Radioactivity – inverse square law, absorption, and rates

1. Radioactivity – inverse square law, absorption, and rates a presentation for Physics 125 related to Lab # 10

2. Inverse square law for radiation • Particles and photons emitted by radioactive nuclei continue moving until they are absorbed by some material. • The particles (or photons) that leave the sample will continue to move out in a radial direction. • The number that pass through an imaginary sphere of radius r in some interval of time is the same for different radii, but the area of the sphere depends on the radius. A = 4pr2 • Therefore, the number per unit area per unit time depends on the radius.

3. Inverse square law

4. Relation between source activity and intensity at various radii • Suppose that S is the number of particles emitted by the source per unit time. • This S might be due to an activity which is written as S decays per second. (in Bq) • At a distance r, the particles pass through a sphere of radius r and area A = 4pr2 • Intensity is I = S/A, the number of particles, per unit time, per unit area. • Then I = S/4pr2 • I decreases as the inverse square of the radius.

5. Intensity for r, 2r, 3r, etc. • If the intensity is I1 at a radius r1, then if we double the radius to 2r1, the intensity will be ¼ as much, because the area is now 2x2 = 4 times as much. • More generally, if we compare the intensity at radii r1 and r2, then we get a ratio of intensities that depends on the square of the radii: • I2 / I1 = (r1/r2)2 • For example, if we compare radii of r and 3r, the intensities have a ratio of 1/9 = (1/3)2

6. Linear plot for inverse square law. We can plot I vs. r on a linear graph, but this is not always useful if the range is too large. I = 1/r2 I(1) = 1 o I(2) = 1/4 I(3) = 1/9 o o r

7. Linear plot for 10000.r -2 over range 1 to 100. not very useful !

8. Log-log plot for inverse square law. • If the intensity is a function of radius that is a power law rm (for example, inverse square is a power law, since I = a.rm where m = -2), • then, we can plot I vs. r on a log-log graph. • Applying the logarithm to both sides of the equation: and using log(ab) = log(a) + log(b) • log (I) = log(a.rm) = log(a) + m . log(r) • and if y = log(I), x = log(r), and log(a) = b, we have the eq. of a straight line y = m . x + b

9. log-log plot of I = 10000 r -2 y = log10(I) I 4 3 2 1 0 r x = log10(r) 0 1 2

10. Analysis of I = 10000 r -2 How does this equation produce a straight line on the log-log plot? log(I) = log(10000.r-2) = log(10000) + log(r-2) = log(104) + (-2).log(r) Now define y = log(I) and x = log(r) and then y = m.x + b with b = log(104) = 4 (the intercept) m = -2 is the slope

11. We study this in laboratory # 10. We use laboratory equipment to study the distance dependence of the radiation from a small source. This is examined experimentally using log-log graph paper. We also examine the use of shielding materials, which requires semi-log paper to plot the absorption of gamma rays.

12. 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).

13. Exponential absorption of X-rays I = Io at detector, with no absorber

14. Exponential absorption of X-rays With absorber in place, I = Io exp(- m x)

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

16. Graph of the exponential exp(x) exp(x) exp(0) = 1 + x

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

18. 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

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

20. Graphs of linear attenuation coefficient m The linear attenuation coefficient m can be obtained from tables, or from automated databases such as the NIST database: http://www.nist.gov/pml/data/ffast/index.cfm which produced this graph for lead (Pb):

21. Tables of linear attenuation coefficient m Data for Z = 82, E = 2 - 433 keV E (keV )µ Total (cm-1) 2.0004844E+00 1.3412E+04 2.0104868E+00 1.3272E+04 2.0205393E+00 1.3133E+04 2.0306420E+00 1.2996E+04 … 4.479101E+01 1.1677E+02 4.788159E+01 9.8248E+01 5.118542E+01 8.2776E+01 5.471721E+01 6.9836E+01 5.849270E+01 5.8933E+01 … 3.544049E+02 3.1633E+00 3.788588E+02 2.7826E+00 4.050001E+02 2.4588E+00 4.329451E+02 2.1827E+00 The NIST database produces this table of m for lead (Pb):

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

23. Shielding of X-rays and gamma rays • To reduce the intensity of radiation from a source, we can use an absorber in the path of the radiation. This is called shielding. • To minimizeI = Io exp(- m x) we want to increase m or x. Then the exponential will be smaller, and I will be smaller for constant Io . • To increase the absorption coefficient m we need to increase the density of the shielding. • To increase the value of x we must use thicker shielding.

24. Shielding of charged particles (alpha and beta particles) • The absorption of charged particles is quite different from the absorption of X-rays (or g). • Charged particles lose kinetic energy continuously, instead of being absorbed in one single event like photons, and they also can scatter (change direction). • The result is a range, a distance that only a small number of particles reach. • Beyond the range, there is zero intensity.

25. Range of alpha and beta particles • The range of alpha particles is a few centimeters in air and much less in solids. However, an alpha particle can cause thousands of ionization events as it slows down. • Alphas may be completely absorbed by a single sheet of paper or by your skin. Ingestion or inhalation of alpha emitters is the main danger with these sources. • Beta particles can travel a few meters in air or a few millimeters in organic materials, depending on their kinetic energy. One cm of polymer will usually stop beta particles. To store beta-emitting liquid solutions, we can use plastic bottles with thick walls. However, betas can easily pass through skin or gloves. Some lab accidents have occurred where a person got liquid on a glove and did not wash it off or take the glove off immediately.