Physics · 2025
JEE Main · 7 April 2025, Shift 2 · Q43
A photoemissive substance is illuminated with a radiation of wavelength λ_i so that it releases electrons with de-Broglie wavelength λ_e. The longest…
A photoemissive substance is illuminated with a radiation of wavelength $\displaystyle \lambda_i$ so that it releases electrons with de-Broglie wavelength $\displaystyle \lambda_{\mathrm{e}}$. The longest wavelength of radiation that can emit photoelectron is $\displaystyle \lambda_{\mathrm{o}}$. Expression for de-Broglie wavelength is given by : (m : mass of the electron, h : Planck's constant and c : speed of light)
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\dfrac{1}{\lambda_i}-\dfrac{1}{\lambda_{\mathrm{o}}}\right)}}$
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.