From a metallic surface photoelectric emission is observed for frequencies $v_1$ and…

From a metallic surface photoelectric emission is observed for frequencies $v_1$ and $v_2\left(v_1>v_2\right)$ of the incident light. The maximum values of the kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1: \mathrm{x}$. Hence the threshold frequency of the metallic surface is
  1. $\frac{v_1-v_2}{x}$
  2. $\frac{v_1-v_2}{x-1}$
  3. $\frac{xv_1-v_2}{x-1}$
  4. $\frac{xv_2-v_1}{x-1}$

Solution

Using Einstein's photoelectric equation, $\begin{aligned} & \mathrm{E}_{\mathrm{k}}=\mathrm{hv}-\phi_0 \\ & \mathrm{E}_{\mathrm{k}}=\mathrm{hv}-\mathrm{h} v_0 \\ & \therefore \quad \mathrm{E}_{\mathrm{K}_1}=\mathrm{h}\left(\mathrm{v}_1-\mathrm{v}_0\right) \text { and } \mathrm{E}_{\mathrm{K}_2}=\mathrm{h}\left(\mathrm{v}_2-\mathrm{v}_0\right) \\ & \text { Given } \frac{E_{K_1}}{E_{K_2}}=\frac{1}{x} \\ & \Rightarrow \frac{v_1-v_0}{v_2-v_0}=\frac{1}{x} \\ & \left(v_1-v_0\right) x=v_2-v_0 \\ & v_1 x-v_0 x=v_2-v_0 \\ & \therefore \quad v_1 x-v_2=v_0 x-v_0 \\ & v_1 x-v_2=v_0(x-1) \\ & \therefore \quad v_0=\frac{v_1 x-v_2}{x-1} \\ & \end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 1)

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