Question

55 keV x-ray photons are incident on a target. At what scattering angle do the scattered photons have an energy of 50 keV?

Short answer

Scattering angle is equal to 86^o.

Step-by-step solution

Step 1. Given information is :Energy of x-ray photons = 55 keV 

We need to find out the scattering angle for which the energy is 50 keV.

Step 2. Using the expression of Compton scattering.

$$\begin{gathered} ∆\lambda ={\lambda }_s-{\lambda }_i=\frac{h}{mc}(1-\cos \theta ) \\ E=\frac{hc}{\lambda } \\ \lambda =\frac{hc}{E} \\ \frac{hc}{E_s}-\frac{hc}{E_i}=\frac{h}{mc}(1-\cos \theta ) \\ (1-\cos \theta )=m{c}^2\left(\frac{1}{E_s}-\frac{1}{E_i}\right) \\ \cos \theta =1-m{c}^2\left(\frac{1}{E_s}-\frac{1}{E_i}\right) \\ \theta =arc\cos \left(1-m{c}^2\left(\frac{1}{E_s}-\frac{1}{E_i}\right)\right) \\ \theta =arc\cos \left(1-(9.11\times {10}^{-31}kg){(3\times {10}^{8}m/s)}^2\left(\frac{1}{50\times 103\times 1.6\times 10-19V}-\frac{1}{55\times 103\times 1.6\times 10-19V}\right)\right) \\ \theta =86° \end{gathered}$$