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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C(2mE)12

1c2mE12

1cE2m12

E2m12(c being velocity of light)

answer is C.

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Detailed Solution

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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