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

Answer 1

According to Einstein's photoelectric equation, the stopping potential $\left(V_0\right)$, incident wavelength $\left(\lambda \right)$ and the threshold wavelength $\left({\lambda }_0\right)$ is related as,

$e{V}_0=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$

For first case,

$eV=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$

$\Rightarrow eV=hc\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)---\left(1\right)$

For second case,

$\frac{eV}{4}=hc(\frac{1}{3\lambda }-\frac{1}{{\lambda }_0})$

$\Rightarrow \frac{eV}{4}=hc\left(\frac{{\lambda }_0-3\lambda }{3\lambda {\lambda }_0}\right)---\left(2\right)$

Dividing $\left(1\right)\text{by}\left(2\right)$

$\Rightarrow 4=\frac{3({\lambda }_0-\lambda )}{{\lambda }_0-3\lambda }$

$\Rightarrow {\lambda }_0=9\lambda$

Hence, $9$ is the correct answer.