The de Broglie wavelength of an electron moving with a velocity of `1.5xx10^(8)ms^(-1)` is equal to that of a photon find the ratio of the kinetic energy of the photon to that of the electron.

Answer

D

Explanation

Speed of photon `(c)=3xx10^(8) m//sec`.Let `lambda` be the wavelength of the photon.The de Broglie wavelength of the electron `h/(mv)`.Given `lambda=l/(mv)`,Now `"energy of photon"/"K.E. of electron"=(hv)/(1/2) mv^(2)=(2hc)/(mv^(2)lambda)`
=`(2c)/v(therefore lambda=h/(mv)` & `v=c/lambda)=(2xx3xx10^(8))/(1.5xx10^(8))=4`