Atomic and Nuclear Physics
Chapter 1: The structure of the atom
Chapter 2: X-rays
Chapter 3: Radioactivity
History of the atom
Brief history of the atom
• 600 BC: Greek Philosopher Democritus asked himself that if one breaks a piece of matter in half and then
breaks it in half again, how many times must it be broken such that the piece cannot be broken any more. He
called the remaining smallest matter that is indivisible an ‘’Atom”
• Aristotle dismissed the idea of atom as worthless. The idea of the atom was not pursuit until 1800
• 1800: John Dalton performed some experiments that pointed out that indeed matter consist of elementary
particle (atom). He could not determine its structure, but the evidence pointed to something fundamental.
• 1897: J. J. Thomson discovered the electron and proposed a model for the structure of the atom. Knowing
that the electron had a negative charge he then proposed the structure of atom as having a positive charge
matter and electron attached to it. His model looked like raisins stuck on the surface of a lump of pudding.
The structure of the atom
There is a nice and rich history concerning the experiments and postulations leading to an understanding of the
structure of an atom. However, we are not going to go through the history, instead, we will briefly talk about
an experiment that lead to the discovery of a structure of an atom as it is known today. It involves firing of
alpha particles from a radioactive source/material.
Radiate Energy: Radioactive decay
P=6 N=8
Unstable-has lots of
energy-Radioactive
E=6
The structure of the atom
1911: Ernest Rutherford fired alpha-particle through the atom so as to determine its internal structure. He used
radon as the source of alpha particles and fired them to a gold foil and put a fluorescent screen behind which
show alpha particles on impact.
Expectation
Radioactive
source
- --- --
- -- -- - - -- ---- - - - - - - -- -- -- -- - - - - - - -- - - --- -- -- -- ---- ---- --
Observation
− particle – Helium nucleus
Proposed
structure of an
atom
Rutherford’s structure of the atom:
Consist of a nucleus of the radius 10-15 m
surrounded by electron forming an atom
of radius 10-10 m.
The nucleus
•
•
•
•
The nucleus contains protons and neutrons - almost same size
Protons are positive, neutrons have no charge
The mass of each proton and neutron is about 1840 times larger than that of an electron
A name give to particles in inside a nucleus, i.e. proton or neutron is the nucleon
• Define a mass number (A) to be a number of nucleons (proton+neutron) contained in a nucleus
• And atomic number (Z) as
(i) the ordinal number of the element in the periodic table
(ii) the number of protons in the nucleus
(iii) the number of electrons surrounding the nucleus in the ‘normal’ atom.
Any nucleus can be fully described by writing
where A and Z are mass and atomic numbers respectively. For example in a chemical symbol
A = 27 and Z=13 which implies there
• 13 protons and
• 14 neutron obtain from (27 – 13) i.e. Neutron no. = A-Z
,
The isotopes
An isotope is a form an element can take in which is a given element can have the same number of
protons (and therefore same Z) but different number of neutrons (and thus different A)
Examples of isotopes of carbon
Examples of isotopes of neon
has 10 protons and 10 neutrons
has 10 protons and 11 neutrons
has 10 protons and 12 neutrons
Atomic mass
The atomic mass (M), also know as atomic weight, is the mass of an atom or molecule. It is normally
expressed in atomic mass unit (u). Atomic mass expresses a mass of a given atom on a scale in which a
mass of a
atom is ascribed a value of 12 atomic mass unit (u) i.e.
mass of a
atom = 12 u which implies that
1 atomic mass unit (1 u) =
of the mass of a
atom.
Let us define an Avogadro’s number (NA) 6.022 × 1023 as the number of atoms in one gram-atomic
mass of any element. Informally consider one gram-atomic mass as saying 1g×magnitude of atomic
mass. Thus
6.022 × 1023 atoms of
→ 12 grams conversely
contains → 6.022 × 1023 atoms or
has a mass of →
g
6.022
But mass of a
atom = 12 u, then
12 grams of
1 atom of
12 u →
6.022
g
g
1u→
6.022
1 u = 1.66 × 10 = 1.66 × 10
The nucleus
Example 1.3: Estimating the number and size of atoms
Taking the atomic mass of copper as 64 and its density as 8930 kg/m3, estimate
(a) the number of copper atoms in a volume of 1 cubic metre, and hence
(b) the diameter of a copper atom.
