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When 1 \mathrm{~cm} thick surface is illuminated with light of wavelength \lambda, the stopping potential is \mathrm{V}. When the same surface is illuminated by light of wavelength 2\lambda , the stopping potential is\mathrm{V}/3. Threshold wavelength for metallic surface is:

Option: 1

\frac{4\lambda }{3}


Option: 2

4\lambda


Option: 3

6\lambda


Option: 4

\frac{8\lambda }{3}


Answers (1)

best_answer

According to the question,
\mathrm{eV}=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right) \quad \quad \quad(i)
and \frac{\mathrm{eV}}{3}=\mathrm{hc}\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right) \quad \quad \quad(ii)

Dividing eq. (i) by eq. (ii), we get

\begin{aligned} & 3=\frac{\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)}{\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)} \\ & \text { or } 3\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)=\frac{1}{\lambda}-\frac{1}{\lambda_0} \\ & \text { or } \frac{3}{2 \lambda}-\frac{1}{\lambda}=\frac{3}{\lambda_0}-\frac{1}{\lambda_0}\\ & \text{or }\frac{1}{2 \lambda}=\frac{2}{\lambda_0} \end{aligned}

Threshold wavelength for metallic surface \lambda_0=4 \lambda

Posted by

Suraj Bhandari

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