Question

the wavelength of photon and the de Broglie wavelength of an electron has the same value show that the energy of the photon is 2lambda mc÷h times the kinetic energy of electron

Answer 1

The wavelength of photon and the De Broglie wavelength of an electron has the same value show that the energy of the photon is 2lambda mc÷h times the kinetic energy of electron as shown below

1. De-Broglie wavelength of an electron is given as
2. λ=h/mv here λ is the De-Broglie wavelength,h is Planck constant ,m is mass nad v is the velocity
3. Energy of photon is given by
4. Ep= hc/λ
5. kinetic energy of electron is $\frac{1}{2} mv^{2}$
6. The ratio of the energy of photon to the kinetic energy of electron is
7. Ep/Ee=(hc/λ)/$\frac{1}{2} mv^{2}$ =2hc/λmv^2 now putting the value of v =h/mλ
8. Ep/Ee=2λmc/h
9. the energy of the photon is Ep = Kinetic energy of electron x 2λmc/h