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.
1
Expert's answer
2011-01-20T12:21:44-0500
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.

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

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