JEE Main 2014 · 11 Apr · Shift 1 · Physics Q26

← JEE Main 2014 · 11 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 26 of 90

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°. 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)

0.16

0.22

0.25

0.28

Answer and worked solution

Official answer: C

The scattered photon has $\lambda'=4\lambda$ . Take the incident photon along $x$ , momentum $\frac h\lambda$ . The scattered photon has $\frac h{4\lambda}$ at $60^\circ$ . The electron takes the difference:
$p_x=\frac h\lambda\left(1-\frac18\right)=\frac{7h}{8\lambda},\qquad p_y=-\frac h{4\lambda}\cdot\frac{\sqrt3}2=-\frac{\sqrt3h}{8\lambda}.$
$\tan\phi=\dfrac{\sqrt3}7\approx0.25$ .

Topics:Photon momentum and radiation pressure Conservation of linear momentum

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 26