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When a metallic surface is illuminated with monochromatic light of wavelength \lambda , stopping potential for photoelectric current is 3V_{0} . When the same metallic surface is illuminated with a light of wavelength 2\lambda, the stopping potential is V_{0} . The threshold wavelength for the surface is-

Option: 1

6\lambda


Option: 2

4\lambda


Option: 3

4\lambda /3


Option: 4

8\lambda


Answers (1)

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3 \mathrm{~V}_0=\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\mathrm{hc}}{\mathrm{e} \lambda_0}                            .....................\left ( 1 \right )
V_0=\frac{h c}{2 \mathrm{e} \lambda}-\frac{\mathrm{hc}}{\mathrm{e} \lambda_0}                                .....................\left ( 2 \right )
\begin{aligned} & \frac{\mathrm{eq}^{\mathrm{n}}(1)}{\mathrm{eq}^{\mathrm{n}}(2)} \Rightarrow \frac{3 \mathrm{~V}_0}{\mathrm{~V}_0}=\frac{\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\mathrm{hc}}{\mathrm{e} \lambda_0}}{\frac{\mathrm{hc}}{2 \mathrm{e} \lambda}-\frac{\mathrm{hc}}{\mathrm{e} \lambda_0}} \\ & \frac{3}{2} \frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{3 \mathrm{hc}}{\mathrm{e} \lambda_0}=\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\mathrm{hc}}{\mathrm{e} \lambda_0} \\ & \frac{3}{2 \lambda}-\frac{3}{\lambda_0}=\frac{1}{\lambda}-\frac{1}{\lambda_0} \\ & \frac{2}{\lambda_0}=\frac{1}{2 \lambda} \\ & \lambda_0=4 \lambda \end{aligned}

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Divya Prakash Singh

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