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A photon of wavelength $λ = \frac{4 h}{3 m c}$ encounters a stationary electron and is scattered directly backwards. What is the final momentum of the electron after the collision? Here $m$ is the mass of electron and $c$ is speed of light in vacuum

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$m c$

$\frac{2 m c}{3}$

$\frac{m c}{2}$

$\frac{3 m c}{4}$

answer is A.

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Detailed Solution

$\frac{h}{λ} = - \frac{h}{λ'} + P_____(1)$

$\frac{h C}{λ} = - \frac{h C}{λ'} + \frac{{(P)}^{2}}{2 m}_____(2)$

$(1) \times C + (2) \Rightarrow \frac{2 h C}{λ} = P C + \frac{P^{2}}{2 m}$ from given data, $\frac{h}{λ} = \frac{3 m C}{4}$ $∴ \frac{3 m C^{2}}{2} = P C + \frac{P^{2}}{2 m}$

On solving this quadratic equation, we get $P = m C$

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