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PhysicsModern PhysicsPhotoelectric effectEasy2 minPYQ_2020
PhysicsEasysingle choice

Light of wavelength500 nmis incident on a metal with work function2.28 eV.The de Broglie wavelength of the emitted electron is:

Options:

Answer:
B
Solution:

Given : work functionϕof metal=2.28 eV

Wavelength of lightλ=500 nm=500×10−9 m

KEmax=hcλ−ϕ

KEmax=6.6×10−34×3×1085×10−7×1.6×10-19−2.82

KEmax=2.48−2.28=0.2 eV

λmin=hp=h2m(KE)max

=203×10−342×9×10−31×0.2×1.6×10−19

λmin=259×10−9

=2.80×10−9 nm∴ λ≥2.8×10−9 m

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect
⏱ 2mℹ️ Source: PYQ_2020

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