Backscattering of gamma rays due to lead

Question

I am conducting an investigation of backscattering in gamma ray spectroscopy.

A backscattering factor $F_b$ is defined as: $$ F_b = \frac{N_b}{N_i}\times 100\% $$

where
$N_b =$ Number of photons counted with source backing
$N_i =$ Number of photons counted without source backing

I want to investigate how the backscattering peak changes when I place a lead slab underneath my gamma source (the detector is above the source) vs when there is no lead slab underneath. If I were to try this with different gamma ray sources, should I expect to find an energy dependence of $F_b$ for lead?

Answer 1

X-ray scattering is mostly governed by Compton scattering for which the cross-section is given (at lowest order) by the well known Klein-Nishina formula

$$\frac{d\sigma}{d\cos\theta} = \frac{\pi\hbar^2\alpha^2}{m_ec^2}\left(\frac{E^\prime_\gamma}{E_\gamma}\right)^2\left(\frac{E_\gamma}{E^\prime_\gamma}+\frac{E^\prime_\gamma}{E_\gamma}-\sin^2\theta\right) $$

where $E_\gamma$ is the energy of the incoming photon, $E_\gamma^\prime$ is the energy of the outgoing photon and $\theta$ is the scattering angle of the photon. This is just, in some sense, the probability that an incoming photon with energy $E_\gamma$ outgoes at an angle $\theta$ from the scattering with energy $E_\gamma^\prime$. The two energies are correlated by the kinematics as for the following formula

$$E_\gamma^\prime = \frac{E_\gamma}{1+\hbar E_\gamma(1-\cos\theta)/(m_ec^2)} $$

Since the cross-section depends on the energy you should expect different $F_b$ at different photon's energies.