Sargent rule

In nuclear physics , the Sargent rule (also called the rule) provides a relationship between the maximum energy released during beta decay and the decay constant . For some nuclides there are large deviations from the Sargent rule due to selection rules . It was named after its discoverer, the Canadian nuclear physicist Bernice Weldon Sargent . ${\displaystyle Q^{5}}$ ${\displaystyle Q}$ ${\displaystyle \Gamma }$

Empirical discovery

In the 1930s, Sargent dealt with the emission spectrum of beta decay of various substances. He found out that the decay constant is proportional to the fifth power of the maximum kinetic energy of the electron: ${\displaystyle \Gamma }$

${\displaystyle \Gamma \sim Q^{5}.}$

The theoretical explanation was later found by Enrico Fermi .

Derived from Enrico Fermi

The starting point for the theoretical derivation is provided by Fermi's Golden Rule in combination with Fermi's Theory of Weak Interaction .

Fermi's golden rule

${\displaystyle \Gamma _{N\rightarrow P}={\frac {2\pi }{\hbar }}|{\mathcal {M}}_{N\rightarrow P}|^{2}\int \limits _{M_{e}c^{2}}^{Q}dE{\frac {d\rho }{dE}}}$

with Planck's quantum of action , the transition matrix element or the probability amplitude for the weak decay of a neutron into a proton and the phase space factor . The electron mass multiplied by the vacuum speed of light represents the lower integration limit. The upper integration limit ensures that the entire spectrum is taken into account. ${\displaystyle \hbar }$ ${\displaystyle {\mathcal {M}}_{N\rightarrow P}}$ ${\displaystyle \rho }$ ${\displaystyle M_{e}}$ ${\displaystyle Q}$

Transition matrix element ℳ _{N → P}

There are two possible transitions for beta decay, the Fermi transition, which originates from the vector part of the weak interaction, and the Gamow-Teller transition, which has its origin in the axial vector coupling.

${\displaystyle |{\mathcal {M}}_{N\rightarrow P}|^{2}=(g_{V}^{2}+3g_{A}^{2})/V^{2}}$

With the axial transition, all spin degrees of freedom are important. This is taken into account by the factor 3 before the axial coupling . With vector coupling, on the other hand, the spin degrees of freedom are not relevant, since it is a pure vector without an axial component (angular momentum are axial vectors). is the volume of the spatial space in which the decay takes place. ${\displaystyle g_{A}^{2}}$ ${\displaystyle g_{V}^{2}}$ ${\displaystyle V}$

Phase space and density of states

Because of the large mass of the core , it can be approximately excluded from the kinematic considerations. The final state is then described by a two-particle phase space (electron and anti- neutrino ). One element of the phase space has the volume. The integrals over the spatial space can be carried out directly and together provide a factor . In order to execute the integrals in momentum space , the dispersion relations for the electron and the neutrino ( here assumed to be massless ) are required: ${\displaystyle d^{3}x_{e}d^{3}x_{\nu }d^{3}p_{e}d^{3}p_{\nu }/(2\pi \hbar )^{6}.}$ ${\displaystyle V^{2}}$

${\displaystyle p_{e}^{2}dp_{e}={\frac {1}{c^{3}}}E_{e}{\sqrt {E_{e}^{2}-M_{e}^{2}c^{4}}}dE_{e},}$

${\displaystyle p_{\nu }^{2}dp_{\nu }={\frac {1}{c^{3}}}E_{\nu }^{2}dE_{\nu }.}$

By normalizing to the rest energy of the electron, the phase space integration can be applied to the form

${\displaystyle \int \limits _{M_{e}c^{2}}^{Q}dE{\frac {d\rho }{dE}}=V^{2}{\frac {M_{e}^{5}c^{4}}{4\pi ^{4}\hbar }}f({\mathcal {E}}_{0})}$

with and ${\displaystyle {\mathcal {E}}_{0}=Q/M_{e}c^{2}~~~~}$ ${\displaystyle ~~~~f({\mathcal {E}}_{0})=\int \limits _{1}^{{\mathcal {E}}_{0}}d{\mathcal {E}}{\mathcal {E}}{\sqrt {{\mathcal {E}}^{2}-1}}({\mathcal {E}}_{0}-{\mathcal {E}})^{2}}$

to be brought. If one also takes into account the relationship between the decay constant and the lifetime , one obtains together with the matrix element ${\displaystyle {\mathcal {M}}_{N\rightarrow P}}$

${\displaystyle {\frac {1}{\tau }}={\frac {M_{e}^{5}c^{4}}{2\pi ^{3}\hbar ^{7}}}(g_{V}^{2}+3g_{A}^{2})\cdot f({\mathcal {E}}_{0}).}$

Sargent rule

For large energy ( ) the following applies approximately ${\displaystyle Q\gg M_{e}c^{2}}$

${\displaystyle f({\mathcal {E}}_{0})\approx {\frac {{\mathcal {E}}_{0}^{5}}{30}}}$

and thus

${\displaystyle {\frac {1}{\tau }}={\frac {1}{\hbar ^{7}c^{6}}}\cdot (g_{V}^{2}+3g_{A}^{2})\cdot {\frac {Q^{5}}{60\pi ^{3}}}.}$

literature

• Particles and Cores , Povh, Rith, Scholz, Zetsche, 8th edition: pages 232ff., 2009