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(B) Initial momentum of the photon is $P_i = \frac{h}{\lambda}$.Final wavelength $\lambda' = \lambda + \Delta \lambda = \lambda + 3\lambda = 4\lambda$

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(B) Initial momentum of the photon is $P_i = \frac{h}{\lambda}$.Final wavelength $\lambda' = \lambda + \Delta \lambda = \lambda + 3\lambda = 4\lambda$.Final momentum of the photon is $P_f = \frac{h}{4\lambda} = \frac{P_i}{4}$. Let $P = P_f = \frac{h}{4\lambda}$,then $P_i = 4P$.Using conservation of momentum: $\vec{P}_i = \vec{P}_f + \vec{P}_e$,where $\vec{P}_e$ is the momentum of the recoiled electron.$\vec{P}_e = \vec{P}_i - \vec{P}_f$.Taking the direction of incident photon as the $x$-axis:$\vec{P}_i = 4P \hat{i}$.$\vec{P}_f = P \cos 60^o \hat{i} + P \sin 60^o \hat{j} = P(\frac{1}{2}) \hat{i} + P(\frac{\sqrt{3}}{2}) \hat{j}$.$\vec{P}_e = (4P - \frac{P}{2}) \hat{i} - \frac{\sqrt{3}P}{2} \hat{j} = \frac{7P}{2} \hat{i} - \frac{\sqrt{3}P}{2} \hat{j}$.The angle $\phi$ of the recoiled electron is given by $	an \phi = |\frac{P_{ey}}{P_{ex}}| = \frac{\sqrt{3}P/2}{7P/2} = \frac{\sqrt{3}}{7}$.$	an \phi = \frac{1.732}{7} \approx 0.247 \approx 0.25$.

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