Question

A photon of wavelength $4\mathring{\text{A}}$ strikes an electron at rest and scattered at an angle of ${120}^{\circ }$ to the original direction. The new wavelength $\left(\text{in}\mathring{\text{A}}\right)$ of the photon after collision is :
[Take $h=6.63\times {10}^{-34}\text{Js};m_0=9.1\times {10}^{-31}\text{kg}$]

• A. $0.48$
• B. $2.42$
• C. $7.64$
• D. $4.03$

Answer 1

The correct option is C $7.64$

Given:

$\lambda =4\mathring{\text{A}};\theta ={120}^{\circ }$

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

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

${\lambda }^{\prime }-4\times {10}^{-10}=\frac{6.63\times {10}^{-34}(1-\cos {120}^{\circ })}{9.1\times {10}^{-31}\times 3\times {10}^8}$

${\lambda }^{\prime }-4\times {10}^{-10}=\frac{9.945\times {10}^{-34}}{27.3\times {10}^{-23}}[\because \cos {120}^{\circ }=-\frac{1}{2}]$

${\lambda }^{\prime }-4\times {10}^{-10}=0.364\times {10}^{-11}$

${\lambda }^{\prime }=(4+0.0364)\times {10}^{-10}=4.03\mathring{\text{A}}$

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