1. Radioactivity – review of laboratory results For presentation on May 2, 2008 by Dr. Brian Davies, 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

6. Inverse square law

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

8. Log-log plot for inverse square law. We can plot I vs. r on a log-log graph. We expect that I = S/4  r 2 log ( I ) = log(S/4  r 2 ) = 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.

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 = I o exp(-  x) = e -  x where I o 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.

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

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 (x ½ ) = I o exp(-  x ½ ) = ½ I o we solve to find the half-thickness x ½. exp(-  x ½ ) = ½ and  x ½ = 0.693 so x ½ = 0.693 / 

15. 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

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