Subatomic Physics

506 Liquid Drop Model, Fermi Gas Model, Heavy Ions
16.2 The Fermi Gas Model
The semiempirical binding-energy relation obtained in the previous section is based
on treating the nucleus like a liquid drop. