Subatomic Physics

11.1. The Continuous Beta Spectrum 333
Fig. 11.1 is thus qualitatively explained. To avoid the problems posed by electrons
inside nuclei, Fermi postulated that the electron and the neutrino were created in
the decay, just as a photon is created when an atom or a nucleus decays from an
excited to the ground state or two photons are created in the decay of the neutral
pion.
Fermi did not simply speculate how beta decay could occur; he performed the
computations to find the expressions for the electron spectrum and the decay probability.
His original treatment (2) is above our level, and it has to be watered down
here. In the present section, we shall show that even a crude approach reproduces
the shape of the beta spectrum. Since the interaction responsible for beta decay
is weak, perturbation theory can be used, and the transition rate is given by the
golden rule, Eq. (10.1),
dwβα = 2π
� |〈β|Hw|α〉| 2 ρ(E).
Here Hw is the Hamiltonian responsible for beta decay, and we have written dwβα
rather than wβα in order to indicate that we are interested in the transition rate
for transitions with electron energies between Ee and Ee + dEe. We first consider
the density-of-states factor ρ(E). Three particles are present in the final state, and
ρ(E) is given by Eq. (10.34) as
ρ(E) =
V 2
(2π�) 6
d
dEmax
�
p 2 e dpe dΩep 2 ¯ν dp¯ν dΩ¯ν. (11.1)
V is the quantization volume. Since the final results are independent of this volume,
it is set equal to 1. The differentiation d/dEmax requires a word of explanation.
Emax is constant, and it thus appears at first sight that d/dEmax should
vanish. However, it has the meaning of a variation; (d/dEmax) � ··· indicates how
the integral changes under a variation of the maximum energy.
To evaluate ρ(E), we must first decide what we are interested in. Figure 11.1
shows the number of electrons emitted with an energy between Ee and Ee + dEe.
To calculate the corresponding transition rate, Ee and consequently also pe are kept
constant. The d/dEmax in Eq. (11.1) then does not affect the terms relating to the
electron, and Eq. (11.1) becomes
ρ(E) = dΩe dΩ¯ν
(2π�) 6 p2e dpep 2 dp¯ν
¯ν
dEmax
. (11.2)
The next step is simplified by the fact that the nucleon in the final state is much
heavier than either lepton and therefore receives very little recoil energy. To a good
approximation, electron and neutrino share the total energy:
Ee + E¯ν = Emax. (11.3)
2 E. Fermi, Z. Physik 88, 161 (1934); translated in The Development of Weak Interaction
Theory (P. K. Kabir, ed.), Gordon and Breach, New York, 1963. See also L. M. Brown and H.
Rechenberg, Am. J. Phys. 56, 982 (1988).