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Question

Physics Question on de broglie hypothesis

The linear momentum of a 3 MeV3\, MeV photon is

A

0.01 eV s m−10.01\,eV\,s\,m^{-1}

B

0.02 eV s m−10.02\,eV\,s\,m^{-1}

C

0.03 eV s m−10.03\,eV\,s\,m^{-1}

D

0.04 eV s m−10.04\,eV\,s\,m^{-1}

Answer

0.01 eV s m−10.01\,eV\,s\,m^{-1}

Explanation

Solution

Energy of a photon, E=3 MeV=3×106eVE=3\,MeV=3\times10^{6}eV Linear momentum of the photon, P=EcP=\frac{E}{c} where cc is the speed of light in vacuum P=3×106 eV3×108 m s−1=10−2 eV s m−1P=\frac{3\times10^{6}\,eV}{3\times10^{8}\,m\,s^{-1}}=10^{-2}\,eV\,s\,m^{-1} =0.01 eV s m−1=0.01\,eV\,s\,m^{-1}