Question #6a4bb

1 Answer
Oct 23, 2017

Energy of a photon of wavelength #lambda_p# is given by the expression

#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 #lambda#. Re-writing above with the help of (1) we get

#KE_e=(h/lambdaxxE_p/c)/(2m)#
#=>KE_e=(hE_p)/(2lambdamc)#

#=>E_p=(2lambdamc)/hxxKE_e#