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Question

By using $\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\mathrm{V}_{0}$

$\Rightarrow \frac{\mathrm{hc}}{\mathr

Vedclass · Accepted Answer

$4\,\lambda $

Vedclass · Answer

By using $\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\mathrm{V}_{0}$

$\Rightarrow \frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=4.8$               .........$(i)$

and $\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_{0}}\right)=1.6$              ...........$(ii)$

From equation $(i)$ by $(ii),$ $\frac{\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)}{\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_{0}}\right)}=\frac{4.8}{1.6}$

$\Rightarrow \lambda_{0}=4 \lambda$

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

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