Question

A photon of wavelength $\lambda$ falls on a metal surface of work function $\varphi$. If stopping voltage for ejected photoelectron is $V_0$, then deBroglie wavelength of ejected photoelectron is

• A. $\frac{h\sqrt{\lambda }}{\sqrt{2mhc-2m\varphi \lambda }}$
• B. $\frac{h}{\sqrt{me{V}_0}}$
• C. $\frac{h}{\sqrt{2m(hc-\varphi )}}$
• D. $\lambda$

Answer 1

The correct option is A $\frac{h\sqrt{\lambda }}{\sqrt{2mhc-2m\varphi \lambda }}$

${\lambda }_e=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$,
where $e{V}_0=E=\frac{hc}{\lambda }-\varphi$

So, ${\lambda }_e=\frac{h}{\sqrt{2me{V}_0}}=\frac{h}{\sqrt{2m(\frac{hc}{\lambda }-\varphi )}}$