91影视

The change in wavelength in Compton Effect is given by,

$\mathit{螖}\mathit{位}\mathbf{=}\frac{\mathbf{h}}{\mathbf{m}\mathbf{c}}\left(1-cos蠒\right)$

Here,$\mathit{螖}\mathit{位}$ is the Compton shift, h is the Planck鈥檚 constant, m is the mass of the electron, c is the speed of light, and$\mathit{蠒}$ is the scattering angle.

Consider the following equation.

$螖位=\frac{h}{mc}\left(1-\cos 蠒\right)$

The mass of the proton is $m_p=1.67脳{10}^{-27}\text{kg}$, and the mass of electron is $m_e=9.11脳{10}^{-31}\text{kg}$.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 $螖位$is inversely proportional to the mass of the particle.

$$\begin{gathered} 螖位鈭�\frac{1}{m} \\ \frac{螖{位}_p}{螖{位}_e}=\frac{m_e}{m_p} \\ \frac{螖{位}_p}{螖{位}_e}<1 \\ 螖{位}_p<螖{位}_e \end{gathered}$$

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

$$\begin{gathered} K鈭�螖位 \\ \frac{K_p}{K_e}=\frac{螖{位}_p}{螖{位}_e} \\ \frac{K_p}{K_e}<1 \\ {K}_p<K_e \end{gathered}$$

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.