Mass and energy
In 1905 Albert Einstein discovered that every object which has mass has energy associated with it. The
energy (E) associated with an object of a mass (m) is given by
=
!
where ! is a speed of light. Thus mass and energy are like two sides of the same coin. The energy that
can be release when 1 kg of combustible material e.g. coal is burnt is given by
= !
= 1 × 3 × 10#
'()
= 9 × 10 % & × .%×
+ ,
= 2.5 × 10 /0 = 25 × 101 /0
Energy can be expressed in units of Joules (J) or electronvolt (eV).
An eV is defined as the energy acquired when an electron (or any particle with a charge equal to that of
the electron) is moving freely through a potential difference of 1 V i.e.
/ = 23foranelectronwithq=e=1.6 × 10
/ = 3=1.6 × 10 1 × 13
3=1.6 × 10 1 &
1
Mass and energy
We have seen that the relationship between the (u) and (kg) is given by
1u = 1.66×10-27kg.
According to Einstein’s mass-energy theorem, the energy associated with 1u is given by
= !
= 1.66 × 10
× 3 × 10#
= 1.49 × 10 &
This energy can be expressed in terms of 3 knowing that 1 3 = 1.60 × 10
i.e.
= 1.49 × 10 & × .%×CDEFG,
= 9.3 × 10# 3 ×
≈ 930J 3
Thus the energy associated with 1u ≈ KLMNOP
HCD
+ CD
1 &
Nuclear energy
Protons within a nucleus are at very short distances from one another, they exert very large repulsive electric forces
on each other. Two neighbouring protons separated by 2 × 10−15 m experience a repulsive electric force of
R
(1.6 × 10 1 )
1
= 9 × 10
= 58N
Q=
(2 × 10 V )
S
Obviously some extra force must be present in the nucleus to prevent it from instantaneously bursting apart. This
attractive force is known as the strong nuclear force. This force binds a nucleon to its neighbour and it acts on a
short range of an order of 10−14 metre.
When protons, neutrons and electrons are put together to form an atom, energy is given off as the nucleus is bound
together by the strong nuclear force. This loss in energy shows up as a loss in mass. This implies that the mass and
energy of the atom is less than that of the individual protons, electron and neutrons.
The difference in mass is termed a loss mass or mass defect while the difference in energy is termed binding
energy. The binding energy can also be understood as the energy that must be supplied to separate the nucleus into
individual protons and neutrons. Its magnitude is given by
YZ[\Z[ [ S ] = ^__ `__ × 930J 3
And the binding energy per nucleon the energy required to separate one nucleon from the nucleus. It is given by
YZ[\Z[ [ S ]a S[b! ^[ =
YZ[\Z[ [ S ]
[^. ^c[b! ^[_
Nuclear energy
Example 1.6: Binding energy of
Given that
Mass of an atom of 1#f
Mass of neutron
Mass of proton without electron
Mass of electron
Mass of proton + mass of electron
Le
K d
238.0508 u
1.008 66 u
1.007 28 u
5.48 × 10-4 u
1.007 83 u
Determine the binding energy and the binding energy per nucleon of
1
#
f(1 u = 934 MeV).
Binding energy per nucleon
Splitting of heavy atoms is achieved by firing neutrons into the nuclei. The nuclei then become
unstable, and split, giving off large amounts of energy. Nuclei on splitting give off neutrons, which
enter more nuclei, causing them to split in turn, and so on. Such a runaway reaction is known as a
chain reaction.
Chain reaction
1
0
n +
235
92
U
→
Ba +
141
56
92
36
Kr
+ 301 n + Q
Run wild → Atomic bomb
Chain reaction
1
0
n +
235
92
U
→
Ba +
141
56
92
36
Kr
+ 301 n + Q
Controlled → Nuclear reactor
X-rays
• X-rays are electromagnetic waves which have shorter wavelength (0.1-10 Å) and thus have higher
energy and can penetrate through some materials
• Applications of X-rays includes
• Medical use
• Industrial use
• Research
• The X-ray instrument consist of the source of X-ray beam (i.e. X-ray tube), the screen/film and the
object to be viewed in between.
