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  • If wavelength of the light incident on a photoelectric cell be reduced from  <img alt="\lambda_1$ to $\lambda_2 A^0" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clambda_1%24%20

If wavelength of the light incident on a photoelectric cell be reduced from  \lambda_1$ to $\lambda_2 A^0 , then what will be the change in the cut off potential?

Option: 1

\frac{h c}{e} \cdot\left(\lambda_1-\lambda_2\right)


Option: 2

\frac{h c}{e^2}\left(\lambda_1-\lambda_2\right)


Option: 3

\frac{h c}{e}\left(\frac{\lambda_1-\lambda_2}{\lambda_1 \lambda_2}\right)


Option: 4

\frac{2 h c}{e} \cdot\left(\frac{\lambda_1-\lambda_2}{\lambda_1 \lambda_2}\right)


Answers (1)

best_answer

Let the work function of the surface be \phi. if \nu be the frequency of the light falling on the surface, then according to Einstein's photo electric equation, the maximum kinetic energy K \cdot E_{\text {max }} of the emitted electron is given by
K \cdot E_{\text {max }} =h \nu-\phi=\frac{h c}{\lambda}-\phi

we know that    K \cdot E_{\text {max }}=e V_0$ $ where \ V_0=c u t \ off \ \ potential
\begin{aligned} e V_0 & =\frac{h c}{\lambda}-\phi \ \ \text { or } \ V_0=\frac{h c}{e \lambda}-\frac{\phi}{e} . \\ \Rightarrow V_0 & =V_{02}-V_{01} \\ & =\frac{h c}{e}\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_1}\right)=\frac{h c}{e} \cdot \frac{\left(\lambda_1-\lambda_2\right)}{\lambda_1 \lambda_2} . \end{aligned}

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