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Subatomic Physics

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336 The Weak Interaction<br />

total transition rate wβα and the mean life τ can be obtained from Eq. (11.7) by<br />

integration over the momentum:<br />

w = 1<br />

τ =<br />

1<br />

2π3c3�7 |〈N ′ eν|Hw|N〉| 2<br />

� pmax<br />

dpeF (∓,Z,Ee)p<br />

0<br />

2 e(Emax − Ee) 2 . (11.8)<br />

For very large energies, where Emax ≈ cpmax, andforsmallZ, whereF ≈ 1, the<br />

integral becomes<br />

� pmax<br />

0<br />

dpep 2 e(Emax − Ee) 2 � 1<br />

30c 3 E5 max. (11.9)<br />

While this relation is useful for estimates, accurate values of the integral are<br />

needed for a meaningful treatment of the data. Fortunately the integral has been<br />

tabulated (3) as<br />

� pmax<br />

dpeF (∓,Z,Ee)p<br />

0<br />

2 e(Emax − Ee) 2 = m 5 ec 7 f(Emax). (11.10)<br />

The factor m 5 e c7 has been inserted in order to make f dimensionless. With<br />

Eqs. (11.10) and (11.8), the matrix element becomes<br />

|〈N ′ eν|Hw|N〉| 2 = 2π3<br />

fτ<br />

�7 m5 . (11.11)<br />

ec4 If τ is measured and f computed (3) then the square of the matrix element can be<br />

obtained from Eq. (11.11). It is customary to use ft 1/2 and not fτ in tabulations.<br />

ft 1/2 is called the comparative half-life. The name stems from the fact that all<br />

beta-decaying states would have the same value of ft 1/2 if all matrix elements were<br />

equal. Nature provides an enormous range of values of ft 1/2, fromabout10 3 to<br />

10 23 sec. If such a variation were caused by the fact that the weak interaction, Hw,<br />

were not universal but would change from decay to decay, an understanding of the<br />

weak processes would be hopeless. It is assumed that Hw is the same for all decays<br />

and that the nuclear wave functions that enter the calculation of 〈N ′ eν|Hw|N〉 are<br />

responsible for the variations. The most fundamental decays have the “best” wave<br />

functions and give rise to the largest matrix elements. A few cases are listed in<br />

Table 11.1.<br />

With ft 1/2 =(ln2)fτ (Eq. (5.34)) and with the numerical values of the constants,<br />

Eq. (11.11) becomes<br />

|〈N ′ eν|Hw|N〉| 2 = 43 × 10−6 MeV 2 fm 6 sec<br />

. (11.12)<br />

ft1/2(in sec)<br />

Now consider the decay of the neutron. With the value of ft 1/2 given in Table 11.1,<br />

the magnitude of the matrix element of Hw becomes<br />

|〈pe¯ν|Hw|n〉| ≈ 2 × 10 −4 MeV fm 3 . (11.13)

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