Question

What is the energy loss of a photon of a certain wavelength after scattering?

A photon of wavelength 29 pm is scattered by a stationary electron. What is the maximum possible energy loss of the photon?
(m = 9.11 × 10-31 kg, h = 6.626 × 10-34 J · s, c = 3.00 × 108 m/s)

The answer is 6.1keV but I am unsure as to how they got this answer.

Answer 1

Max energy is passed onto electron when the photon bounces straight back, ie `theta = pi`.

See calculation

Explanation:

`Delta lambda = 2.43 " pm" underbrace((1 - cos theta))\_("When is this a max to maximise " Delta lambda)`

`Delta lambda_("max") = 2.43 * color(red)(2) " pm" = 4.86 " pm"`

KE gained by electron is energy lost by photon:

• `K_("max") = 1240 " eV nm " (1/(29*10^3) - 1/((29 + 4.86)*10^3) ) = 6137 " eV"`