Fermi/Coulomb correction factor for beta decay

[rrg92]

Hi all,

Fermi introduced a multiplicative correction factor in order to correct for the effect of the Coulomb field on the resulting energy distribution of ejected beta particles (i.e., either electrons or positrons) following a beta decay reaction. The result is that positrons are accelerated as they exit the nucleus, while electrons are slowed down. My question is, why does this correction factor have to be a multiplicative term, as opposed to simply subtracting the energy lost or gained for the electron or positron respectively? Is there a way that I can derive the momentum-dependent energy loss/gain, and arrive at the final corrected energy distribution?

Kind regards

 

[AndyW]

,Hello,

Thank you for your question. The reason why the correction factor introduced by Fermi is a multiplicative term is because the energy loss or gain for beta particles is directly related to their momentum. This means that the amount of energy lost or gained will vary depending on the initial momentum of the particle.

To understand this further, we need to look at the concept of momentum conservation. In a beta decay reaction, a neutron is transformed into a proton and a beta particle (either an electron or a positron). This means that the total momentum of the system must be conserved before and after the decay. The Coulomb field within the nucleus affects the momentum of the beta particle as it exits, resulting in a change in its energy.

Now, if we were to simply subtract the energy lost or gained by the beta particle, we would not be taking into account its change in momentum. This is why the correction factor is a multiplicative term, as it accounts for the momentum-dependent energy loss or gain.

To derive this correction factor, you can use the concept of energy-momentum relation, which states that the energy of a particle is equal to its momentum times the speed of light. By considering the change in momentum of the beta particle, you can arrive at the final corrected energy distribution.

I hope this helps to answer your question. Please let me know if you have any further inquiries.