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# 5.6: Example of Golden rule - beta decay


A nucleus decays via the reaction $$n \rightarrow p e^− \bar{\nu}.$$ to form a electron and antineutrino, releasing energy $$E_0$$.

The simplest form for the matrix element describing nuclear $$\beta$$-decay is given by the so-called Fermi ansatz $$V_{mk} = G_FM/\Omega$$ where $$\Omega$$ is the normalisation volume for the wavefunctions, $$|M|^2 \approx 1$$ is the wavefunction overlap between initial and final nuclear states and $$G_F$$ is a constant.

We can work in the COM reference frame, so the kinetic energy of the nucleus is zero. Momentum is conserved, so the final state has nuclear, electron and neutrino momentum $${\bf P + p + q}=0$$ while the energy released goes into the electron and neutrino, which for simplicity we treat as massless: $$E_0 = E_e + qc$$ The proton and neutron are heavy compared with the electron and neutrino. Given that momentum must be conserved, the kinetic energy must be concentrated in the lighter particles.

The density of final states for the electron is given by the phase space volume

$dn = \frac{d^3{\bf p}d^3 {\bf r}}{(2\pi \hbar)^3} \nonumber$

with a similar expression for the neutrino. Number of states in a volume of phase space is given by the number of electron states, times the number of neutrino states, provided energy is conserved:

$dn = \frac{d^3{\bf p}d^3 {\bf r}}{(2\pi \hbar)^3} \frac{d^3{\bf q}d^3 {\bf r}}{(2\pi \hbar)^3} \delta (E_e + qc − E_0) \nonumber$

Using the relativistic relation $$E^2 = p^2 c^2 + m^2 c^4$$ implies $$\frac{dp}{dE} = \frac{E}{pc^2}$$

the normalisation volume is just $$\int d^3 {\bf r} = \Omega$$, and rotational invariance gives $$d^3 {\bf p} = 4 \pi p^2dp$$.

All of which which simplifies the integral to

$dn = \frac{\Omega^2}{4\pi^4 \hbar^6 c^6} E_e \sqrt{E^2_e − m^2_e c^4} E^2_{\nu} \delta (E_e + E_{\nu} − E_0)dE_edE_{\nu} \nonumber$

where $$E_e$$ is the electron energy and $$E_{\nu}$$ is the neutrino energy. What can actually be measured is the electron energy, so we integrate over the neutrino energies,

$\frac{dn}{dE_e} = \frac{\Omega^2}{4\pi^4\hbar^6 c^6}E_e \sqrt{E^2_e − m^2_e c^4} (E_0 − E_e)^2 \nonumber$

This is the distribution of electron energies from beta decay: the rate fo emission of electrons at a particular energy is given by the Golden Rule

$R = \frac{2\pi}{\hbar} \frac{G^2_F M^2}{4\pi^4 \hbar^6 c^6} E_e \sqrt{E^2_e − m^2_e c^4} (E_0 − E_e)^2 \nonumber$

Figure $$\PageIndex{1}$$ shows the simplest case of beta-emission: neutron decay. Conservation laws tell us that the electron energy must lie between its rest mass (0.51MeV) and the total energy available (0.7823MeV). But the entire shape can be deduced from geometry.

5.6: Example of Golden rule - beta decay is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.