A photo-emissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases…

A photo-emissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectron is $\lambda_0$. Expression for de-Broglie wavelength is given by :
( $\mathrm{m}:$ mass of the electron, $\mathrm{h}:$ Planck's constant and $c$ : speed of light)
  1. $\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}$
  2. $\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h} \lambda_0}{2 \mathrm{mc}}}$
  3. $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}$
  4. $\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h} \lambda_{\mathrm{i}}}{2 \mathrm{mc}}}$

Solution

$\begin{aligned} & \mathrm{K}. \mathrm{E}=\mathrm{E}-\mathrm{W} \\ & \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK} \cdot \mathrm{E}}}, \mathrm{E}=\frac{\mathrm{hc}}{\lambda_{\mathrm{i}}}, \mathrm{W}=\frac{\mathrm{hc}}{\lambda_0} \\ & \frac{\mathrm{~h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2}=\frac{\mathrm{hc}}{\lambda_{\mathrm{i}}}-\frac{\mathrm{hc}}{\lambda_0} \\ & \lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}\end{aligned}$

Asked in: JEE Main 2025 (07 Apr Shift 2)

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