Question

When a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2\lambda$, the stopping potential is $\frac{V}{4}$. The threshold wavelength for metallic surface is times $\lambda$.

Answer 1

Answer: 3

Using Einstein's photoelectric equation
Case I:
$eV=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$
$=hc\left[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right]$....(i)
Case II:
$e\frac{V}{4}=\frac{hc}{2\lambda }-\frac{hc}{{\lambda }_0}$
$\therefore eV=\frac{4hc}{2\lambda }-\frac{4hc}{{\lambda }_0}$
$=4hc\left[\frac{1}{2\lambda }-\frac{1}{{\lambda }_0}\right]$ ..... (ii)
Equating (i) and (ii)
$hc\left[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right]=4hc\left[\frac{1}{2\lambda }-\frac{1}{{\lambda }_0}\right]$
$\frac{1}{\lambda }-\frac{1}{{\lambda }_0}=\frac{4}{2\lambda }-\frac{4}{{\lambda }_0}$
$\frac{4}{{\lambda }_0}-\frac{1}{{\lambda }_0}=\frac{2}{\lambda }-\frac{1}{\lambda }$
$\frac{3}{{\lambda }_0}=\frac{1}{\lambda }$
$\therefore {\lambda }_0=3\lambda$