In an x-ray tube an electron moving initially at a speed of slows down in passing near a nucleus. A single photon of energy 43.8 keV is emitted. We have to find the final speed of the electron. (Relativity must be taken into account; energy imparted to the nucleus may be ignored.)

Solution:             Click For PDF Version

The relativistic energy of an electron moving with speed is given by the relation

.

The energy of the electron when it is moving with speed of will be

The energy of the electron after it has emitted a photon of energy 43.8 keV will be (neglecting energy imparted to the nucleus)

.

The speed of the electron after it has emitted the electron will therefore be

Therefore,

.