Question #6a4bb
1 Answer
Oct 23, 2017
Energy of a photon of wavelength
#E_p=(hc)/lambda_p# ......(1)
de Brogile wavelength of an electron is expressed as
#lambda_e=h/p_e#
#=>p_e=h/lambda_e# ........(2)
Kinetic energy equation for a particle can also be written as
#KE=p^2/(2m)#
Using (2), for electron
#KE_e=(h/lambda_e)^2/(2m)#
Given that wavelength of a photon and the de Brogile wavelength of an electron has same value. Let it be equal to
#KE_e=(h/lambdaxxE_p/c)/(2m)#
#=>KE_e=(hE_p)/(2lambdamc)#
#=>E_p=(2lambdamc)/hxxKE_e#