Answer to Question #1285 in Quantum Mechanics for Patricia

Question #1285
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.
Expert's answer
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 = Ee + W2, where W2 = hf2 (h - is Plank's constant, f - frequency), f2 = 1.64x1019 Hz.
Ee = W1 - hf2.
The momentum of the photon can be found
p = hf/c = W/c, where c- is the speed of the light.

W1/c = pe + hf2/c; pe = W1/c - hf2/c

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