we get
AA∗
1
k22
=
+
e2k2 L
FF∗
4 16k12
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Further, since vIII+ = vI+ (because the wave numbers are the same in both regions and thus the group
velocity of the de-Broglie waves is the same), we have
#
16
e−2k2 L
T =
k2 2
4 + ( k1 )
"
where
p
2m(U − E)
k2 =
~
Now consider the quantity
k2
.
k1
Substituting their values, we see that
k2
U −E
=
k1
E
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(E.12)
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Thus this quantity is slowly varying compared to the exponential. Also, the quantity in front of the
exponential in Eq E.12 is of order 1. Thus to a good approximation, we can write
R
E.3
T = e−2k2 L
(E.13)
Tunnel Effect with Nuclear Potential Barrier: Geiger-Nuttall Law
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All this analysis is standard for a rectangular barrier in quantum mechanics. However, for the alpha
particle, it does not face a rectangular barrier. Instead, the barrier is like in Figure E.2 as we have seen
in Section 2.2.1, Figure 2.1.
b
Figure E.2: Energy diagram for Alpha decay
2
La
2Ze
is the electric potential energy of the alpha particle of charge 2e at a distance
In the figure, U = 4π
0r
2Ze2
r from the nucleus of charge Ze. R = 4π
is the distance where E, the energy of the alpha particle
0E
is equal to the potential energy U . R0 is the radius of the nucleus. Z is the atomic number of the
daughter nuclei.
To solve for the transmission probability in this case is somewhat complicated. However, we will do a
simple analysis to get an expression for the transmission probability which will allow us to determine λ
as a function of energy as in Eq. E.1. We first write Eq E.13 as
ln T = −2k2 L
or
310
(E.14)
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L
ln T = −2
R
k2 (r)dr = −2
0
k2 (r)dr
(E.15)
R0
Thus,
p
1/2
1/2 2m(U − E)
2m
2Ze2
=
−E
k2 =
~
~2
4π0 r
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But at r = R, U = E and thus
2Ze2
4π0 R
E=
and therefore
2mE
k2 =
~2
1/2 1/2
R
−1
r
(E.16)
With this k2 , we can now evaluate the integral in Eq E.15.
R
ln T = −2
k2 (r)dr
R0
1/2 R 1/2
1/2
R
dr
−1
r
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2mE
= −2
~2
2mE
= −2
~2
R0
"
R0
R arccos
R
1/2
R0
−
R
1/2 R0
1−
R
1/2 #
(E.17)
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b
But we have assumed that the barrier is sufficiently wide and hence R R0 and therefore
R0
arccos
R
1/2
1/2
π
R0
≈ −
2
R
and
1/2
R0
1−
≈1
R
We finally get
2mE
ln T = −2
~2
1/2
1/2 #
π
R0
R
−2
2
R
Substituting the value of R in this expression, we get
311
"
(E.18)
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1/2
e2 h m i1/2
4e m
1/2 1/2
Z R0 −
ZE −1/2
ln T =
~ π0
~0 2
(E.19)
Putting in the numbers for the various constants, we get
R
1/2
log T = 1.29Z 1/2 R0 − 1.72ZE −1/2
(E.20)
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where the nuclear radius R0 is in fermis and the energy E of the alpha particle is in MeV. Finally, we
know that the decay constant is related to the transmission probability Eq E.1 as
λ = νT =
Substituting, we get
R
v
T
2R0
v
Z
1/2
log λ = log
+ 1.29Z 1/2 R0 − 1.72 √
2R0
E
(E.21)
This is the famous Geiger-Nuttall Law relating the decay constant λ with the alpha particle energy
E. This law is usually written in the form
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Z
ln λ = −a1 √ + a2
E
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where a1 and a2 are two constants depending on the nuclei in question. This law has been confirmed
experimentally as can be seen in Figure E.3. In this figure, the log of the half life of various alpha
emitters is plotted against E −1/2 . We see that as expected, it is a straight line.
