Question

An electron of mass m is initially moving with a constant speed v , where v << c . Express all algebraic answers in terms of the given quantities and fundamental constants.

(a) Determine the kinetic energy of the electron.

(b) Determine the de Broglie wavelength of the electron.

The electron encounters a particle with the same mass and opposite charge (a positron) moving with the same speed in the opposite direction. The two particles undergo a head-on collision, which results in the disappearance of both particles and the production of two photons of the same energy.

(c) Determine the energy of each photon.

(d) Determine the wavelength of each photon.

(e) Explain why there must be two photons produced instead of just one.

Answer 1

(a) K = ½ mv² 

(b) p = h / λ … mv = h/ λ … λ = h/mv

(c) The total energy involved here is both the kinetic energy of the particles as well as the energy released from the annihilation of mass (mc²). It’s important to remember that the particles were moving before hand so we cannot simply ignore this kinetic energy as part of the total energy.

E_before = KE_{electron + positron} + rest mass energy of the two particles … 2( ½ mv²) + 2 mc²

After the annihilation, there are two photon particles and they will split this total before energy in half so the energy of each photon is given by mv² + mc² …. And since in the problem it says v << c this means the rest mass term will dominate the energy so the approximate photon energy is simply E = mc². You can leave out the kinetic energy term in this derivation but in reality this is part of the energy so you should at least state in the problem that you are omitting it because it is small relative to the rest mass energy.

(d) E = hc / λ … mc² = hc / λ … λ = h/mc

(e) Two photons must be produced in order to conserve momentum. Before the collision, there was zero net momentum since the same mass and velocity particles were moving in opposite directions, so after the collision there must also be zero net momentum. Since after the collision a photon will be moving off in one direction and has a mass equivalent and momentum, there much be another photon moving in the other direction with opposite momentum to conserve the net momentum of zero.