Question

When radiation of wavelength $\lambda$ is incident on a metallic surface, the stopping potential is 4.8 V. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 V. Then the threshold wavelength for the surface is :

• A. 2$\lambda$
• B. 4$\lambda$
• C. 6$\lambda$
• D. 8$\lambda$

Answer 1

The correct option is B 4$\lambda$

$4.8\times e=\frac{hc}{\lambda }-\varphi ---------1)$

$1.6\times e=\frac{hc}{2\lambda }-\varphi ------2)$

From 1 and 2
so, $3(\frac{hc}{2\lambda }-\varphi )=\frac{hc}{\lambda }-\varphi$

$\frac{3hc}{2\lambda }-3\varphi =\frac{hc}{\lambda }-\varphi$

$\frac{hc}{2\lambda }=2\varphi$

$\frac{hc}{{\lambda }_T}=\frac{hc}{4\lambda }=\varphi$

So, threshold wavelength is $4\lambda .$

Hence, the answer is (B).