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Lab Manual for Nuclear Physics
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Shobhit Mahajan
Figure E.3: Geiger-Nuttall Law
§(Source:
http://www.open.edu/openlearn/science-maths-technology/science/physics-and-astronomy/scattering-and-
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b
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tunnelling/content-section-5.2 )
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Appendix F
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Fermi’s Theory of Beta Decay
The properties of Beta decay can be understood by using quantum mechanics. In particular, one uses
Fermi’s Golden Rule to find the transition rate between the initial and final states of the nucleus which
are involved in Beta decay. Thus, we need to first see how we can obtain Fermi’s Golden Rule from
elementary quantum mechanics.
F.1
Fermi’s Golden Rule
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Consider a system with Hamiltonian H0 . We assume that the the eigenvalues Ei and the eigenfunctions
ui (x) can be determined. That is
H0 ui (x) = Ei ui (x)
(F.1)
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b
Given this, we know that any state ψ of the system will evolve as
ψ(x, t) =
X
ci (0)e−iωi t ui (x)
(F.2)
i
Of course if the state is an eigenfunction of the Hamiltonian, it will be stationary, that is have no time
dependence. In our case, we take the initial state of the system, that is the state at t = 0 to be an
eigenfunction.
ψ(x, 0) = ui (x)
Next we consider a time independent perturbation being applied to the system. Thus, if our initial state
was the ground state of an atom, we could think of a momentary pulse of laser light on the system. Of
course, the perturbation in general can be time dependent but we will consider only time independent
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perturbations here. The perturbation can be written as a potential term V̂ and the total Hamiltonian
of the system thus becomes
H = H0 + V̂
(F.3)
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To solve for the state of the system, we would need to find the eigenvalues and eigenfunctions of this
composite Hamiltonian. This is not always easy to do. Thus we adopt a different strategy. We assume
that the state of the system with this new Hamiltonian can be expressed in terms of the
eigenfunctions ui (x) of the original, unperturbed Hamiltonian H0 . That is we guess a solution
of the form
ψ 0 (x, t) =
X
ci (t)e−iωi t ui (x)
(F.4)
i
Notice that Eq F.4 is almost the same as Eq F.2 except for the fact that now the coefficients are time
dependent. This is because of the perturbation that we have introduced.
The task then is to find the time dependent coefficients ci (t). We use the Schrodinger equation
∂ψ 0
= (H0 + V̂ )ψ 0
(F.5)
∂t
and substitute Eq F.4 in Eq F.5. Using the fact that the ui (x) are eigenfunctions of H0 and that
Ei = ~ωi , we get
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i~
X
i~ċi e−iωi t ui (x) =
X
i
ci (t)e−iωi t V̂ [ui (x)]
(F.6)
i
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b
Notice that the LHS of the above expression has the time derivative of the coefficients ci and the RHS
has the perturbing potential V̂ . We now take the inner product of both sides with uj (x) and use the
orthogonality of the eigenfunctions to get
X
i
But
+∞
i~ċi e−iωi t
+∞
u∗j (x)ui (x)dx =
X
ci e−iωi t
i
−∞
+∞
u∗j (x)ui (x)dx = δij
−∞
and therefore
315
−∞
u∗j (x)V̂ [ui (x)]dx
(F.7)
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+∞
X
i
u∗j (x)V̂ ui (x)dx = i~ċj e−iωj t
i~ċi e−iωi t
(F.8)
−∞
We define
+∞
u∗j (x)V̂ [ui (x)]dx
V̂ij =
Eq F.7 becomes
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−∞
ċj = −
or
iX
cj (t) = −
~ k
iX
ck (t)ei(ωj −ωk )t V̂jk
~ k
(F.9)
t
0
ck (t0 )ei(ωj −ωk )t V̂jk dt0 + cj (0)
(F.10)
0
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This is now the formal solution to the problem. Once we have the coefficients cj (t), we can substitute
them in Eq F.4 and find the time evolution of the state of the system in the presence of the perturbing
potential. However, to actually solve Eq F.10 is not easy. Therefore we make a crucial approximation. We assume that the perturbing potential is such that its effect on the system
is slow. That is the coefficient ck (t) on the RHS of Eq F.10 doesnt change with time and
therefore can be replaced with its initial value at t = 0. This then allows us to rewrite Eq F.10
as
t
0
ei(ωj −ωk )t V̂jk dt0 + cj (0)
(F.11)
0
b
iX
cj (t) = −
ck (0)
~ k
La
But our initial assumption was that the unperturbed state of the system initially, that is the state at
t = 0 is an eigenfunction of the original unperturbed Hamiltonian, that is ψ(x, 0) = ui (x). Therefore
the coefficients ck (0) take a fixed value, that is ck (0) = 0 for all k 6= i. We are now left with
i
cj (t) = −
~
i
= −
~
t
0
ei(ωj −ωi )t V̂ji dt0
0
t
0
ei∆ωj t V̂ji dt0
0
where
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(F.12)
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∆ωj = ωj − ωi
This integral can be solved easily and we get
cj (t) = −
1
V̂ji 1 − ei∆ωj t
~∆ωj
(F.13)
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Once we know the coefficients, we can find the quantity of interest, that is the probability of transition
of the system from an initial state ψ(x) = ui (x) to the final state which we assume to be also an
eigenfunction of the original Hamiltonian H0 . For this, we simply need to take the modulus squared of
the coefficients.
