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 the metallic surface is :

• A. $5\lambda$
• B. $\frac{5}{2}\lambda$
• C. $3\lambda$
• D. $4\lambda$

Answer 1

Answer: (C)

$eV=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$ ....(i)
$eV/4=\frac{hc}{2\lambda }-\frac{hc}{{\lambda }_0}$ .....(ii)
From equation (i) and (ii)
$\Rightarrow 4=\frac{\frac{1}{\lambda }-\frac{1}{{\lambda }_0}}{\frac{1}{2\lambda }-\frac{1}{{\lambda }_0}}$ On solving ${\lambda }_0=3\lambda$