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

Option: 1

2 \lambda


Option: 2

4 \lambda


Option: 3

6 \lambda


Option: 4

8 \lambda


Answers (1)

best_answer

\begin{aligned} &\text{From equation,}\\ & \mathrm{eV}_0=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \\ & 4.8=\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\\ &\text{and } 1.6=\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)\\ &\text{From eqs. (i) and (ii), we get}\\ &\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}=\lambda_0=4 \lambda \end{aligned}

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