X-ray tube
• Accelerating voltage (20-60 kV) which provides very high V between the Metal plate (anode) and
heated filament (cathode)
• The filament is heated and excites e-’s
• Accelerating voltage accelerates the e-’s at the very high speed towards the metal target
• Accelerates e-’s incident on the metal plate excite a huge number of e-’s in the metal plate
• When e-’s move back to ground states, they emit photons/radiations with wavelength in the range
(0.01-1 nm) i.e. X-rays
X-ray emission spectrum
X-ray spectrum, i.e. the intensity of X-rays as a function
of wavelength consist of three features:
• Continuous background spectrum
• Characteristic line spectrum
• ‘Cut-off’ wavelength
Continuous spectrum
Accelerating e-’s excite a huge number of outer e-’s. When
they fall back to ground states they emit photons (X-rays)
consisting of a wide rage of g‘s. A plot of the intensity of
X-rays as function of g produces a continuous spectrum
Characteristic line spectrum
• Accelerating e- may penetrate through a cloud of e- and excites
one of the inner e-’s
• An e- in the higher orbital falls back to fill the vacancy, and
emits a photon with g corresponding to that of the characteristic
line spectrum
• K indicate that the ejected e- was in the K-shell
X-ray emission spectrum
Cut-off wavelength/frequency
Accelerating e- may penetrate through a cloud of e-‘s without colliding with any eand come very close to the nucleus The energy of an e- is given by
= 3
• Attractive force between the e- and the nucleus will decelerate the e-.
Consequently, the e- looses energy in form of a photon. This photon is called a
‘bremstrahlung photon’ and has energy given by
a = 0c
where 0 is the Plank’s constant and a value of 6.63 × 10 &_
•
If all the energy ( 3) acquired by the electron in speeding through the potential difference V goes into creating a single
photon, then
Since cjkl =
m
nopqErss
then
0!
gmtu
`i
= 3
= 3
vww
0!
3
From this we can see that cut-off g depends
on the accelerating V
gmtu
0c
vww
=
c
`i
=
0
3
X-ray emission spectrum
Diagnostic X-ray in medicine
A potential difference of 87.0 kV is applied between the filament and the target in the x-ray tube used at the
local clinic to look at the broken bones. What are the shortest wavelength x-rays produced by this tube?
Diffraction of X-rays by crystals
• When a beam of X-rays is incident on a crystal, a
diffraction pattern produced by a crystal can be used to
infer the atomic structure of the crystal.
• Suppose a beam of X-rays of certain g is incident on a
crystal at a glancing angle x i.e. the angle between
the incident beam and the surface of the crystal. If
the path difference between the beams is equal to an
integral number of wavelengths then there will be
constructive interference i.e. the receiver will detect
more X-rays. Mathematically
2i = 2\_Z[y = [g
This equation is known as Bragg’s law.
θ
θ θ
d
x
x
x = d sin θ
Diffraction of X-rays by crystals
Example 2.1: Voltage of an X-ray tube and atomic spacing
(a) What is the least voltage that must be applied across an X-ray tube in order to produce an X-ray
beam of wavelength 0.40 × 10 −10 m?
(b) What is the spacing of the atomic planes in a crystal which produce a second-order diffracted beam
for an angle of incidence of 60° with the above beam?
Radio activity
Previously we noted that for bigger nuclei (Z > 92), the electrostatic repulsive force exceeds strong nuclear
force. The result is the spontaneous disintegration/decay of the nucleus i.e. undergo radioactivity
Radioactivity is the spontaneous disintegration of atoms of a given element into atoms of another element,
accompanied by the emission of radioactive ‘rays’ i.e. alpha (z), beta ({) and gamma (|) rays.
• These particle are invisible so their presence can be
detected using an experiment shown alongside.
•
In this experiment a radioactive material such as
radium (Ra) is used which emits all three particles is
used.
•
•
When the particle enters the magnetic field they separate
into three paths and they are exposed on the photographic
film in three position.
The right hand rule is used to determine the direction of
the positive particle (z). The direction of { will be
opposite to that of z particle. The } rays are not
affected by magnetic field.
v
F+
Right hand rule
• Thumb-velocity of the particle
• Fingers - direction of magnetic field
• Palm - direction of the force or
direction in which the z particle
moves.
