Lab 13: Nuclear Radiation
Objectives: To investigate beta and gamma radiation from nuclei as a function of distance;
to see that beta radiation is easier to shield than gamma radiation;
to verify that gamma intensity decreases as an exponential with thickness penetrated.
To observe that radiation intensity from a single isotope decreases with time;
to measure the half-life of a short-lived isotope;
to practice calculating the expected activity of old sources, from published half-life.
Apparatus:
beta source, gamma source, Cs-Ba “cow” isotope generator ; card stock , lead sheets
Equipment: Geiger tube with counter and automatic timer
Background: At the center of an atom is its nucleus. The atom’s chemical properties are determined
by its proton number Z, but its nuclear properties also depend on its neutron number N .
Isotopes are distinguished by including their nucleon number A = Z+N as a superscript
at left of the element abbreviation. The protons are held together, despite having huge
positive electric PE (up to 1000[MeV]), because all nucleons attract one another via the
Strong Nuclear Force and via intense Magnetic fields. Nucleons tend to pair up as
magnets, which is why 42He is so stable. For Z ≤ 10, NZ ; but for 10<Z<60, N1⅝Z.
Nuclei with too few neutrons have their protons too close together (high Electric PE);
they become more stable when they lower their PE , by doing one of these processes:
(1) proton becomes neutron when it emits a positron (+ =anti-electron) and a neutrino;
(2) proton becomes neutron when it captures an atomic electron and emits a neutrino;
(3) nucleus emits an entire alpha particle (42He) …{single proton is held magnetically}.
Nuclei with too many neutrons have some at high Energy levels; having fewer protons,
there are empty proton levels at much lower Energy. These nuclei lower their PE by:
(1) neutron becomes proton when it emits a – (electron) and an anti-neutrino.
Much of the PE reduction is carried by the emitted particles as KE; ~1[MeV] is typical.
After a nucleus emits (or absorbs) a charged particle, the remaining nucleons drop to
lowest empty levels by emitting one or more gamma  (photon) … ~ 1[MeV] is typical.
The likelihood of any particular nucleus doing anything depends only on its isotope;
it is isolated from the outside world by the electrons around it, so its timing is random.
The probability that it will decay (per time interval) is called its decay constant “λ” .
The average time they’ll wait before decaying is called their mean lifetime Tmean = 1/λ .
For any isotope, there is a time at which the likelihood to decay becomes one-half;
starting with many of this isotope, by the end of 1 half-life (), ½ of them probably did
decay already. Half of the rest will endure the entire next half-life, and ¼ will make it
thru 2 more, etc; so the average lifetime is longer than a half-life …  =0.693Tmean.
A sample’s decay rate (“activity”) depends on the number of that isotope: ΔN/Δt = N .
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By the end of one half-life, only half as many remain which can do that emission, so
the sample’s decay rate (“activity”) also is reduced by the factor ½.
We detect only radiation that goes into our Geiger tube, whose Area is a fraction of the
4d2 Area surrounding the source; but as we measure our counting rate, we can use
geometry to deduce the sample’s actual decay rate in all directions.
t
t
N t
I
N
 1 

.
 N  N 0    N 0 e  t , so that I  I 0 2  ;
2
tubeArea
4 r
t
2
An alpha is a 42He nucleus, so as a big nucleus emits an  , its Z changes by –2 and its
A changes by –4. Having has mass 4[u], an  is slow (v < c/100) even when first emitted;
as it travels thru matter, its charge 2[e] has time to do Work to atomic electrons there –
they lose all their KE in a very short distance, ~10[m]. We will not be able to detect 
because they can’t get through the “thin end window” on our Geiger-Muller tubes.
A beta is an electron, so as a nucleus emits a – , its Z changes by +1 . With small mass
m1/2000[u] , a – is usually quite fast when first emitted (v>0.9c), so its charge (–e) has
little time to do Work to atomic electrons that it passes by (W=FΔx=F2Δt2/2m), at first.
Once they’re slower, they transfer KE to these atomic electrons quickly, and will shake
off x-rays during their violent accelerations as they stop (Bremsstrahlung = “braking”).
