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Q.

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=hcE

For the electron, E'=p22m⇒p=2mE'

∴ λ'=hp=h2mE'

Since λ'=λ, we can write h2mE'=hcE

⇒E'=E22mc2

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