When a light of wavelength ' $\lambda$ ' falls on the emitter of a photocells, maximum speed of emitted…

When a light of wavelength ' $\lambda$ ' falls on the emitter of a photocells, maximum speed of emitted photoelectrons is ' $\mathrm{V}$. If the incident wavelength is changed to $\frac{2 \lambda}{3}$, maximum speed of emitted photoelectrons will be
  1. less than $(1.5)^{1 / 2} \mathrm{~V}$
  2. $\sqrt{\mathrm{V}}$
  3. greater than $(1.5)^{1 / 2} \mathrm{~V}$
  4. V

Solution

$(\mathrm{K} . \mathrm{E} .)_1=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0$ Multiplying by $3 / 2$ $\begin{aligned} & \frac{3}{2}(\mathrm{~K} . \mathrm{E} .)_1=\frac{3}{2} \frac{\mathrm{hc}}{\lambda_1}-\frac{3}{2} \mathrm{~W}_0 ...(1)\\ & (\mathrm{~K} . \mathrm{E} .)_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0 \\ & (\mathrm{~K} . \mathrm{E} .)_2=\frac{3}{2} \frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0 ...(2) \end{aligned}$ By equation (1) and (2) $\begin{aligned} & (\mathrm{K} \cdot \mathrm{E} \cdot)_2=\frac{3}{2}(\mathrm{~K} \cdot \mathrm{E} \cdot)_1+\frac{1}{2} \mathrm{~W}_0 \\ & \text { or } \frac{1}{2} \mathrm{mv}_2^2>\frac{3}{2}\left(\frac{1}{2} \mathrm{mv}_1^2\right) \\ & \therefore \mathrm{v}_2^2>\frac{3}{2} \mathrm{v}_1^2 \\ & \therefore \mathrm{v}_2>(1.5)^{1 / 2} \mathrm{v}_1 \end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

Practice more Dual Nature of Matter and Radiation questions on Aicharya