Question

A photon of wavelength 5.58 pm scatters at an angle of 122° from an initially stationary, unbound electron. What is the de Broglie wavelength(in pm) of the electron after the photon has been scattered??

Notice: Answer is not (9.29, 2.12, 2.06, nor 4.11)pm

Explanation:

The de Broglie wavelength of a massive particle is related to its momentum in the same way that a photon's momentum is related to its wavelength.

The well-known Compton scattering relationship gives the final wavelength of the photon as a function of the inital wavelength, the scattering angle, Planck's constant, the speed of light, and the mass of the electron.

To determine the momentum of the electron after the scattering event (and therefore its de Broglie wavelength), consider the total momentum of the electron-photon system before and after the photon is scattered. How should they be related? Remember that momentum is a vector quantity.

Answer 1

The change in energy of the photon = kinetic energy of the electron

so, the kinetic energy of the electron is:

now, [E + mc²]² = [pc]² + [mc²]²

=> pc = 5.034 x 10^-14 J

=> p = 1.678 x 10^-22 kgm/s

now use the De-Broglie relation

this is the De-Broglie wavelength of the electron.