Question

A photon of frequency n causes photoelectric emission from a surface with threshold frequency v${}_o$. the de Broglie wavelength $\lambda$ of the photoelectron emitted is given as:

• A. $\Delta$n = $\frac{h}{2m\lambda }$
• B. $\Delta$n = $\frac{h}{\lambda }$
• C. [$\frac{1}{v_o}$ - $\frac{1}{v}$] = $\frac{m{c}^2}{h}$
• D. $\lambda$ = $\sqrt{\frac{h}{2m\Delta n}}$

Answer 1

The correct option is D $\lambda$ = $\sqrt{\frac{h}{2m\Delta n}}$

E${}_T=$ IE + KE

or $E_T=$ Threshold E (or) Work function + KE

or ($hv=hv$${}_0$ $+1/2m{u}^2$)

$hv=hv$${}_0$ + $\frac{1}{2}$mu${}^2$

$\frac{1}{2}$mu${}^2$ $=h(v-v$${}_0$) $=h$$\Delta n$ .... (i)

($\lambda$ $=$ $\frac{h}{mu}$, $u=$ $\frac{h}{m\lambda }$)

Substitute the value of u in equation (i)

$\frac{1}{2}m\times$ $\frac{h^2}{m^2{\lambda }^2}$ $=h$$\Delta n$

$\therefore \lambda =$ $\sqrt{\frac{h}{2m\Delta n}}$