Question

When gamma rays are incident on matter, the intensity of the gamma rays passing through the material varies with depth x as I(x) = Ioe^(-ux) , where Io is the intensity of the radiation at the surface of the material (at x = 0) and u is the linear absorption coefficient. For 0.400 MeV gamma rays in lead, the linear absorption coefficient is 1.59cm^-1 .
(a) Determine the "half-thickness" for lead, that is, the thickness of lead that would absorb half the incident gamma rays.
(b) What thickness reduces the radiation by a factor of 10^4?

Answer 1

a) According to the definition of intensity, we have the formula for the intensity of radiation as

$I = \frac{E}{At}$

From this formula, intensity I is directly proportional to the energy E of the radiation.

When gamma rays are incident on matter, the intensity of the gamma rays passing through the material varies with depth x as

$I(x) = I_0e^{-μx}$

Where $I_0$ is the intensity of the radiation at the surface of the material, and μ is the absorption coefficient for lead.

$I_0 = 0.400 \;MeV \\ I = \frac{I_0}{2} \\ I(x) = I_0e^{-μx} \\ \frac{I_0}{2} = I_0e^{-μx} \\ \frac{1}{2} = e^{-μx} \\ = e^{-(1.59 \;cm^{-1})x} \\ -(1.59 \;cm^{-1})x = ln(\frac{1}{2}) \\ = -0.69315 \\ x = \frac{-0.69315}{-1.59 \;cm^{-1}} \\ = 0.44 \;cm$

b) It is given that

$I = \frac{I_0}{10^4} \\ = 10^{-4}I_0 \\ I(x) = I_0e^{-μx} \\ 10^{-4}I_0 = e^{-(1.59 \;cm^{-1})x} \\ -(1.59 \;cm^{-1} x) = ln(10^{-4}) \\ = -9.21034 \\ x =\frac{-9.21034}{-1.59 \;cm^{-1}} \\ = 5.793 \;cm$