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Radioactivity – review of laboratory results
For presentation on May 2, 2008 by
Dr. Brian Davies, WIU Physics Dept.
Plateau voltage for Geiger tube
• We chose the plateau voltage by using a constant source and varying the voltage that was applied to the Geiger tube.
• Threshold – minimum voltage to get counts. • Plateau – approximately constant counts. • Avalanche region – get excess counts and
nonlinear behavior. • Most tubes had a plateau near 750 volts.
Data for a Geiger tube, obtained 7/1/04
2300
2200
2100
2000
1900
1800
Cou
nts
per
min
ute
(cpm
)
900850800750700Geiger tube voltage (V)
Plateau for geiger tube
Data for a Geiger tube, obtained 7/1/04
2500
2000
1500
1000
500
0
Rat
e (c
pm)
900850800750700650600Geiger tube voltage (V)
Plateau for geiger tube
Log-log plot for inverse square law.
• We can plot I vs. r on a log-log graph.
• We expect that I = S/4r2
• log (I) = log(S/4r2) = log(S/4) - 2 log(r)
• We expect the slope to be m = -2
• However, the detector is a volume, and parts of it are at different distance from the source, so we do not get a perfect inverse square, but usually a lower power for m.
Rate vs. distance, Co-60 source Linear plot of data obtained 7/1/04
1000
800
600
400
200
0
Rat
e (c
pm)
30252015105Distance (cm)
Co-60 source at various distances
Rate vs. distance, Co-60 source Log-log plot of data obtained 7/1/04
10
2
4
6
8100
2
4
6
81000
Rat
e (c
pm)
3 4 5 6 7 8 910
2 3
Distance (cm)
Co-60 source at various distances
Absorption of X-rays and gamma rays
• X-rays and gamma rays can be very penetrating. • Scattering of photons is not very important. It is
more probable for the photon to be absorbed by an atom in the photoelectric effect.
• The photon is absorbed with some probability as it passes through a layer of material. This results in an exponential decrease in the intensity of the radiation (in addition to the inverse square law for distance dependence).
Exponential absorption of X-rays
The exponential decrease in the intensity of the radiation due to an absorber of thickness x has this form:
I = Io exp(- x) = e - x
where Io is the intensity without the absorber,
and I is the intensity with the absorber, and
and is the linear absorption coefficient.
depends on material density and X-ray energy.
Graph of the exponential exp(-x)
exp(-x)
x
+ exp(-1) = 1/e = 0.37
exp(-0.693) = 0.5 = ½ +
exp(0) = 1+
Half-thickness for absorption of X-rays
For a particular thickness x ½ the intensity is decreased to ½ of its original magnitude. So if
I(x½) = Io exp(- x ½) = ½ Io
we solve to find the half-thickness x ½.
exp(- x ½) = ½ and x ½ = 0.693
so x ½ = 0.693 /
Calculation of half-thickness
To calculate x ½ we need to know .
As an example, for X-rays of energy 50 keV,
= 88 cm-1 (for Pb) and x ½ = 0.693/
x ½ = 0.693 / (88 cm-1) = 0.0079 cm
But for hard X-rays with energy 433 keV (Co-60),
= 2.2 cm-1 (for Pb) and we find:
x ½ = 0.693 / (2.2 cm-1) = 0.31 cm
Half-thickness data from ORTEC-online. (link)
X
X Gamma rays from Co-60
X
Rate vs. shielding thickness, Co-60 source Linear plot of data obtained 7/1/04
1200
1000
800
600
400
200
0
Rat
e (c
pm)
1.00.80.60.40.20.0Shielding thickness (cm)
Lead shielding, Co-60 source
Half-thickness is about 0.6 cm
Rate vs. shielding thickness, Co-60 source Semi-log plot of data obtained 7/1/04
5
6
7
8
9
1000
Rat
e (c
pm)
1.00.80.60.40.20.0Shielding thickness (cm)
Lead shielding, Co-60 sourceToo high due to beta?
Too high due to scattered gamma?
Range of alpha and beta particles
• The range of alpha particles is a few centimeters in air and much less in solids.
• Beta particles can travel a few meters in air or a few millimeters in organic materials.
• One cm of polymer will usually stop beta particles.
• Our experiment used high-density polyethylene, often denoted as HDPE.
Rate vs. shielding thickness, Sr-90 source Linear plot of data obtained 7/1/04
7000
6000
5000
4000
3000
2000
1000
0
Rat
e (c
pm)
1.00.80.60.40.20.0Shielding thickness (cm)
Polyethylene shielding, Sr-90 source
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