Radio activity
Further experiments proved that
• z - particle is Helium (He) nucleus
• { - particles is an electron
• | - ray is an electromagnetic radiation
Note that some radioactive substances emit positive β-particles i.e. same mass but opposite
charge to the negative β-particles. These positive electrons are called positrons.
Radio activity decay types
Unstable nucleus can under radioactive decay in three ways i.e.
• Alpha decay
• Beta decay
• Electron emission
• Positron emission
• Electron capture
• Gamma decay
Alpha (z) decay
• Is a process where a large nucleus decays into a nucleus of another element and emits an
zparticle.
• A parent nucleus is converted into a daughter nucleus.
• The parent nucleus will loss 2 proton and 2 neutrons i.e. the atomic number (Z) will
decrease by 2 and the mass number (A) by 4
• Alpha decay is a consequence of the instability of large nucleus
1
#
V
f=
…† =
…
…
…
…
+ € → ‚ = 90,
= 231 →
+ € → ‚ = 80,
= 210 →
1
#
V
f=
…† =
1
#
„0 + €
€ + €
Radio activity decay types
Beta ({) decay
Electron emission
• A parent nucleus is converted into a daughter nucleus by losing an electron.
• An beta particle or an electron is represented as MO.
• This decay results in the increase in atomic number (Z) while the mass number (A) stay the same
1
V
f=
…
…
+
→ ‚ = 93,
= 235 →
1
V
f=
1
V
a+
Positron emission
• A parent nucleus is converted into a daughter nucleus by losing a positron.
• An beta particle or a positron is represented as is represented as ˆMO
• This decay results in the decrease in atomic number (Z) while the mass number (A) stay the same
#
\=
+…
… → ‚ = 1,
= 0 →
#
\=
+ˆ
Gamma (|) decay
• The nucleus which underwent radioactive decay can be left in an excited state. When
the nucleus go back to ground state, a photon is released.
• The emitted photon is in the | ray portion of the electromagnetic spectrum, thus is
called a | ray.
1
V
∗
f =
1
V
f+ |
Decay constant and half-life
• Suppose a sample of radioactive material contains number of atoms. After some time ∆t, a certain
number of atoms ∆ decays.
• The number of nuclei ∆ that decay in the time interval ∆Š is proportional to a product of ∆Š and the
total number of radioactive nuclei i.e.
∆ ∝ ∆Š
• Introducing a constant of proportionality g which is the decay constant, the number nuclei that has
decayed is given by
∆ = −g ∆Š
• Large values of g indicate large decay rate while small values of g indicate small decay rate. The
minus sign indicates that decreases with time.
• Let v be the number of atoms present initially, then decreases with time according to the equation
(result quoted without proof)
= v nu
where e is the base of natural logarithms (= 2.718 28 . . .).
Decay constant and half-life
•
decreases with time approximately as in the sketch
opposite—an exponential decay. The half-life („) is
the time for to reduce to F v , i.e. the time for half
the atoms to decay.
• „ varies tremendously, e.g. the half-life of # …^ is
10−4 s whereas the half-life of ##f 4.5 × 109 years!
• Note that from = v nu , when t equals the halflife T,
= F v thus
Œr
=
v
n•
where T is half-life
2= ˆn•
g„ = ln2
[2 0.693
„=
=
g
g
Decay constant and half-life
The rate at which the radioactive nucleus decays with time is called the activity (A) i.e. the number of
disintegration per second or decays per second and is given by
Δ
†bŠ∆ = −g ∆Š ∴ = g
ΔŠ
The SI unit of the activity the becquerel (Bq) where 1 Bq = 1 disintegration/s or 1 decay/s. Activity can also be
expressed in curie (Ci) where 1 Ci = 3.7×1010 Bq
=−
Form = v nu
= g
=
The following equations
=
v
=
nu
and
=
nu
=
•‘
u
•
v
nu
v
=(
nu
v
•‘
if we let g
v
=
v
nu
can be further simplified noting that g =
)
v
v
1
=
2
v
1
=
2
u ڥ
u ڥ
Similarly
u ڥ
=2
u ڥ
= 2
u ڥ
•‘
•
.
