Question:

When radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is $V$. When the same surface is illuminated with radiation of wavelength

$3 \lambda$, the stopping potential is $\frac{V}{4} .$ If the theshold wavelength for the metallic surface is $n \lambda$ then value of $n$ will be_______

Solution:

$(9)$

When radiation of wavelength $A, \lambda_{A}$ is used to illuminate, stopping potential $V_{A}=V$

$\frac{h c}{\lambda}=\phi+e V$              ...(1)

When radiation of wavelength $B, \lambda_{B}$ is used to illuminate,

stopping potential, $V_{B}=\frac{V}{4}$

$\frac{h c}{3 \lambda}=\phi+\frac{e V}{4}$    ...(2)

From eq. (i) - (ii),

$\frac{h c}{\lambda}\left(1-\frac{1}{3}\right)=\frac{3}{4} e V$

$\Rightarrow \frac{h c}{\lambda} \frac{2}{3}=\frac{3}{4} \mathrm{eV} \Rightarrow \mathrm{eV}=\frac{8}{9} \frac{h c}{\lambda}$

$\frac{h c}{\lambda}=\phi+\frac{8}{9} \frac{h c}{\lambda}$

$\therefore \phi=\frac{h c}{9 \lambda}=\frac{h c}{n \lambda}$, so, $n=9$