Show that the angles of scatter of the photon and electron in the Compton effect are related by the following formula:

$\mathit{c}\mathit{o}\mathit{t}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{=}\left(1+\frac{h}{mc\lambda }\right)\mathit{t}\mathit{a}\mathit{n}\mathit{\varphi }$

The Compton effect is described as the increase in the X-ray wavelength and also other electromagnetic radiations which are scattered by the electrons. Hence, the Compton effect helps in absorbing the radiant energy in matter.

The equation of the x-component of the scattering formula is expressed as:

$\frac{h}{\hat{\lambda }}-\frac{h}{{\hat{\lambda }}^{\text{'}}}\cos \theta ={\gamma }_{\alpha }{m}_{n}u\cos \varphi$ …(i)

Here,$h$is the Planck’s constant,$\lambda$and${\lambda }^{\text{'}}$are the initial and the final wavelength,$\theta$is the angle subtended by the gamma photon, $y_u$is the gamma photon,$m_e$is the mass of the electron, is$\eta$the speed of the photon and$\varphi$is the angle subtended by the electron.

The equation of the y-component of the scattering formula is expressed as:

localid="1657607359918" $\frac{h}{{\lambda }^{\text{'}}}\sin \theta ={\gamma }_u{m}_{e}u\sin \varphi$ …(ii)

Here, $h$is the Planck’s constant, ${\lambda }^{\text{'}}$is the final wavelength, $\varphi$is the angle subtended by the gamma photon,${\gamma }_u$is the gamma photon,$m_e$is the mass of the electron, $u$is the speed of the photon $\varphi$and is the angle subtended by the electron.

Dividing the equation (ii) by equation (i).

$$\begin{gathered} \frac{{\gamma }_u{m}_{e}u\sin \varphi }{{\gamma }_u{m}_{e}u\cos \varphi }=\frac{\frac{h}{{\lambda }^{\text{'}}}\sin \theta }{\frac{h}{\lambda }-\frac{h}{{\lambda }^{\text{'}}}\cos \theta } \\ \tan \varphi =\frac{\frac{h}{{\lambda }^{\text{'}}}\sin \theta }{\frac{h}{\lambda }-\frac{h}{{\lambda }^{\text{'}}}\cos \theta } \\ \tan \varphi =\frac{\sin \theta }{\frac{{\lambda }^{\text{'}}}{\lambda }-\cos \theta } \end{gathered}$$

…(iii)

The equation of the scattering formula of Compton is expressed as:

localid="1657607417620" $$\begin{gathered} {\lambda }^{\text{'}}-\lambda =\frac{h}{m_{e}c}(1-\cos \theta ) \\ {\lambda }^{\text{'}}=\frac{h}{m_{e}c}(1-\cos \theta )+\lambda \\ \frac{{\lambda }^{\text{'}}}{\lambda }=\frac{\frac{h}{m_{e}c}(1-\cos \theta )+\lambda }{\lambda } \end{gathered}$$ …(iv)

Here,$h$is the Planck’s constant, $\lambda$and${\lambda }^{\text{'}}$are the initial and the final wavelength, $\theta$is the angle subtended by the gamma photon,$m_e$is the mass of the electron and$c$is the velocity of light.

Substitute the value of the equation (iv) in the equation (iii).

localid="1657608169168" $$\begin{gathered} \tan \varphi =\frac{\sin \theta }{\frac{\frac{h}{m_{e}c}(1-\cos \theta )+\lambda }{\lambda }-\cos \theta } \\ =\frac{\sin \theta }{\frac{h}{\lambda m_{e}c}+1-\left(1+\frac{h}{\lambda m_{e}c}\right)\cos \theta } \\ =\frac{\sin \theta }{\left(1+\frac{h}{\lambda m_{e}c}\right)(1-\cos \theta )} \\ \left(1+\frac{h}{\lambda m_{e}c}\right)\tan \varphi =\frac{\sin \theta }{(1-\cos \theta )} \end{gathered}$$

Using the formula of double angle in the above equation.

$$\begin{gathered} \left(1+\frac{h}{\lambda m_{e}c}\right)\tan \varphi =\frac{\sin \theta }{(1-\cos \theta )} \\ =\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{1-\left(1-2{\sin }^2\frac{\theta }{2}\right)} \\ =\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}} \\ =\cot \frac{\theta }{2} \end{gathered}$$

Thus, the formula of the angles of scatter of the photon and electron in the Compton effect is proved.