Question

A photoelectric material having work-function ${\varphi }_0$ is illuminated with light of wavelength $\lambda$ $(\lambda <\frac{hc}{{\varphi }_0})$. The fastest photoelectron has a de Broglie wavelength ${\lambda }_d$. A change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta {\lambda }_d$ in ${\lambda }_d$. Then the ratio $\Delta {\lambda }_d/\Delta \lambda$ is proportional to

• A. $\frac{{\lambda }_d^3}{{\lambda }^2}$
• B. $\frac{{\lambda }_d^2}{{\lambda }^2}$
• C. $\frac{{\lambda }_d}{\lambda }$
• D. $\frac{{\lambda }_d^3}{\lambda }$

Answer 1

The correct option is A

$\frac{{\lambda }_d^3}{{\lambda }^2}$

According to photoelectric equation,
we know that kinetic energy of fastest photoelectron is
$\frac{hc}{\lambda }-{\varphi }_0=\frac{1}{2}m{v}^2=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }_d^2}$

Diffrentiating both sides we get,

$\frac{-hc}{{\lambda }^2}\times \Delta \lambda =\frac{-h^2}{m{\lambda }_d^3}\times \Delta {\lambda }_d$
$\therefore$ we get the proportionality of

$\frac{\Delta {\lambda }_d}{\Delta \lambda }\propto \frac{{\lambda }_d^3}{{\lambda }^2}$