We emit – from 146C that we obtained (mixed with 126C) from recently-living plants;
they got it from air’s CO2 during photosynthesis. It takes 5700 years for half the 146C to
become 147N, so paper is recently-living; most plastic is made from fossil hydrocarbons
(petroleum) so it has no 14C left in it … CO2 from cars is different than CO2 from cows.
A gamma is a high-Energy photon, so a nucleus which emits a  just lowers its PE .
With wavelength ~1[pm], most gamma penetrate farther than a longer-wave x-ray
before being absorbed. Some get absorbed while traversing a material with thickness x.
That fraction does not depend on  intensity; matter can’t be made transparent to .
For a specific gamma Energy (photon ), each material has a thickness in which half
those gammas will be absorbed: its half-thickness x½ . The intensity I decreases
exponential with thickness: I  I 0 e   x  I 0 2  x / x½ , with x½ = 0.693/μ .
A Geiger tube is used with (+)HV on its thin central wire, surrounded by neutral gas.
Nuclear radiation passing thru the gas, knocks some electrons off their gas atoms
(ionizing them); as the electrons rush quickly toward the central wire, their KE grows
high enough to ionize more gas atoms, etc ; the free electrons might double 25 times
before they arrive at the thin central wire and cause the current pulse that gets counted.
(The (+) ions drift slowly outward to the tube wall, where they’re given an electron.)
Geiger tubes operated this way count ionizing radiation efficiently, but the pulse height
(charge) is no longer proportional to the Energy that the radiation delivered in the tube.
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Activity 1. Find your Geiger Tube Plateau Voltage Get a radiation source with moderate activity.
Remove the Geiger-Muller tube from its protective plastic jar, carefully remove the red
screen end and slip it into the tube holder. Connect the tube’s BNC to your controller’s
“GM” HV supply. Turn on the tube’s power supply, and cycle its display to “HV” ;
adjust to about 500[V]. Set the “Time” to count at about 50[s] . Place the source on a
shelf near the top of the holder; reset the counts, and count for the 50[s]. Adjust the
Voltage to about 600[V], and count for the same time. If the count increased more than
about 15% for that 20% voltage increase, then you’re not on a plateau yet, so repeat at
700[V] … but the expected variation in these random event counts is N = N . Get
help from the instructor if you haven’t found a plateau by 900[V]. Set your voltage
about 50[V] above the “knee”, for the rest of the lab. Counting time: ________ ___
voltage
counts N
±N
N/N
V/V
------
-------
Activity 2. Counting Background Radiation Return the source to very far away from your GM tube.
Take 9 background counts, each 50[s] duration, and note each one’s expected variation.
Notice how the higher counts have higher expected variation in count, but have lower
expected variation in % . If none of your 9 are bizarre (as in a blumder), average them;
note that the expected variation is Ntotal , before dividing by 9. Circle the background
runs that are farther from your average than this expected deviation; about 1/3 of the
runs should be “borderline extremely deviant”. Counting time: _____ ___
counts N
±N
counts N
±N
total N↓
±
average background per 50s =>
±
(Try your last “background run” with a hand under the GM tube window…)
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Activity 3. Beta Source
204
81Tl
→
_____
_____[
] +
–1β
(KEmax =763 keV) + ____
A. Variation with Distance Use the vernier calipers’ ears (not jaws) to measure the GM
tube’s inside diameter _______ ____ , and compute its entrance Area : ________ ___.
Obtain a beta source; record the info that is written on it:
Place the beta source in the tray in the bottom slot, far from the GM tube. Measure the
distance from the source to the lower end of the GM tube, and compute the Area of the
entire Gaussian shell at that distance; enter these in the table below. Count for 50[s].
Use the ratio of Areas to compute the total activity [emissions/sec] from this sample.
Repeat for several distances, from bottom slot to top slot.
Slot # Distance
Counts
± counts
Area
Activity
±
Are all your calculated total Activity values about the same?
If not, is there a general trend in these values?
How does your total Activity compare with the nominal value recorded on the holder?
How many half-lives ago was the Activity value written there?
What should its Activity be now?
Are any of your count rates slow enough that we need to subtract the background rate?
