When a photosensitive surface is irradiated by lights of wavelengths ' $\lambda_1$ ' and ' $\lambda_2$ ',…

When a photosensitive surface is irradiated by lights of wavelengths ' $\lambda_1$ ' and ' $\lambda_2$ ', kinetic energies of the emitted photoelectrons is ' $E_1$ ' and ' $E_2$ ' respectively. The work function of the photosensitive surface is
  1. $\frac{\left(E_2 \lambda_2-E_1 \lambda_1\right)}{\left(\lambda_2-\lambda_1\right)}$
  2. $\frac{\left(E_1 \lambda_1+E_2 \lambda_2\right)}{\left(\lambda_2-\lambda_1\right)}$
  3. $\frac{\left(E_1 \lambda_1-E_2 \lambda_2\right)}{\left(\lambda_2-\lambda_1\right)}$
  4. $\frac{\left(E_2 \lambda_2+E_1 \lambda_1\right)}{\left(\lambda_1-\lambda_2\right)}$

Solution

From Einstein's photoelectric equation, $\begin{aligned} & \mathrm{E}_1=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0 \\ \therefore \quad & \mathrm{E}_1 \lambda_1=\mathrm{hc}-\mathrm{W}_0 \lambda_1 \\ & \mathrm{E}_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0 \\ \therefore \quad & \mathrm{hc}=\mathrm{E}_1 \lambda_1+\mathrm{W}_0 \lambda_1...(i) \end{aligned}$ $\begin{array}{ll} \therefore \quad & \mathrm{E}_2 \lambda_2=\mathrm{hc}-\mathrm{W}_0 \lambda_2 \\ & \Rightarrow \mathrm{hc}=\mathrm{E}_2 \lambda_2+\mathrm{W}_0 \lambda_2...(ii) \end{array}$
From equations (i) and (ii), $\begin{array}{ll} & \mathrm{E}_1 \lambda_1+\mathrm{W}_0 \lambda_1=\mathrm{E}_2 \lambda_2+\mathrm{W}_0 \lambda_2 \\ \therefore \quad & \mathrm{E}_1 \lambda_1-\mathrm{E}_2 \lambda_2=\mathrm{W}_0\left(\lambda_2-\lambda_1\right) \\ \therefore \quad & \mathrm{W}_0=\frac{\mathrm{E}_1 \lambda_1-\mathrm{E}_2 \lambda_2}{\lambda_2-\lambda_1} \end{array}$

Asked in: MHT CET 2024 (10 May Shift 2)

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