The wavelenths `gamma`of a photon and the de-Broglie wavelength of an electron have the same value.Find the ratio of energy of photon to the kinetic energy of electron in terms of mass m, speed of light and planck constatn

Options

A. `(gammamc)/(h)`
B. `(hmc)/(gamma)`
C. `(2hmc)/(gamma)`
D. `(2gammamc)/(h)`

Answer

D

Explanation

The de Brogli wavelength
`h=(h)/(mv)rArrv=(h)/(mlambda)`
Energy of photon
`E_(p)=(hc)/(lambda)(since lambda is same)`
Energy of Photon
Kinetic energy of photon `=(E_(p))/(E_(e))=(hc//lambda)/((1)/(2)1mu^(2))=(2hc)/(lambdamv^(2))`
substituting value of V from Eq (i) we get
`(E_(p))/(E_(e)) =(2hc)/(lambda_(m))(h)/(mlambda))^(2)=(2lambdamc)/(h)`