1. λ = (2m/hc)λ_d² 2. λ_d = (2mc/h)λ² 3. λ = (2mc/h)λ_d² 4. λ = (2h/mc)λ_d²

An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λ_d, then :

1. \(\lambda = \left(\frac{2m}{hc}\right)\lambda_d^2\)

2. \(\lambda_d = \left(\frac{2mc}{h}\right)\lambda^2\)

3. \(\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2\)

4. \(\lambda = \left(\frac{2h}{mc}\right)\lambda_d^2\)

Doubtly Asked 30 Oct 2026 NEET UG 2021 Practice test neet Save WhatsApp Report

Best answer

Answer: 3. \(\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2\)

Key idea: with negligible work function, the photoelectron's kinetic energy equals the photon energy.

\(\dfrac{hc}{\lambda}=\dfrac{p^2}{2m}\Rightarrow p^2=\dfrac{2mhc}{\lambda}\).
\(\lambda_d=\dfrac hp\Rightarrow\lambda_d^2=\dfrac{h^2}{p^2}=\dfrac{h^2\lambda}{2mhc}=\dfrac{h\lambda}{2mc}\).
So \(\lambda=\dfrac{2mc}{h}\lambda_d^2\).

Option 3.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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