B. Variation with Shielding Move the beta source down to a about 3 slots from the top.
Count for 50[s] with zero note-cards (only air) to shield the GM tube from the source.
(data table next page) If your slot has about 30[mm] air between it and the detector, the
effective air thickness is mass density times the geometric thickness 3[mg/cm2].
Ten (10) of these note-cards had mass 14.6[g]; each has Area 7.5[cm]×12.6[cm] ;
compute the (avg) effective thickness of each card-stock layer: ___________ _____ .
Count for 50[s] with 1 note-card layer between the source and detector, then 2, 3, 4 etc.
For each card layer, compute the fraction of impinging betas that got through it.
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slot # _____ distance _______ ____ count duration ______ ____
N cards thickness
Counts
±
0
Fraction
±
----
----
Are any of your count rates slow enough that we should worry about background?
If so, take the beta source back to its lead pig, and do a background count with all the
note-cards still in place.
Are the transmission fractions about the same for each card? Comment:
Activity 4. Ba* Daughter 30[yr] parent:
→ 13756Ba* + –1β (KE<512 keV) + ___
The parent is chemically bonded in a porous ceramic matrix. When it beta decays, the
Barium daughter can be easily “leached” out of the ceramic with weak HCl solution;
all of the long-lived parent isotope is supposed to remain in the ceramic.
_____
____[___]
The “new” proton in the Barium daughter is made at a higher Energy level than the
lowest empty proton level. It takes a while (much longer than ½[ns]) but the proton
eventually lowers its PE and occupies that lowest energy level.
A. Half-Life of Excited Daughter
137
56Ba*
→
_____
____[___]
+  (662 keV)
Here we want to count for 50[s], record data and reset during 10[s], then count again,
so your data runs have uniform 60[s] (=1 [minute]) intervals, for 12 minutes. We also
want a clock-time at the start of data collection, since we’ll take one or two more data
points at the bottom of this table (next page), after completely doing Activity 5.
Because we will end up with few counts, we will eventually subtract the background
counts from each of the 50[s] run counts … but we’ll first check that our HCl did not
increase the background very much by washing out some of the Cesium parent.
When you’re ready, with clear lab bench, obtain a couple drops of acid-washed barium.
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slot # ___ run1 start time ____: _____: ____
run #
raw counts
counts - bkgd
run7 start time ____: _____: ____
run #
Count duration ____ ____
Estimate the half-life for this Ba* excited state?
raw counts
counts - bkgd
run12 end time ____: _____: ____
run13 start time ____ : _____: ____
Place Ba sol’n safely out of the way for 1[ks]. 13 counts ______ 14 counts ______
If higher than your act.3 empty background: 15 counts ______ => avg Cs bkgd _____
Activity 5. Gamma Penetration Through Lead
Obtain a (hot) gamma source; record the info that is written on it.
Place in slot #3 with one lead sheet above it, to stop the betas.
Use a micrometer to measure each lead sheet’s thickness, before inserting it
count duration ______ ____ Area ratio _______
sheet
Δx [mm]
raw counts
raw-bkgd
count rate
±
Σ(Δx)
1
When you’re done here, return the gamma source to its pig, and take 2 runs with Ba*.
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Lab 13 Summary
Name :_______________________
Lab Partner : _____________
1. Activity 2 showed that random variation was about √N … what does that imply about a sports
team or player who has recently had 2 wins and 4 losses?
2. in Activity 3, what thickness had the greatest fraction of impinging betas stopped?
How far would such betas penetrate a patient’s tissues, before depositing their KE?
3. Graph Activity 4’s counts-background data vs time (in middle of the count interval).
Attach the graph, and identify the half-life and the mean lifetime. List them here:
How many Ba* (total) were in your drop of liquid?
4. For Activity 5, graph the source’s count rate as a function of total lead thickness.
It should look like an exponential decrease with thickness; attach the graph,
and identify the “half-thickness” and the mean penetration thickness. List them here:
Identify the “0” intercept on this graph, which is the count rate of “only gammas” from this
source. Show how to use the Area ratio to infer the total gamma activity from this source.
Use the source’s production date and half-life to calculate its expected activity now.
How does that compare with the inferred total gamma activity just computed above?
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