Beta Decay — Question 6

Question 6

A simplified beta spectrum illustrates continuous energy sharing

In an explicitly ultrarelativistic, massless-electron approximation, neglect Coulomb effects and take the electron kinetic-energy density $f(T)=C T^2(Q-T)^2$ for $0<T<Q$.

1. Normalize f and calculate the mean electron energy.

2. Find the most probable energy and explain the symmetry that fixes the mean.

Question 6 - Solution

A simplified beta spectrum illustrates continuous energy sharing

(a) Method and calculation

$$1=C\int_0^QT^2(Q-T)^2dT=CQ^5/30\Rightarrow\boxed{C=30/Q^5}.$$$$\langle T\rangle=\frac{30}{Q^5}\int_0^QT^3(Q-T)^2dT=\boxed{Q/2}.$$

(b) Interpretation and further calculation

$$f\prime(T)=2CT(Q-T)(Q-2T)=0\Rightarrow\boxed{T_{\mathrm{mode}}=Q/2}.$$$$\boxed{f(T)=f(Q-T)}\quad\text{makes the distribution symmetric about }Q/2.$$

Physical check

Real low-energy beta spectra require the electron rest mass and Coulomb corrections. This model isolates continuous sharing and phase-space weighting.