Question

A photoelectric material having work-function ${\varphi }_0$ is illuminated with light of wavelength $\lambda \left(\lambda <\frac{hc}{{\varphi }_0}\right).$ 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)

$\frac{P^2}{2m}=\frac{hc}{\lambda }-{\varphi }_0$
$\frac{h^2}{2m\cdot {\lambda }_D^2}=\frac{hc}{\lambda }-{\varphi }_0$
$-\frac{h^2}{m}\frac{1}{{\lambda }_D^3}d{\lambda }_D=\frac{hc}{{\lambda }^2}d\lambda$
$\frac{\Delta {\lambda }_D}{\Delta \lambda }\propto \frac{{\lambda }_D^3}{{\lambda }^2}$