P (i → j) = |cj (t)|2
(F.14)
The probability of transition is then given by
4|V̂ji |2
P (i → j) = 2
sin2
~ ∆ωj2
∆ωj t
2
(F.15)
We can rewrite this in another way for reasons which will become apparent soon.
2
|V̂ji |
~2

∆ω t
sin2 2 j


∆ωj2
22
(F.16)
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P (i → j) =

Now we are interested in the behaviour of our system at large times, that is t → ∞. In this limit, the
function sinx x becomes very narrow until we can approximate it with a delta function. The exact limit
can be taken and we get
2π|V̂ji |2 t
δ(∆ωj )
(F.17)
~2
This allows us to find the transition rate which is simply the probability of transition per unit time as
R
La
b
P (i → j) =
Wij =
2π|V̂ji |2
δ(∆ωj )
~2
(F.18)
This is Fermi’s Golden Rule which provides a reasonably accurate way of finding the transition rates.
This form of the Golden Rule is obviously valid if the transition is from an initial state of definite energy
to another one of definite energy. In the cases of interest to us, namely beta and gamma decay, the
final particles (electron or photon) are free and therefore have a continuous range of energies. In that
case we need to modify the Golden Rule as state above to take into account this. To do this, recall that
in the derivation above, we used a delta function for the difference between the initial and final energy,
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that is δ(∆ωj ) where ∆ωj = ωj − ωi . If the final state does not have a definite energy, then we need
to consider transitions not to a definite final state but instead to states in a narrow interval centered
on Ej , that is Ej ± dE. Obviously the transition rate will be proportional to the number of states to
be found in this energy interval. This is related to the density of final states, ρ(E) by dN = ρ(E)dE.
Thus we replace the delta function by the density of states centered around Ej = Ei . In doing this,
remember that
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δ(∆ωj ) = ~δ(~∆ωj ) = ~δ(Ej − Ei )
by the properties of delta function. Thus the final result has only one ~ in the denominator. We can
finally write the Golden Rule as
R
Wij =
2π|V̂ji |2
ρ(Ej )
~
(F.19)
where it is understood that the density of states needs to be evaluated for Ej = Ei .
F.2
Fermi’s Theory of Beta Decay
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The accurate description and theory explaining beta decay of course would mean that we use quantum
field theory to analyse the process. In this, the particles in question, namely the electron and the
anti-neutrino are described by field operators. The interaction, as we have already seen which is
responsible for Beta Decay is the weak interaction. We will not attempt to do a full quantum field
theoretic analysis of the process. Instead we will try to see if we can use Fermi’s Golden Rule to get
some results.
R
La
From Eq F.19, we see that the transition rate will be
W =
2π| < ψi |V̂ |ψf > |2
ρ(Ef )
~
(F.20)
where V̂ is the interaction Hamiltonian relevant to the process. Thus we need two things to use Fermi’s
Golden Rule: the matrix element, < ψi |V̂ |ψf > and the density of final states ρ(Ef ). Let us look at
the Matrix element first.
