Question:
A certain metallic surface is illuminated by monochromatic radiation of wavelength $\lambda$. The stopping potential for photoelectric current for this radiation is $3 \mathrm{~V}_{0}$. If the same surface is illuminated with a radiation of wavelength $2 \lambda$, the stopping potential is $\mathrm{V}_{0}$. The threshold wavelength of this surface for photoelectric effect is_________$\lambda$.
Solution:
$\mathrm{KE}=\frac{\mathrm{hc}}{\lambda}-\phi \mathrm{hc}$
$\mathrm{e}\left(3 \mathrm{~V}_{0}\right)=\frac{\mathrm{hc}}{\lambda_{0}}-\phi$ ..........(i)
$\mathrm{eV}_{0}=\frac{\mathrm{hc}}{2 \lambda_{0}}-\phi$ ..............(ii)
Using (i) & (ii)
$\phi=\frac{h c}{4 \lambda_{0}}=\frac{h c}{\lambda_{t}}$
$\lambda_{t}=4 \lambda_{0}$