Question

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

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

Answer 1

Answer: (B)

By using $\frac{hc}{e}\left(\frac{1}{\lambda }-\frac{1}{{\left(\lambda \right)}_0}\right)=V_0$
$\Rightarrow \frac{hc}{e}\left(\frac{1}{\lambda }-\frac{1}{{\left(\lambda \right)}_0}\right)=4.8$ $...(i)$
and $\frac{hc}{e}\left(\frac{1}{2\lambda }-\frac{1}{{\left(\lambda \right)}_0}\right)=1.6$ $...(ii)$
Fromequation $(i)$ by $(ii)$ , $\frac{\left(\frac{1}{\lambda }-\frac{1}{{\left(\lambda \right)}_0}\right)}{\left(\frac{1}{2\lambda }-\frac{1}{{\left(\lambda \right)}_0}\right)}=\frac{4.8}{1.6}\Rightarrow {\left(\lambda \right)}_0=4\lambda$ .