Question

If $p$ is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $\lambda$, then for $1.5p$ momentum of the photoelectron, the wavelength of the light should be:
(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)

• A. $\frac{1}{2}\lambda$
• B. $\frac{3}{4}\lambda$
• C. $\frac{2}{3}\lambda$
• D. $\frac{4}{9}\lambda$

Answer 1

Answer: (D)

$hv-\varphi =KE$
$\Rightarrow {\left(\frac{hc}{\lambda }\right)}_{\text{incident }}=KE+\varphi$
$\begin{array}{l}{\left(\frac{hc}{\lambda }\right)}_{\text{incident }}\simeq KE\\ KE=\frac{p^2}{2m}=\frac{hc}{{\lambda }_{\text{incident }}}=\frac{hc}{\lambda }\dots (1)\end{array}$
$\Rightarrow \frac{p^2\times (1.5{)}^2}{2m}=\frac{hc}{{\lambda }^{\prime }}$
divide (1) and (2)
$(1.5{)}^2=\frac{\lambda }{{\lambda }^{\prime }}$
$\Rightarrow {\lambda }^{\prime }=\frac{4\lambda }{9}$