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An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is

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a
C(2mE)12
b
1c2mE12
c
1cE2m12
d
E2m12(c being velocity of light)

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detailed solution

Correct option is C

For electron of energy  E de-Broglie wavelength, λe=hp=h2mEFor photon of energy,E=hv=hcλp⇒λp=hcE∴λeλp=h2mE×Ehc=1cE2m1/2

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