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F.2.1
Lab Manual for Nuclear Physics
Matrix Element
We write the interaction term in terms of field operators which create and destroy particles. In this
case, the initial state is a neutron while the final state has a proton, electron and anti-neutrino. So we
take the interaction term to be
V = gψe† ψν̄†
(F.21)
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where g is the strength of the interaction and the two operators ψe† and ψν̄† create an electron and
antineutrino.With this our matrix element can be written as (where for the moment, we are not explicitly
writing the nuclear initial and final states)
d3 xψp∗ ψe∗ ψν̄∗ ψn
Vif = g
(F.22)
where we have replaced the dagger with a star since we consider our operators to be scalars. Now we
know that the process results in a free electron and a free anti-neutrino. Thus as an approximation we
can take the operators to be plane waves. With this approximation, we get
Vif = g
3
d
i~ke ·~
x
∗e
xψp √
~
eikν̄ ·~x
√ ψn
V
V
(F.23)
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√
Please remember that the V in the denominator above refers to the volume that we need to
normalise the plane wave states and not the potential.
Now we can make a further approximation. Typical kinetic energies of the electron are in the MeV range.
Thus if we take the kinetic energy of the electron to be 1 MeV, we can easily find the corresponding k.
To do this, note that the kinetic energy of the electron, Te is given by
p
p2e c2 + m2e c4 − me c2
b
Te =
La
Solving for pe , with Te = 1 MeV and me = 0.51 MeV c−2 , we get
pe ≈ 1.4 MeV c−1
Thus by de-Broglie relation and the definition of the wave number, we get
pe
≈ 0.007 fermi−1
h
We can thus easily see that for nuclear dimensions,
ke =
ke r 1
With this, we see that
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eike r = 1 + ike r −
ke2 r2
+ ··· ≈ 1
2
We thus get for the matrix element
g
Vif =
V
d3 xψp∗ ψn =
g
Mf i
V
(F.24)
R
F.2.2
W =
Density of Final States
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where we have introduced a function Mf i which is some complicated function which takes into account
all the details of the nuclear states. We also know that there would be some electromagnetic interaction
between the electron and the proton which we have neglected above. So to take care of that, we replace
|Mf i |2 by |Mf i |2 F (Z, Q) where the function F (Z, Q), called the Fermi function, incorporates the
electromagnetic interactions. This function is tabulated extensively for various values of Q and Z. We
thus have the expression
2π g
|Mf i |2 F (Z, Q)ρ(Ef )
~ V
(F.25)
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The two particles that we see in beta decay are the electron and the anti-neutrino. The density of
final states should therefore reflect this as also the fact that both of them are free particles, that is
they exist in a continuum of possible states. To find the density of final sates, we need to know the
number of states accessible to the electron and the anti-neutrino. Suppose the electron is emitted with
a momentum pe and the anti-neutrino with q. Since we need to know only the shape of the energy
spectrum, the directions of these momenta are not relevant to us. Consider a coordinate system with
p
axes along pex , pey , pez . Then for a specific value of the electron momentum p = p2ex + p2ey + p2ez , the
locus of points is a sphere of this radius. Thus the locus of all the points with momentum between p
and p + dp is a shell with inner radius p and outer radius p + dp with a volume 4πp2 dp.
Recall that phase space is an imaginary space consisting of 3 space dimensions (x, y, z) and 3 momentum
coordinates (px , py , pz ). To specify the particle, say the electron, we need to specify its position and
momentum. Of course, uncertainty principle tells us that
∆x∆px ≈ h
Therefore
∆x∆y∆z∆px ∆py ∆pz ≈ h3
This is basically saying that when I say an electron is at a point in phase space, it is basically in a
unit cell of phase space of “volume” h3 . Now if we say that an electron has momentum between pe and
pe + dpe , then we know that the uncertainty in the momentum of the electron will be
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∆px ∆py ∆pz = 4πp2e dpe
and if we also require that the electron is confined to a volume V , then the corresponding volume in
phase space where the electron could be found will be
4πp2e dpe V
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Now the question is really how many ways can an electron exist in this volume of phase space. The
answer is simply the number of unit cells in this volume. Recall that each unit cell has a volume h3 and
therefore we get
4πp2e dpe V
h3
Similar arguments lead us to the expression for the number of neutrinos.
