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

Explanation

The energy of the photon, `E = hf = (hc)/(lamda)`
The kinetic energy of the electron,
`E'= (1)/(2) mv^(2) = (p^(2))/(2m)` [where, p = mv = momentum]
As `p = (h)/(lamda)`, So, `E' = (h^(2))/(2m lamda^(2))`
`:. (E)/(E') = (hc)/(lamda). (2m lamda^(2))/(h^(2)) = (2 lamda mc)/(h) or, E = (2 lamda mc)/(h) . E'`