Question

The 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 $△\lambda$ results in change $△{\lambda }_d$ in ${\lambda }_d$. then the ratio $△{\lambda }_d/△\lambda$ 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

The correct option is D ${\lambda }_d^3/{\lambda }^2$

solution:

$\frac{hc}{\lambda }-\Phi o=\frac{1}{2}m{v}^2$

$K.E=\frac{p^2}{2m}$

So, $\frac{hc}{\lambda }-\Phi o=\frac{h^2}{2m{\lambda }_d^2}$

$\frac{hc\Delta \lambda }{{\lambda }^2}=\frac{h^2}{2m{\lambda }_d^3}(-\Delta {\lambda }_d)$

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

hence the correct option : D