Decay of a Particle (Relativity)

Though my principle is not to help others do assignments or homework, I don't want people to be misled. Let me just give you some guidelines.

(a)
First, we need to calculate the energy released of the decay in the rest frame of the 14C nucleus:
Energy released = (mass difference between 14C and 14N) x c²

This energy will be the kinetic energy of the electron (KE).
But the total energy of the electron can be written as
γmec² = KE + mec², where γ = (1 - (v/c)²)-1/2
Then, γ and hence v can be found.

(Please note that the above assume the 14N nucleus does not take any speed after the decay, which is a good approximation as the mass of a 14N nucleus is much larger than that of an electron. For a more vigorous calculation, the recoil speed of the 14N nucleus can be determined by the law of conservation of momentum. And the energy released will be taken by the kinetic energy of both the electron and the 14N nucleus.)

(2)
The speed of the electron can be easily found by the formula for the composition of velocity:

w' = (w - v) / ( 1 - w v / c²)

By putting v = 0.8c and w = +/- the speed of electron found in (a), the answers will follow.

thanks so much.

(a) First of all, the mass of an electrion is:
9.1 × 10-31/1.66 × 10-27 = 0.000542168 u
So total mass of N-14 nucleus and an electron = 14.003616170 u
Therefore mass defect of C-14 = 14.003616170 - 14.003241982 = 0.000374188 u
So binding energy = 0.000374188 × 931 × 106 × 1.6 × 10-19 = 5.574 × 10-14 J by taking 1 u = 931 MeV
Hence, the K.E. of the electron is also 5.574 × 10-14 J and its speed, relative to the carbon atom is given by:
0.5 × 9.1 × 10-31 × v2 = 5.574 × 10-14
v = 3.5 × 108 m/s
= 1.17 c
(b) If ejected forward, its speed relative to the lab = 1.17 + 0.8 = 1.97 c
If backward, speed = 1.17 - 0.8 = 3.7 c