When photons of energies twice and thrice the work function of a metal are incident on the metal surface one…

When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other, the maximum velocities of the photoelectrons emitted in the two cases are $v_1$ and $v_2$ respectively. The ratio $v_1: v_2$ is
  1. $\sqrt{2}: 1$
  2. $\sqrt{3}: 1$
  3. $\sqrt{3}: \sqrt{2}$
  4. $1: \sqrt{2}$

Solution

$\begin{array}{ll} & \mathrm{K} \cdot \mathrm{E}_{\max }=\mathrm{h} v-\phi_0 \\ & \text { Given, } \\ & \mathrm{E}_1=2 \phi_0 \text { and } \mathrm{E}_2=3 \phi_0 \\ \Rightarrow & \mathrm{K} \cdot \mathrm{E}_1=2 \phi_0-\phi_0=\phi_0 \\ \Rightarrow & \mathrm{K} \cdot \mathrm{E}_2=3 \phi_0-\phi_0=2 \phi_0 \\ & \text { but, } \mathrm{K} \cdot \mathrm{E}_1=\frac{1}{2} \mathrm{mv}_1^2 \text { and } \mathrm{K} \cdot \mathrm{E}_2=\frac{1}{2} \mathrm{mv}_2^2 \\ \therefore \quad & \frac{\mathrm{K} \cdot \mathrm{E}_1}{\mathrm{~K} \cdot \mathrm{E}_2}=\frac{\mathrm{v}_1^2}{\mathrm{v}_2^2}=\frac{1}{2} \\ \therefore \quad & \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{1}{\sqrt{2}}\end{array}$

Asked in: MHT CET 2023 (09 May Shift 2)

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