Electron-capture / β-decay

Electron-capture / β-decay#

Consider a pair of nuclei, \(X\) and \(Y\) linked together by electon-capture and \(\beta\)-decay:

\[X + e^- \rightarrow Y + \nu_e\]
\[Y \rightarrow X + e^- + \bar{\nu}_e\]

These nuclei have the same mass number, \(A\), but different proton numbers, \(Z\):

  • X: (Z, A)

  • Y: (Z-1, A)

For example, \({}^{23}\mathrm{Na}\) (Z = 11, A = 23) and \({}^{23}\mathrm{Ne}\) (Z = 10, A = 23)

Important

Any nuclear reaction needs to conserve baryon number, lepton number, and charge. The neutrino / anti-neutrinos here ensure that lepton-number is conserved.

The evolution equations for these are:

\[\frac{dY(A)}{dt} = - Y(A) \lambda_\mathrm{ec} + Y(B) \lambda_{\beta}\]
\[\frac{dY{B}}{dt} = + Y(A) \lambda_\mathrm{ec} - Y(B) \lambda_{\beta}\]

where we note that the electron capture rate will have an implicit dependence on \(\rho Y_e\)—the density of electrons.

Threshold#

When an electron is captured by \(A\), it usually results in an excited state of nucleus \(B\). This means that the electron must have an energy of

\[\mathcal{E} = Q_0 + \delta Q\]

where \(Q_0 = \mbox{mass}(A) - \mbox{mass}(B)\) is the mass difference between the nuclei, and \(\delta Q\) depends on the energy level structure of the nucleus.

Note

In an degenerate gas, the most energetic electrons will have \(\mathcal{E} \sim \mathcal{E}_F\) (the Fermi energy).

For a \(\beta\)-decay, it is the same idea. The electron that is emitted needs to have a place to go. For a degenerate gas, all of the states up to \(\mathcal{E}_F\) will be filled, so the electron can only decay when \(Q_0 > \mathcal{E}_F\).

So approximately, we can imagine:

  • \(\mathcal{E}_F > Q_0\) : electron-capture possible

  • \(\mathcal{E}_F < Q_0\) : beta-decay possible

Estimating the crticial density#

We expect the density where \(\mathcal{E}_F \sim Q\) to be high, so we will assume we are relatvistic. Then

\[\mathcal{E}_F = p_F c = \left ( \frac{3}{8\pi} \right )^{1/3} h c n_e^{1/3} \sim Q\]

and taking:

\[n_e = \frac{\rho Y_e}{m_u}\]

we have

\[\rho \sim \frac{8\pi}{3} \frac{m_u}{(hc)^3} \frac{Q^3}{Y_e}\]

Now we can put in numbers. We’ll take:

  • \(Q \sim 4~\mathrm{MeV}\)

  • \(Y_e \sim 1/2\)

import math

h = 6.63e-27
c = 3.e10
m_u = 1.66e-24
Y_e = 0.5
Q = 4 * 1.6e-6  # convert to erg

rho = 8 * math.pi / 3 * m_u / (h * c)**3 * Q**3 / Y_e
rho

926599926.1391867

So we would expect

\[\rho \sim 10^9~\mathrm{g~cm^{-3}} \left (\frac{Q}{4~\mathrm{MeV}}\right )^3 \left (\frac{Y_e}{0.5} \right )^{-1}\]

This means that we don’t expect electron captures until our density is really high. We will need to accrete a while before the electron-captures kick in.

Sources for rates#

Computing the rates requires knowing the detailed nuclear structure of the nuclei involved. This means knowing:

  • the energy level structure

  • the spins for the different states (to see if transitions are allowed)

  • the half life of excited states / transition widths

These are usually computed via theoretical nuclear structure calculations.

The paper The importance of Urca-process cooling in accreting ONe white dwarfs (Schwab et al. 2017) gives a bit of detail into the calculations.

For these labs, we will primarily use the rates from Electron-capture and β-decay Rates for sd-Shell Nuclei in Stellar Environments Relevant to High-density O-Ne-Mg Cores (Suzuki et al. 2016)