Discussion Forum



1)

A photoelectric  material having work-function Φ0 is illuminated with light of wavelength λ  $(\lambda <\frac{hc}{\phi_{0}})$ . The fastest photoelectron has de-Broglie wavelength λd. A change in wavelength of the incident light by Δλ results in a change Δλd in λd

Then, the ratio $\frac{\triangle \lambda_{d}}{\triangle\lambda}$ is proportional to


A) $\frac{\lambda_d^2}{\lambda^{2}}$

B) $\frac{\lambda_d}{\lambda^{}}$

C) $\frac{\lambda_d^3}{\lambda}$

D) $\frac{\lambda_d^3}{\lambda^{2}}$

Answer:

Option D

Explanation:

According to photoelectric effect equation

$KE_{max}=\frac{hc}{\lambda}-\phi_{0}$

$\frac{p^{2}}{2m}=\frac{hc}{\lambda}-\phi_{0}$

                                                            [KE= P2/2m]

$\frac {\left(\frac{h}{\lambda_{d}}\right)^{2}}{2m}=\frac{hc}{\lambda}-\phi_{0}$

Assuming small changes, differentiates in both sides

$\frac{h^{2}}{2m}(-\frac{2d\lambda_{d}}{\lambda_d^3})=-\frac{hc}{\lambda^{2}} d\lambda$

$\frac{d\lambda_{d}}{d\lambda}\alpha \frac{\lambda_d^3}{\lambda^{2}}$