Question

Show that $\Delta \mathbf{E}/\mathbf{E}$, the fractional loss of energy of a photon during a collision with a particle of mass m, is given by

$\frac{\mathbf{\Delta }\mathbf{E}}{\mathbf{E}}\mathbf{=}\frac{\mathbf{h}{\mathbf{f}}^{\mathbf{\text{'}}}}{\mathbf{m}{\mathbf{c}}^{\mathbf{2}}}\left(1-cos\varphi \right)$
where E is the energy of the incident photon, $\mathit{f}\mathbf{\text{'}}$is the frequency of the scattered photon, and $\mathit{\varphi }$is defined as in Fig. 38-5.

Short answer

The derivation of the required formula that is $\frac{\Delta E}{E}=\frac{h{f}^{\text{'}}}{m{c}^2}\left(1-cos\varphi \right)$.

Step-by-step solution

Determine the Compton scattering

Compton scattering is the process in which photons interact with matter. There are other three competing processes of competing. This involves X-ray scattering.

Fig.38-5

Determine the derivation for the equation

From the Compton scattering of light by electrons.

Solve as:

$$\begin{gathered} \frac{∆E}{E}=\frac{∆\lambda }{\lambda +∆\lambda } \\ \frac{∆E}{E}=\frac{\left({\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$mc$}\right.}\right)\left(1-\cos \varphi \right)}{\lambda \text{'}} \\ \lambda +∆\lambda =\lambda \text{'}=\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$f\text{'}$}\right. \\ \frac{∆E}{E}=\frac{hf\text{'}}{m{c}^2}\left(1-\cos \varphi \right) \end{gathered}$$

So this is the derivation of the required formula that is $\frac{\Delta E}{E}=\frac{h{f}^{\text{'}}}{m{c}^2}\left(1-cos\varphi \right)$.