Let K be the kinetic energy that a stationary free electron gains when a photon scatters from it. We can plot K versus the angle $\mathit{\varphi }$at which the photon scatters; see curve 1 in Fig. 38-21. If we switch the target to be a stationary free proton, does the end point of the graph shift (a) upward as suggested by curve 2, (b) downward as suggested by curve 3, or (c) remain the same?

The change in wavelength in Compton Effect is given by,

$\mathit{\Delta }\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\mathbf{m}\mathbf{c}}\left(1-cos\varphi \right)$

Here,$\mathit{\Delta }\mathit{\lambda }$ is the Compton shift, h is the Planck’s constant, m is the mass of the electron, c is the speed of light, and$\mathit{\varphi }$ is the scattering angle.

Consider the following equation.

$\Delta \lambda =\frac{h}{mc}\left(1-\cos \varphi \right)$

The mass of the proton is $m_p=1.67\times {10}^{-27}\text{kg}$, and the mass of electron is $m_e=9.11\times {10}^{-31}\text{kg}$.Clearly, the mass of proton is greater than the mass of electron.

$$\begin{gathered} m_p>m_e \\ {m}_e<m_p \\ \frac{m_e}{m_p}<1 \end{gathered}$$

From the Compton equation, the shift $\Delta \lambda$is inversely proportional to the mass of the particle.

$$\begin{gathered} \Delta \lambda \propto \frac{1}{m} \\ \frac{\Delta {\lambda }_p}{\Delta {\lambda }_e}=\frac{m_e}{m_p} \\ \frac{\Delta {\lambda }_p}{\Delta {\lambda }_e}<1 \\ \Delta {\lambda }_p<\Delta {\lambda }_e \end{gathered}$$

The kinetic energy of the recoiled particle is proportional to the Compton shift.

$$\begin{gathered} K\propto \Delta \lambda \\ \frac{K_p}{K_e}=\frac{\Delta {\lambda }_p}{\Delta {\lambda }_e} \\ \frac{K_p}{K_e}<1 \\ {K}_p<K_e \end{gathered}$$

Since the recoiled kinetic energy of the proton is less than the recoiled kinetic energy of electron. So, the end point of the kinetic energy versus scattering angle of the graph shifts downwards as suggested by curve (3).

Therefore, the option (b) is correct.