Examples
Example 3.1: Activity of a radioactive isotope
A solution containing a radioactive isotope emits β-particles with a half-life of 12 days surrounds
a Geiger counter which records 480 counts per minute. What counting rate will be obtained
(a) 48 days later
(b) 50 days later?
Example 3.2: Determining the half-life
from a graph
Determine the half-life of the element whose
decay curve is shown in the figure.
Examples
Example 3.3: The fraction of a radio-active substance left after a time t
For T = 5 years, calculate the fraction left after 17.25 years.
Example 3.4: Estimating the mass of a radioactive source
Estimate the mass of # …^ needed for a radioactive source of strength 2 × 108 Bq, if its half-life
is 138 days.
Tutorials
3. How many atoms are there in 1 kg of gold, if its atomic mass is 197 u?
7. The most energetic X-rays from a 15 kV tube were found to give 3rd order diffraction maxima when
beamed onto a crystal face at an angle of incidence of 60⁰. Calculate the spacing of the atomic planes
responsible.
9. Complete the following nuclear reaction equations given that there are two reaction products in each case
(neglect neutrinos and energy released):
1“ → … ` + ⋯
VY
+ ⋯ → •Z + …€
11. Cobalt-60 has a half life of approximately 5 years. If there was one microgram present in a sample
originally, how many atoms would have decayed after
(a) 5 years
(b) 20 years
(c) 21 years?
12. A strontium-90 source (half-life 27 years) has an activity of 2 × 105 Bq. How many atoms does it
contain? Hence estimate the mass of strontium-90 present. What will be its activity 81 years later
(a) in becquerels,
(b) in microcuries?
Carbon dating
Cosmic rays are high energy radiation from outer space consisting of, among other particles,
neutrons. When they reach the earth’s atmosphere, the neutrons induce the following reaction:
[+
→ a+ %
• The reaction of carbon with oxygen results in
the formation of – .
• The carbon atom exist in the following three
isotopes: % , % and % where 99% of
carbon is % and 1% is % and % .
• Among these isotope % and % are stable
while % is unstable and thus undergoes
radioactivity. The half life of % is about
5700 years. This element is use the determine
the age of any object (e.g. plants, animals or
rocks) containing % , in a process called
carbon dating.
• This process uses the fact that the activity of
in the atmosphere, plants and animals,
%
while they are still living, is constant at
about 15.3 decays per min.
Carbon dating
When the plant or animal dies, the take-in of
exponentially as shown in the graph below.
%
stops and the activity A decrease with time
By measuring the residual activity of plant or animal remains, the time since death can be determined
from
=
Š = −„
v
uڥ
nu ^S
˜™( ⁄ r )
˜™( )
=
r
^S
Š=„
log F
r
Carbon dating
Example 3.5: Carbon dating
A sample from an excavated wooden utensil gives 10 counts per minute per gram of carbon present. The
corresponding count for present-day living organic matter is 15.3. Estimate how long ago the utensil was
made. (Take the half-life of carbon-14 as 5730 years.)
Radioactive series
Often one radioactive substance (the ‘parent’) decays into another (the ‘daughter product’), which in turn decays into
another. This constitutes a radioactive series.
Most radioactive elements fall into one of four series, in each of which there is decay (at various rates) from one end to
the other.
Age of the earth about 4.5×109 years
Detection of radiation - Gieger counter
Gieger counter consists of a cylindrical tube containing a gas, a wire in the
middle of the tube, voltage source (100-200V) (with negative connected to the
cylindrical tube and positive to the wire), and a counter.
•
•
•
When a charged particle enter through a thin wall it ionizes few atoms of the gas.
Ejected electrons are accelerated toward the wire in the middle. On their way they
strike and ionize more atoms. A huge no. of electrons flow toward the anode and
produce a current pulse.
The voltage pulse across the R is amplified and sent to an electronic counter or a
loud speaker. So each detection of the particle is heard as a click.
After each count the Geiger counter resets itself in100 to 200 ms.
It is normally used for β-particles (or for α-particles if provided with a special, very thin, end-window which the
α’s can penetrate). Its efficiency for γ-rays is very low.
