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_______
$(9)$
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, $n=9$
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