dNe =
4πp2ν dpν V
h3
We can rewrite these expressions in a more useful form as
4πV
dNe =
p2e dpe
3
(2π~)
dNν =
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Similarly for the neutrino we get
dNν =
4πV
(2π~)3
p2ν dpν
(F.26)
(F.27)
Thus we write the number of states in a small energy volume as
(F.28)
b
dN = dNe dNν
La
where we have for simplicity use the symbol for neutrino instead of anti-neutrino. We also know that
the total kinetic energy available in the reaction, that is the Q value is shared between the electron and
neutrino.
Q = Te + Tν
We will take the neutrino to be massless and therefore
Tν = pν c
The electron kinetic energy is simply
Te = E − me c2 =
p
p2e c2 + m2e c4 − me c2
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(F.29)
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Now
pν =
Tν
Q − Te
=
c
c
Therefore for a fixed value of Te , we have
dQ
c
The density of final states for a fixed electron energy is thus given by
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dpν =
dNν
dTν
2 2
dpν
16π V 2
=
pe dpe p2ν
6
(2π~)
dTν
ρ(pe )dpe = dNe
(F.30)
Now from the relationship between the kinetic energy and momentum for the neutrino, we know that
Tν = pν c
and
pν =
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Therefore we get
(Q − Te )
c
i2
p
V2 h
2
2
2
2
4
ρ(pe )dpe = 4 6 3 Q − ( pe c + me c − mc ) p2e dpe
4π ~ c
But
La
Therefore
p
p2e c2 + m2e c4 − mc2
b
Te =
(F.31)
pe dpe =
Te + me c2
dTe
c2
Substituting in Eq F.31, we get
R
F.2.3
ρ(pe )dpe =
p
V2 2 2
Q
−
T
( Te2 + 2me c2 Te )(Te + me c2 )
e
4π 4 ~6 c6
Decay Rate
We are now finally ready to find the decay rate using Fermi’s Golden Rule. We have
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(F.32)
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2π
|Vif |2 ρ(E)
~
Notice that as far as the shape of the final energy spectrum is concerned, all the factors in the decay
rate that are independent of the momentum can be collected into some constant, say C. This will
include the nuclear matrix element Mf i etc since we assume them to be independent of the electron
momentum. Then the number of electrons between momentum p and p + dp will be
W =
But we know that
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Npe dpe ≈ Cp2e p2ν dpe
pν =
Substituting, we get
R
(Q − Te )
c
Npe dpe ∼ Cp2e (Q − Te )2
and
R
2
p
2
2
2
2
4
Q − pe c + me c − mc
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Npe dpe ∼
Cp2e
(F.33)
(F.34)
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These functions can be plotted for any value of Q. We see that for pe = 0 and for Te = Q, the function
vanishes. A plot of for the shape for Q = 2.5 MeV is given in Fig F.1.
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Beta Spectrum
5
Beta Spectrum
4.5
4
3.5
2.5
2
1.5
1
0.5
0
0
0.5
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N(T)
3
1
1.5
2
2.5
T(MeV)
Figure F.1: Beta spectrum plot for Q = 2.5 MeV
The complete beta spectrum thus contains three contributions:
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1. The statistical factor coming from the density of final states. This as we have seen is proportional
to p2e [Q − Te ]2 .
2. The contribution coming from the nuclear Coulomb field, that is the electromagnetic interaction
of the nucleus with the beta particle. This as we saw above is proportional to the Fermi function
F (Z, Q).
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3. The contribution coming from the nuclear matrix element, Mf i which includes the effects of the
particular initial and final nuclear states of the particular decaying nucleus and could also contain
some momentum dependent factors.
