Question

Determine the most probable interaction for the following: a. 30 kev photon on calcium b. 100 kev photon on iodine c. 300 kev photon on water d. 10 Mev photon on lead

Answer 1

There are basically three types of photon interaction :

• Photoelectric Effect (which is proportional to Z⁴ / E³) - This interaction holds predominance for relatively low photon energies E and relatively high atomic number Z. It results in a scattered photon and electron.
• Compton Scattering (which is proportional to Z) - This interaction holds predominance for gamma-ray energies, e.g. in radioistopes. It results in an electron.
• Pair Production (which is proportional to Z²) - This interaction holds predominance for photon energies higher than 1.022 MeV. It results in a positron and an electron.

a) Photon Energy = E_P = 30 kev

Atomic number of Calcium = Z_C = 20

Assuming the most probable interaction to be Photoelectric Effect, the interaction is,

I_C = Z_C⁴ / E_P³ = (20⁴ / 30³) = 5.92 (approx)

b) Photon Energy = E_P = 100 kev

Atomic number of Iodine = Z_I = 53

Considering the photon is striking the iodine crystal, we assume Compton Scattering to be the most probable interaction, the dependency of which is Z_I, i.e., 53.

c) Photon Energy = E_P = 300 kev

Atomic number of Water = Z_W = 2 * (Atomic Number of Hydrogen) + 1 * (Atomic Number of Oxygen)

Or, Z_W = (2 * 1)+ (1* 8) = 10

Since the atomic number of water is relatively low, we can assume the particles of matter are loosely bound. Therefore again, Compton Scattering is assumed to be the most probable interaction the probability of which is Z_W = 10.

d) Photon Energy = E_P = 10 MeV

Atomic number of Lead = Z_L= 82

Since the photon energy is very high (more than 1.022 MeV), we infer that Pair Production is the most probable interaction in this case. The variation is given by Z_L² = 82² = 6724.