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

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

Solution

$\begin{aligned} & \mathrm{E}_1=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W} \\ & \mathrm{E}_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W} \\ & \mathrm{E}_1 \lambda_1+\mathrm{W} \lambda_1=\mathrm{E}_2 \lambda_2+\mathrm{W} \lambda_2 \\ & \mathrm{~W}=\frac{\mathrm{E}_1 \lambda_1-\mathrm{E}_2 \lambda_2}{\lambda_2-\lambda_1} \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

Practice more Dual Nature of Matter questions on Aicharya