Thus we can write the final beta spectrum as
Npe ∝ p2e (Q − Te )2 F (Z, Q)|Mf i |2
(F.35)
We can rewrite this as
R
s
(Q − Te ) ∝
Npe
2
pe F (Z, Q)
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(F.36)
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Note that if we plot the RHS of Eq F.36 against the electron energy, we get a straight line and the value
of the x intercept will give us the Q value or the end-point energy of the beta particles. This is the
famous Fermi-Kurie plot shown in Figure F.2. The Fermi-Kurie plot is a good way to test Fermi’s
theory of Beta Decay. It also provides a very accurate way to determine the end-point energy.
Figure F.2: Fermi-Kurie plot
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§(http://physics-database.group.shef.ac.uk/phy303/phy303-4.html )
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Appendix G
G.1
Introduction
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Semi-Classical Theory of Gamma Decay
We have already seen in Section 2.2.3 that gamma radiation comes when a nucleus de-excites from an
excited state to a lower energy state. We have also seen that the typical energies of the gamma rays is
of order MeV. To understand the gamma emission, we can use a semi-classical approximation alongwith
Fermi’s Golden Rule as described in Section F.1. The transition rate from Fermi’s Golden Rule is given
by Eq F.19 as
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Wij =
where
2π|V̂ji |2
ρ(Ej )
~
(G.1)
Vji =< j|V̂ |i >
G.2
La
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that is the matrix element of the interaction potential V̂ with the initial and final states and ρ(Ej ) is
the density of final states as discussed in Section F.2.2. Thus we see that if we need to apply this to
gamma decay, we need two ingredients, the density of final states and the matrix element.
Density of Final States
We have already seen in the case of beta decay that the number of electrons in a volume of phase space
is given by Eq F.26. Using similar arguments, we consider the nucleus and the gamma radiation as
existing in a box of volume V (or L3 ), we can easily find that the number of final states with the gamma
ray photon momentum between k and k + dk will be
dNs = 4πk 2 dk
V
(2π)3
If we consider a solid angle dΩ instead of the whole sphere, we get
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(G.2)
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dNs = k 2 dkdΩ
V
(2π)3
(G.3)
But we know that for photons, p = ~k and E = pc = ~kc = ~ω. Substituting we get
R
Interaction Hamiltonian
ω2 V
dNs
= 3
dΩ
dE
~c (2π)3
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G.3
ρ(E) =
(G.4)
Since we are dealing with gamma rays, that is electromagnetic radiation, we need to consider the
interaction of a particle with an electromagnetic field. This is expressed in terms of the vector potential
~ associated with the electromagnetic field as
A
V̂ =
e ~ˆ ˆ
A · p~
mc
(G.5)
~ˆ is the quantum mechanical operator associated with the vector
where the particle charge is e. Here A
~ and p~ˆ is the momentum operator for the particle. Now we consider the electromagnetic
potential A
field as a quantum field and therefore expand it in terms of creation and destruction operators, âk and
â†k . These can be thought of as operators which create and annihilate photons of momentum k.
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s
~ˆ =
A
X
k
2π~c2 i~k·~r
~
âk e + â†k e−ik·~r ~k
V ωk
(G.6)
where ~k is the polarisation of the electromagnetic wave.
G.3.1
Dipole Approximation
La
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With the interaction potential now in this form, we can evaluate the transition rate using Fermi’s Golden
Rule now. We have
2π|V̂ji |2
ρ(Ej )
~
In our case, we note that in gamma decay, our initial state has no photon (there is only the excited
nucleus) and the final state has one photon with an energy E = ~ω = ~kc. If we substitute the
~ˆ from Eq G.6 in the expression for V̂ in Eq G.5, we see that only one term will be
expression for A
non-zero in the evaluation of the matrix element. This will be the term with one â†k since this is the
only one which can connect an initial state of no photon with a final state of one photon. Thus
Wij =
e
Vji =
mc
s
D
E
2π~c2
~
~k · p~ˆe−ik·~r
V ωk
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(G.7)
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The expectation value of the momentum operator p~ˆ is obviously between the initial and final states.
