Subatomic Physics

334 The Weak Interaction
For a massless neutrino, E¯ν = p¯νc, and for constant Ee,
so that
dp¯ν
dEmax
= 1
c
dE¯ν
dEmax
= 1
c ,
ρ(E) = dΩe dΩ¯ν
(2π�) 6 c p2 e p 2 ¯ν dpe. (11.4)
As written, ρ(E) is the density-of-states factor for a transition in which the electron
has a momentum between pe and pe + dpe and is emitted into the solid angle dΩe.
With Eq. (11.3), p 2 ¯ν is replaced by (Emax −Ee) 2 /c 2 . Moreover, if the matrix element
〈β|Hw|α〉 is averaged over the angle between the electron and the neutrino, dwβα
can be integrated over dΩedΩ¯ν and with Eq. (11.4) the result is
dwβα =
1
2π 3 c 3 � 7 |〈pe− ¯ν|Hw|n〉| 2 p 2 e (Emax − Ee) 2 dpe. (11.5)
This expression gives the transition rate for the decay of a neutron into a proton,
an electron, and an antineutrino, with the electron having a momentum between pe
and pe + dpe. Does the expression agree with experiment? Since at this point we
know nothing about the matrix element, the simplest approach is to assume that
it is independent of the electron momentum and to see how the other factors in
Eq. (11.5) fit the observed beta spectra. In principle, then, a function
p 2 e (Emax − Ee) 2 dpe
could be fitted to the experimental data. There exists an easier way: Equation (11.5)
is rewritten into the form
�
dwβα
p2 e dpe
�1/2 �
=const. |〈pe−¯ν|Hw|n〉| 2
�1/2 (Emax − Ee). (11.6)
If the expression on the left-hand side is determined experimentally and plotted
against the electron energy Ee, a straight line results if the matrix element is
momentum-independent. Such a plot is called a Fermi or Kurie plot. Figure 11.2
shows the Kurie plot for the neutron decay. It is indeed a straight line over most of
the energy range. The deviation at the low-energy end was caused by experimental
difficulties in this early experiment: The electron counter had a window 5 mg/cm 2
thick, and it absorbed low-energy electrons. (See Fig. 3.8 and Eq. (3.7).) The
number of electrons shown in Fig. 11.2 is not corrected for this loss.
The technique just described can be applied to beta decays other than that
of the neutron with a small modification. If a nucleus decays by beta emission,
the charged lepton experiences the Coulomb force once it has left the nucleus with
charge Ze. This force will decelerate negative and accelerate positive electrons.