The wavelength `lambda` of a photon and the de-Broglie wavelength of an electron have the same value. Show that the energy of the photon is `(2 lambda mc)/h` times the kinetic energy of the electron, Where m,c and h have their usual meanings.

Explanation

Given,
`lambda_(ph) = lambda_(e) = lambda = h/(mv)`
Also, `E_(ph) =hv =(hc)/lambda`…….(i)
And `E_(e) =1/2 mv^(2) =1/2m(h/(lambdam))^(2)`
`[therefore lambda = h/(mv), therefore v=h/(lambdam)]`
or `E_(e) = h^(2)/(2lambda^(2)m)`……(ii)
Dividing Eq. (i) by Eq. (ii), we get
`E_(ph)/E_(e) = (hc)/lambda xx (2lambda^(2)m)/h^(2) =(2lambdamc)/h`
or `E_(ph) =(2hmc)/h E_(e)`