Question

The wavelength λ of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of a photon in (2λmc/h) times the kinetic energy of electron; where m, c and h have their usual meaning.

Answer 1

Given that λ is the wavelength of the photon.

The de-Broglie wavelength of the electron is \[\lambda_e = \frac{h}{mv}\].

The kinetic energy of the electron is given as 

\[E_e = \frac{1}{2}m v^2 = \frac{1}{2}m(\frac{h}{m \lambda_e} )^2 = \frac{h^2}{2m \lambda_e^2} \]

[∵ according to the question λ = λ_e]

We know that the energy of the photon is

\[E_p = \frac{hc}{\lambda}\].

Taking product of equation (1) with 

\[\frac{2\lambda mc}{h}\]  

we get

\[E_e \times \frac{2\lambda mc}{h} = \frac{h^2}{2m \lambda^2} \times \frac{2\lambda mc}{h} = \frac{hc}{\lambda} = E_p \]\[ \Rightarrow E_p = E_e \times \frac{2\lambda mc}{h}\]

Answer 2

As, `lambda = h/(mv), v = h/(mlambda)` ...(i)

Energy of photon `E = (hc)/lambda`

& Kinetic energy of electron

`K = 1/2mv^2 = 1/2(mh^2)/(m^2lambda^2)` ...(ii)

Simplifying equation (i) & (ii) we get,

`E/K = (2λmc)/h`