An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength λ′, then
λ=3mc2hλ′2
λ=mchλ′2
λ=5mchλ′2
λ=2mchλ′2
A
λ=3mc2hλ′2
B
λ=5mchλ′2
C
λ=2mchλ′2
D
λ=mchλ′2
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Solution
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The correct option is Cλ=2mchλ′2 Kinetic energy of emitted electron = Energy of incident photon i.e. 12mv2=hν
or p22m=hcλ
[∵mv=p,ν=cλ] or p=√2mhcλ
de-Broglie wavelength of emitted electrons
λ′=hp=h√2mhcλ or
λ′=√hλ2mc∴λ=2mchλ′2
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