Radioactivity – inverse square law, absorption, and rates

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.

Inverse square law

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.

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

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 o I(1) = 1 I(2) = 1/4 I(3) = 1/9 o o r

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

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

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

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

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.

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 I = Io at detector, with no absorber

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

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.

Graph of the exponential exp(x) + x

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

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

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

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://physics.nist.gov/PhysRefData/FFast/Text/cover.html which produced this graph for lead (Pb):

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):

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

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 minimize I = 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.

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.

Range of alpha and beta particles The range of alpha particles is a few centimeters in air and much less in solids. Alphas may be completely absorbed by a single sheet of paper or by your skin 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. However, they can easily pass through skin or gloves.