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The threshold frequency for a photosensitive metal is \mathrm{v_0} . When photons of frequency \mathrm{2v_0} are incident on a photosensitive plate, the cut off potential is\mathrm{v_0}. What will be the cut off potential, when light of frequency \mathrm{5v_0} is incident on it?

Option: 1

\mathrm{V_0}


Option: 2

\mathrm{2V_0}


Option: 3

\mathrm{4V_0}


Option: 4

\mathrm{5V_0}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \mathrm{hv}-\phi_{\mathrm{o}}=e \mathrm{v}_{\mathrm{o}} \\ & \phi_{\mathrm{o}}=\mathrm{h} \mathrm{v}_{\mathrm{o}} \end{aligned}}

\mathrm{\text { As when } v=2 v_0 \text { Then } h\left(2 v_0\right)-h v_0=e v_0}

\mathrm{\begin{aligned} & \Rightarrow h\left(2 v_0\right)-h v_0=e v_0 \\ & \text { Now, e } \cdot v_0^{\prime}=h \times 5 v_0-h v_0=4 h v_0 \\ & \therefore \quad v_0^{\prime}=\frac{4 h}{e} \times \frac{e}{h} \cdot v_0=4 v_0 . \end{aligned}}

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