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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Physics
Chapter
Dual Nature of Matter

JEE Main 2025 · 7 April · Evening Shift · Physics · Dual Nature of Matter

A photo-emissive substance is illuminated with a radiation of...

A photo-emissive substance is illuminated with a radiation of wavelength λi\lambda_i so that it releases electrons with de-Broglie wavelength λe\lambda_e. The longest wavelength of radiation that can emit photoelectron is λ0\lambda_0. Expression for de-Broglie wavelength is given by :
( m:\mathrm{m}: mass of the electron, h:\mathrm{h}: Planck's constant and cc : speed of light)

Options

  1. A\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}
  2. Bλe=hλ02mc\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h} \lambda_0}{2 \mathrm{mc}}}
  3. C\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}
  4. Dλe=hλi2mc\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h} \lambda_{\mathrm{i}}}{2 \mathrm{mc}}}

Correct answer

A. \lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}

Step-by-step solution

K.E=E−Wλe=h2mK⋅E,E=hcλi,W=hcλ0 h22 mλe2=hcλi−hcλ0λe=h2mc(1λi−1λ0)\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}$$
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