When a metallic surface is illuminated with a radiation of wavelength ' $\lambda$ ', the stopping potential…

When a metallic surface is illuminated with a radiation of wavelength ' $\lambda$ ', the stopping potential is ' $V$ '. If' the same surface is illuminated with radiation of wavelength ' $3 \lambda$ ', the stopping potential is " $\left(\frac{\mathrm{V}}{6}\right)$ '. The threshold wavelength for the surface is
  1. $3 \lambda$
  2. $4 \lambda$
  3. $5 \lambda$
  4. $6 \lambda$

Solution

From Einstein's equation, $\mathrm{h} \nu=\mathrm{e} \mathrm{V}_0+\mathrm{h} \nu_0$ $\therefore \quad \frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0}=\mathrm{eV}_0$ case (i) $\lambda=\lambda ; \mathrm{V}_0=\mathrm{V}$ $\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0}=\mathrm{eV}$ case (ii) $\lambda=3 \lambda ; \mathrm{V}_0=\frac{\mathrm{V}}{6}$ $\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0}=\frac{\mathrm{eV}}{6}$ $\begin{aligned} & \text {dividing equation (i) by equation (ii) } \\ & \therefore \quad \frac{\left(\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0}\right)}{\left(\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0}\right)}=6 \\ & \therefore \quad \frac{1}{\lambda}-\frac{1}{\lambda_0}=6\left(\frac{1}{3 \lambda}-\frac{1}{\lambda_0}\right) \\ & \therefore \quad \frac{1}{\lambda}-\frac{1}{\lambda_0}=\frac{2}{\lambda}-\frac{6}{\lambda_0} \\ & \therefore \quad \frac{-1}{\lambda_0}+\frac{6}{\lambda_0}=\frac{2}{\lambda}-\frac{1}{\dot{\lambda}} \end{aligned}$ $\therefore \quad \frac{5}{\lambda_0}=\frac{1}{\lambda} \quad \Rightarrow \lambda_0=5 \lambda$ *

Asked in: MHT CET 2024 (02 May Shift 1)

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