Question

A photon of wavelength $\lambda$ incident on a free electron and is scattered in the backward direction. The fractional shift in its wavelength in terms of the Compton wavelength ${\lambda }_c$ of the electron is-

• A. $\frac{{\lambda }_c}{2\lambda }$
• B. $\frac{2{\lambda }_c}{3\lambda }$
• C. $\frac{3{\lambda }_c}{2\lambda }$
• D. $\frac{2{\lambda }_c}{\lambda }$

Answer 1

The correct option is D $\frac{2{\lambda }_c}{\lambda }$

The change in wavelength after collision $\left(\Delta \lambda \right)$ is,

$\Delta \lambda ={\lambda }^{\prime }-\lambda =\frac{h}{m_{0}c}(1-\cos \theta ).......\left(1\right)$

$\text{Here,}{\lambda }_c=\frac{h}{m_{0}c}=\text{Compton wavelength}$

Therefore, $\left(1\right)$ becomes,

$\Delta \lambda ={\lambda }^{\prime }-\lambda ={\lambda }_c(1-\cos \theta )[\because \theta ={180}^{\circ }]$

$\Delta \lambda ={\lambda }_c[1-(-1\left)\right]=2{\lambda }_c$

$\Rightarrow \frac{\Delta \lambda }{\lambda }=\frac{2{\lambda }_C}{\lambda }$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{D}\right)$ is the correct answer.

Why this question ? Key concept : From the Compton effect experiment, The change in wavelength after collision $\left(\Delta \lambda \right)$ is, $\Delta \lambda ={\lambda }^{\prime }-\lambda =\frac{h}{m_{0}c}(1-\cos \theta )$ $\text{Here,}{\lambda }_c=\frac{h}{m_{0}c}=\text{Compton wavelength}$