Answer to Question #1285 in Quantum Mechanics for Patricia

Question

An x-ray photon of the maximum energy produced by a tube leaves the tube and collides elastically with an electron at rest. As a result, the electron recoils and the x-ray is scattered. The frequency of the scattered x-ray photon is 1.64x10^19Hz. Relativistic effects may be neglected for the electron.
Determine the kinetic energy of the recoiled electron and the magnitude of its momentum.

Answer 1

We need to know the initial energy of the X-ray photon W₁.
Then we can use the Law of Energy conservation and the Law of Momentum conservation:
W₁ = E_e + W₂, where W₂ = hf₂ (h - is Plank's constant, f - frequency), f₂ = 1.64x10¹⁹ Hz.
E_e = W₁ - hf₂.
The momentum of the photon can be found
p = hf/c = W/c, where c- is the speed of the light.

Thus
W₁/c = p_e + hf₂/c; p_e = W₁/c - hf₂/c