A photon of wavelength λ is scattered from an electron, which was at rest. The wavelength shift Δλ is three times of λ and the angle of scattering θ is 60o. The angle at which the electron recoiled is φ. The value of tan φ is: (electron speed is much smaller than the speed of light) (a) 0.16 (b) 0.22 (c) 0.25 (d) 0.28

Answer: (c). h/λh/4λ60°φelectron Δλ = 3λ, so scattered wavelength λ' = 4λ and its momentum is (h)/(4λ). Momentum along x: px = (h)/(λ) - (h)/(4λ)cos 60° = 7/8(h)/(λ) Momentum along y: py = (h)/(4λ)sin 60° = √(3)/8(h)/(λ) tan φ = (py)/(px) = √(3)/7 ≈ 0.25 So the answer is (c) 0.25.

A photon of wavelength λ is scattered from an electron, which was at… | JEE Main 2014 (11 Apr, Shift 1)

Q17JEE Main 2014 · 11 APR · SHIFT 1Difficult

Davisson-Germer, Uncertainty, Compton & X-Rays

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 60o. 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)

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Answer: (c)

$\Delta\lambda = 3\lambda$ , so scattered wavelength $\lambda' = 4\lambda$ and its momentum is $\frac{h}{4\lambda}$ .
Momentum along x: $p_x = \frac{h}{\lambda} - \frac{h}{4\lambda}\cos 60^\circ = \frac{7}{8}\frac{h}{\lambda}$
Momentum along y: $p_y = \frac{h}{4\lambda}\sin 60^\circ = \frac{\sqrt{3}}{8}\frac{h}{\lambda}$
$\tan\phi = \frac{p_y}{p_x} = \frac{\sqrt{3}}{7} \approx 0.25$
So the answer is (c) 0.25.

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