For a photosensitive material, work function is ' $\mathrm{W}_0$ ' and stopping potential is ' V '. The…

For a photosensitive material, work function is ' $\mathrm{W}_0$ ' and stopping potential is ' V '. The wavelength of incident radiation is ($\mathrm{h}=$ Planck's constant, $c=$ velocity of light, $e=$ electronic charge)
  1. $\frac{\mathrm{h}^2 \mathrm{c}^2}{\mathrm{~W}_0+\mathrm{eV}}$
  2. $\frac{\mathrm{hc}}{\mathrm{W}_0}$
  3. $\frac{\mathrm{hcV}}{\mathrm{W}_0}$
  4. $\frac{\mathrm{hc}}{\mathrm{W}_0+\mathrm{eV}}$

Solution

From Einstein's photoelectric equation, we can write, $\begin{array}{ll} & \mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}_0 \\ \therefore \quad & \frac{\mathrm{hc}}{\lambda}=\mathrm{W}_0+\mathrm{eV} \\ \therefore \quad & \lambda=\frac{\mathrm{hc}}{\mathrm{~W}_0+\mathrm{eV}} \end{array}$

Asked in: MHT CET 2024 (04 May Shift 1)

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