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 $\frac{h c}{e}\left(\frac{1}{\lambda } - \frac{1}{\left(\lambda \right)_{0}}\right)=V_{0}$
$\Rightarrow \frac{h c}{e}\left(\frac{1}{\lambda } - \frac{1}{\left(\lambda \right)_{0}}\right)=4.8$ $...\left(\right.i\left.\right)$
and $\frac{h c}{e}\left(\frac{1}{2 \lambda } - \frac{1}{\left(\lambda \right)_{0}}\right)=1.6$ $...\left(\right.ii\left.\right)$
Fromequation $\left(\right.i\left.\right)$ by $\left(\right.ii\left.\right)$ , $\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 $ .