Question
A photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda\left(\lambda<\frac{\mathrm{hc}}{\phi_0}\right)$. The fastest photoelectron has a de Broglie wavelength $\lambda_d$. A change in wavelength of the incident light by $\Delta \lambda$ results in change $\Delta \lambda_{\mathrm{d}}$ in $\lambda_{\mathrm{d}}$. then the ratio $\Delta \lambda_{\mathrm{d}} / \Delta \lambda$ is proportional to

Answer

$\frac{\mathrm{P}^2}{2 \mathrm{~m}}=\frac{\mathrm{hc}}{\lambda}-\phi_0$

$\frac{\mathrm{h}^2}{2 \mathrm{~m} \cdot \lambda_{\mathrm{D}}^2}=\frac{\mathrm{hc}}{\lambda}-\phi_0$

$-\frac{\mathrm{h}^2}{\mathrm{~m}} \frac{1}{\lambda_{\mathrm{D}}^3} \mathrm{~d} \lambda_{\mathrm{D}}=\frac{\mathrm{hc}}{\lambda^2} \mathrm{~d} \lambda$

$\frac{\Delta \lambda_{\mathrm{D}}}{\Delta \lambda} \propto \frac{\lambda_{\mathrm{D}}^3}{\lambda^2}.$

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