Discussion Forum

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 λ_d. A change in wavelength of the incident light by Δλ results in a change Δλ_d in λ_d

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

Answer:

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

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

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