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

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{4 hc }{2 \lambda}-\frac{4 hc }{\lambda_{0}} $
$=4 hc \left[\frac{1}{2 \lambda}-\frac{1}{\lambda_{0}}\right]$ ..... (ii)
Equating (i) and (ii)
$hc \left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]=4 hc \left[\frac{1}{2 \lambda}-\frac{1}{\lambda_{0}}\right]$
$ \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} $
$\therefore \lambda_{0}=3 \lambda$