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)

$3 \lambda$

Answer 2

$eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ ....(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 $\lambda_{0} = 3 \lambda$