An electron of mass m and a photon have the same deBroglie wavelength. If energy of the photon is E, then energy of the electron is (C = speed of light)
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a
E2/2mC2
b
E2/mC2
c
2m2C4E
d
E2/4mC2
answer is A.
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Detailed Solution
For the photon E=hcλ⇒λ=hcEFor the electron, E'=p22m⇒p=2mE'∴ λ'=hp=h2mE'Since λ'=λ, we can write h2mE'=hcE⇒E'=E22mc2