Question

A photon of wavelength $\lambda$ is scattered from an electron, which was at rest. The wavelength shift $\Delta \lambda$ is three times of $\lambda$ and the angle of scattering $\theta$ is ${60}^{\circ }$. The angle at which the electron recoiled is $\varphi$. The value of tan $\varphi$ is : (electron speed is much smaller than the speed of light)

• A. $0.28$
• B. $0.22$
• C. $0.25$
• D. $0.16$

Answer 1

The correct option is C $0.25$

Initial momentum of photon$=\frac{h}{{\lambda }_i}$
scattered momentum of photon$=\frac{h}{{\lambda }_f}$
Initial momentum of electron$=0$
Final momentum of electron$=p$
Momentum conservation for horizontal component, $\frac{h}{{\lambda }_i}+0=\frac{h}{{\lambda }_f}\cos \theta +p\cos \varphi ....\left(1\right)$
for vertical component,$0=\frac{h}{{\lambda }_f}\sin \theta -p\sin \varphi ....\left(2\right)$
Given, $\theta ={60}^0,{\lambda }_f-{\lambda }_i=3{\lambda }_i,{\lambda }_f=4{\lambda }_i$
so from (1) and (2), $p\cos \varphi =\frac{7h}{8{\lambda }_i},p\sin \varphi =\frac{\sqrt{3}h}{8{\lambda }_i}$
$\tan \varphi =\frac{\sqrt{3}}{7}=0.25$.