Question

A photoelectric material having work-function $\phi_{0}$ is illuminated with light of wavelength $\lambda\left(\lambda<\frac{ hc }{\phi_{0}}\right) .$ The fastest photoelectron has a de Broglie wavelength $\lambda_{d}$. A change in wavelength of the incident light by $\Delta \lambda$ results in change $\Delta \lambda_{ d }$ in $\lambda_{ d }$. then the ratio $\Delta \lambda_{ d } / \Delta \lambda$ is proportional to

• A. $\lambda_{ d } / \lambda$
• B. $\lambda_{ d }^{2} / \lambda$
• C. $\lambda_{ d }^{3} / \lambda$
• D. $\lambda_{ d }^{3} / \lambda^{2}$

Answer 1

Answer: (D)

$\frac{ P ^{2}}{2 m }=\frac{ hc }{\lambda}-\phi_{0}$
$\frac{ h ^{2}}{2 m \cdot \lambda_{ D }^{2}}=\frac{ hc }{\lambda}-\phi_{0}$
$-\frac{ h ^{2}}{ m } \frac{1}{\lambda_{ D }^{3}} d \lambda_{ D }=\frac{ hc }{\lambda^{2}} d \lambda$
$\frac{\Delta \lambda_{ D }}{\Delta \lambda} \propto \frac{\lambda_{ D }^{3}}{\lambda^{2}}$