Now we can try to simply this. Recall that
[p~ˆ2 , ~rˆ] = −2i~p~ˆ
Therefore
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im p~ˆ2 ˆ
i ˆ2 ˆ
[p~ ,~r] =
[
,~r]
p~ˆ =
2~
~ 2m
But we know that the nuclear Hamiltonian Hn is
Hn =
p~ˆ2
+ V̂n (~rˆ)
2m
ˆ and
where V̂n (~rˆ) is the nuclear potential term. Since this only a function of r, it will commute with~r
so we see that
im
im p~ˆ2
[
+ V̂n (~rˆ), ~rˆ] =
[Hn , ~rˆ]
p~ˆ =
~ 2m
~
This is an enormous simplification since we know that the initial and final states in the matrix element
are eigenstates of the nuclear Hamiltonian. Thus
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i
im
im h
< j|p~ˆ|i >=
< j|[Hn , ~rˆ]|i >=
< j|Hn~rˆ|i > − < j|~rˆHn |i >
~
~
But since the states are eigenstates of Hn , we can simplify this to
ˆ >= im (Ej − Ei ) < j|~rˆ|i >= imωk < j|~rˆ|i >
< j|~p|i
~
where ~ωk = Ej − Ei .
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Thus
e
Vji =
mc
s
D
E
2π~c2
−i~k·~
r
ˆ
imωk~k · ~re
V ωk
(G.8)
(G.9)
(G.10)
ˆ −~k·~r . To simplify this further, let us consider a
We still need to evaluate the expectation value of ~re
typical gamma ray of energy 1 MeV. The wavelength of this radiation is ∼ 10−13 m. Nuclear size we
know is of the order of 1 fermi or ∼ 10−15 m. Thus the quantity kr 1. This allows us to approximate
the exponential by its first term. In this approximation,which is called the dipole approximation,
we thus have
R
r
Vji =
D E
2π~e2 ωk
ˆ
~k · ~r
V
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The reason for calling this the dipole approximation is obvious now since the dipole operator e~rˆ is
what is relevant as can be seen from Eq G.11.
G.4
Transition Rate & Lifetime
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The transition rate can now be calculated since we have the density of final states (Eq G.4) and the
matrix element (EqG.11). Then the transition rate is given by
E
2π|V̂ji |2
ω3 D
ˆ
W ≡ λ(E1) =
ρ(Ej ) =
| ~k · ~r |2 dΩ
(G.12)
~
2πc3 ~
Here E1 signifies that this is only the dipole term.
But we know that the angle between the polarization vector and the dipole moment vector is π/2 − θ.
To see this, consider a dipole aligned along the z axis. Now the position vector of a point (r, θ, φ), will
~ field direction will be perpendicular to the direction of
have an angle of θ with the z axis. But the E
propagation and therefore the angle between it and the z axis is π2 − θ. Please do not confuse between
the position vector ~r and the dipole operator e~rˆ though they look the same. The angle θ is between
~ is precisely the direction of the polarization vector ~k .
the two in our example. The direction of the E
That is, the angle between ~k and ~rˆ is also π2 − θ. Therefore,
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D E D E
~k · ~rˆ = ~rˆ sin θ
Then the transition rate is simply
e2 ω 3 D ˆE 2 2
| ~r | sin θdΩ
2πc3 ~
To get the total transition rate, we need to integrate over all directions. Then since
we get
R
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λ(E1) =
2π
π
sin2 θ sin θdθ =
dφ
0
(G.13)
8π
3
0
λ(E1) =
4e2 ω 3 D ˆE 2
| ~r |
3c3 ~
(G.14)
This is the transition rate. We can get a reasonable estimate of this quantity. We know that ω = E~ .
We can also approximate the value of the expectation value of the position operator by the nuclear
radius, Rn ∼ r0 A1/3 where r0 ∼ 1.25 fermi. Putting it all together we get
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R
λ(E1) ≈
4e2 E 3 2 2/3
r A
3~(~c)2 0
(G.15)
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If we put in representative numbers for the quantities above, E = 1 MeV and A = 65, we get λ(E1) ≈
1.5 × 1015 s−1 or its reciprocal the time to be τ ∼ 10−15 seconds.
330