Question

When a metallic surface is illuminate 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. $4\lambda$
• B. $5\lambda$
• C. $\frac{5}{2}\lambda$
• D. $3\lambda$

Answer 1

The correct option is D $3\lambda$

We know that:

$eV=\frac{hc}{\lambda }-\varphi$ ........ (1)

$\frac{eV}{4}=\frac{hc}{2\lambda }-\varphi$ ........ (2)

From equations (1) and (2), we get:

$\Rightarrow 4=\frac{\frac{1}{\lambda }-\frac{1}{{\lambda }_0}}{\frac{1}{2\lambda }-\frac{1}{{\lambda }_0}}$

$\Rightarrow {\lambda }_0=3\lambda$