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A photo-emissive substance is illuminated with a radiation of wavelength λ_i so that it releases electrons with de-Broglie wavelength λ_e. The…

JEE Main 2025 — Physics Modern Physics

2025mcqmedium

A photo-emissive substance is illuminated with a radiation of wavelength λi\lambda_iλi​ so that it releases electrons with de-Broglie wavelength λe\lambda_eλe​. The longest wavelength of radiation that can emit photoelectron is λ0\lambda_0λ0​. Expression for de-Broglie wavelength is given by :

( m:\mathrm{m}:m: mass of the electron, h:\mathrm{h}:h: Planck's constant and ccc : speed of light)

Official previous-year question

Held on 7 Apr 2025 · Verified 6 Jul 2026.

Options

  1. A

    λe=h2mc(1λi−1λ0)\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}λe​=2mc(λi​1​−λ0​1​)h​​

  2. B

    λe=hλ02mc\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h} \lambda_0}{2 \mathrm{mc}}}λe​=2mchλ0​​​

  3. C

    λe=h2mc(1λi−1λ0)\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}λe​=2mc(λi​1​−λ0​1​)​h​

  4. D

    λe=hλi2mc\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h} \lambda_{\mathrm{i}}}{2 \mathrm{mc}}}λe​=2mchλi​​​

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