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  • A metallic surface is illuminated with radiation of wavelength  , the stopping potential is <img alt="\math

A metallic surface is illuminated with radiation of wavelength \lambda , the stopping potential is \mathrm{V_{0}}. If the same surface is illuminated with radiation of wavelength \mathrm{2\lambda } , the stopping potential becomes \mathrm{\frac{V_{0}}{4} }. The threshold wavelength for this metallic surface will be

Option: 1

\frac{3}{2}\lambda


Option: 2

4\lambda


Option: 3

3\lambda


Option: 4

\frac{\lambda }{4}


Answers (1)

best_answer

\mathrm{E=K.E+\phi _0}

Now

\frac{\mathrm{hc}}{\lambda}=\mathrm{ev}_0+\phi_0

And \: \: \frac{\mathrm{hc}}{2 \lambda}=\frac{\mathrm{eV}_0}{4}+\phi_0

(2) \times 4

\frac{2 \mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda}=0+\left(4 \phi_0-\phi_0\right)

\begin{aligned} & \frac{\mathrm{hc}}{\lambda}=3 \phi_0 \\ & \frac{\mathrm{hc}}{\lambda}=3 \frac{\mathrm{hc}}{\lambda_0} \\ & \lambda_0=3 \lambda \end{aligned}






 

Posted by

Irshad Anwar

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