When a light of wavelength ′λ′ falls on the surface of a metal with threshold wavelength ′λ′0, an electron of mass '2m' is emitted. The de Broglie wavelength of emitted electron is:
A
[h(λλ0)4mc(λo−λ)]12
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B
[h(λλ0)2mc(λ−λ0)]12
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C
[hλλ02mc]12
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D
[h(λ−λ0)2mc(λλ0)]12
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Solution
The correct option is A[h(λλ0)4mc(λo−λ)]12 Given: Threshold wavelength of the metal surface=λ0Wavelength of the light=λMass of the electron=2m
Kinetic energy (KE) of the emitted electrons is given by: KE=hc(1λ−1λ0)=hc(λ0−λλλ0)
de Broglie wavelength in terms of kinetic energy= h√2KE(2m) Substituting the above value of KE, we get: =h√2×hc(λ0−λλλ0)×2m=[h(λλ0)4mc(λo−λ)]12