Radiation measurement and biological effects
Exposure of your body to radiation may have negative health effects. The effects of radiation depend on the
amount/dose of radiation your body is exposed to and the effectiveness of the those radiations. The amount and the
effectiveness of radiation can be quantified by (1) Absorbed dose (D) and Dose equivalent (H).
(1) The absorbed dose is defined as the amount of energy per unit mass imparted to the body and is given by
š=
, [ S ]Z a`SŠ \
J, `__
The unit if D is the gray (Gy) such that 1 Gy = 1 J/kg. This applies to ionizing radiation of any kind.
(2) Dose equivalent, H, takes into account the fact that the biological damage caused by a given absorbed dose is
different for different types of radiation. For example, 1 Gy of α-radiation is more damaging than 1Gy of γ-radiation.
Thus the dose equivalent is give by the product of the absorbed dose › and the quality factor œ (no units),
mathematically
€ = Rš.
Unit of H is the sievert.
The quality factor is sometimes called the relative biological
effectiveness R.B.E. It is actually a measure of how ionising
the radiation is.
Examples
Example 3.6: Energy due to X-ray radiation
A patient receives an absorbed dose of 1.4 mGy in 0.20 kg of tissue from a dental X-ray machine. What is the
total energy deposited in the tissue?
Example 3.7: Dose equivalent and quality factor
During a period of one week, a laboratory worker receives the following absorbed doses:
γ-rays
protons
slow neutrons
0.15 mGy
0.02 mGy
0.05 mGy
The total dose equivalent is 0.50 mSv. If the quality factor for the γ-radiation and the protons
is 1 and 10 respectively, determine Q for the slow neutrons.
Example 3.8: Absorbed dose
V
^ emits 122 keV |-rays. If a person weighing 60 kg swallows 2.0 µCi of V ^, what will be the absorbed
dose rate, in gray per day, averaged over the whole body? Assume that 50% of the |-ray energy is deposited
in the body. (1 eV ≡ 1.6 × 10−19 J)
Tutorials
17. For diagnostic purposes, radiopharmaceuticals are usually given in millicurie amounts. What is the
mass of 50mCi of 11„! ? The half-life of the technetium isotope is 6 hours.
18. A film badge worn by a radiologist indicates that she has received an absorbed dose of 2.5 × 10-3 Gy.
The mass of the radiologist is 65 kg. How much energy has she absorbed?
19. An absorbed dose of 100 Gy completely destroys living tissues. By how much will this absorbed
dose raise the temperature of such tissue if none of the heat is lost? Take the specific heat capacity of the
issue as 3 × 103J kg-1 K-1.
20. Treating fish (or meat) with 2 kGy kills many of the bacteria present and increases the refrigerated
shelf life five to seven times.
(a) How much energy is absorbed per kilogram of fish?
(b) Neglecting heat losses, estimate the temperature rise. The specific heat capacity of fresh fish is 3.43
kJ kg-1 K-1.
21. What absorbed dose (in grays) of α-particles (Q = 20) causes as much biological damage
as a 60 gray dose of protons (Q = 10)?
Tutorials
22. If someone stands near a radioactive source and receives doses of the following types of radiation: γrays (0.20 mGy, Q = 1), electrons (0.3 mGy, Q = 1), protons (0.04 mGy, Q = 10), and slow neutrons
(0.05 mGy, Q = 2), what is the total biologically equivalent dose (mSv) received?
23. A stewardess flies at an average height of 10 700 m for 20 hours per week. If she receives 7 × 10−3 m
Sv per hour, what is her annual biologically equivalent dose? Compare this with the maximum
permissible dose (above background) for the general population in South Africa of 1mSv/year.
24. In a certain town the average yearly absorbed dose of background radiation consists of 0.25 mGy of
X- and γ-rays plus 0.03 mGy of particles having a quality factor of 10. What is the dose equivalent a
person receives per year on average? Express your answer in sievert. (5.5 × 10−4 Sv)
25. A certain source produces an absorbed dose of 40 Gy per hour in tissue. If the quality factor of the
radiation is 2.0
(a) How much time is required for an absorbed dose of 4 Gy?
(b) How much time is required for a biologically equivalent dose of 4 Sv?