Difference between endpoints and kinetic energy in a beta minus decay

Question

I want to understand the difference between the Q value and Endpoint energy and Kinetic energy of a beta minus decay.

I understood that the Q value is the overall energy of the reaction given by

Q value $= T_e+T_\nu$

where $T_e$ is the KE of the electron and $T_nu$ is the KE of the anti-neutrino. However, Q value is also written as :

Q value$= E_e+E_\nu$ and also $E_o=E_e+E_\nu$

where $E_e$ is the electron energy and $E_\nu$ is the anti-neutrino.

We also know that $E^2= p^2c^2+m^2c^4$ and $E=T_e+m_ec^2$ with $T_e=\sqrt{p^2c^2+m^2c^4}-mc^2$.

So isnt Q value is same as endpoint energy? Second, what is the kinetic energy here, is it the endpoint energy or something else.

Answer 1

The Q value is the same as the endpoint energy for the electron in the limit where the decaying nucleus is infinitely massive. However, if the nucleus has finite mass, then there will be a difference between the Q value and the endpoint energy due to the kinetic energy of the nucleus, which will be approximately

$$ T_\text{nucleus}\approx \frac{(\vec p_e+\vec p_\nu)^2}{2m_\text{nucleus}} $$

In general, unless you are doing high precision work, or dealing with a low energy and low mass decay system such as tritium, the approximation that the electron's maximum energy is equal to the Q value is a good approximation. An electron whose kinetic energy is equal to the Q value would correspond to the case where the electron leaves the daughter nucleus and the neutrino both at rest. But such a decay would not conserve momentum.

In general, the momentum of the decay is shared more or less equally among all three bodies. This means that the majority of the energy is shared between the electron and the neutrino, and that the most probable electron energy is somewhere around half of the Q value. Computing the distribution of electron energies is a non-trivial exercise.