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 \(\phi\). The value of \(\tan\phi\) is : (electron speed is much smaller than the speed of light)

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

Hint:

Use momentum conservation along and perpendicular to the incident photon direction with \(\lambda'=\lambda+\Delta\lambda=4\lambda\).

Answer 1

The Correct Option is C

\(\lambda'=\lambda+3\lambda=4\lambda\).
Along x: \(\frac{h}{\lambda}=\frac{h}{\lambda'}\cos\theta+p\cos\phi\). Along y: \(0=\frac{h}{\lambda'}\sin\theta-p\sin\phi\).
\(\tan\phi=\frac{\frac{1}{\lambda'}\sin\theta}{\frac{1}{\lambda}-\frac{1}{\lambda'}\cos\theta}=\frac{\lambda\sin\theta}{\lambda'-\lambda\cos\theta}\).
\(=\frac{\sin60^\circ}{4-\cos60^\circ}=\frac{0.866}{3.5}=0.247\approx0.25\).
Answer: (3) 0.25