Radioactivity – review of laboratory results

1. Radioactivity – review of laboratory results A discussion of results for Lab # 10 in the Physics 125 course WIU Physics Dept.

2. 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.

3. The applied voltage on the geiger tube collects charge due to ionizing radiation

4. Data for a Geiger tube, obtained 7/1/04

5. Data for a Geiger tube, obtained 7/1/04

6. Inverse square law

7. Linear plot for inverse square law. I = 1/r2 o I(1) = 1 I(2) = 1/4 o I(3) = 1/9 o r

8. Log-log plot for inverse square law. • We can plot I vs. r on a log-log graph. • We expect that I = S/4pr2 • log (I) = log(S/4pr2) = log(S/4p) - 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.

9. Rate vs. distance, Co-60 source Linear plot of data obtained 7/1/04

10. Rate vs. distance, Co-60 source Log-log plot of data obtained 7/1/04

11. 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).

12. 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(- m x) = e - m x where Io is the intensity without the absorber, and I is the intensity with the absorber, and and m is the linear absorption coefficient. m depends on material density and X-ray energy.

13. Graph of the exponential exp(-x) + exp(-x) exp(0) = 1 exp(-0.693) = 0.5 = ½ + + exp(-1) = 1/e = 0.37 x

14. 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(- m x ½) = ½ Io we solve to find the half-thickness x ½. exp(- m x ½) = ½ and m x ½ = 0.693 so x ½ = 0.693 / m

15. Calculation of half-thickness To calculate x ½ we need to know m. As an example, for X-rays of energy 50 keV, m = 88 cm-1 (for Pb) and x ½ = 0.693/m x ½ = 0.693 / (88 cm-1) = 0.0079 cm But for hard X-rays with energy 433 keV (Co-60), m = 2.2 cm-1 (for Pb) and we find: x ½ = 0.693 / (2.2 cm-1) = 0.31 cm

16. Half-thickness data from ORTEC-online X Gamma rays from Co-60 X X

17. Rate vs. shielding thickness, Co-60 source Linear plot of data obtained 7/1/04 Half-thickness is about 0.6 cm

18. Rate vs. shielding thickness, Co-60 source Semi-log plot of data obtained 7/1/04

19. 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.

20. Rate vs. shielding thickness, Sr-90 source Linear plot of data obtained 7/1/04

21. Rate vs. shielding thickness, Sr-90 source Semi-log plot of data obtained 7/1/04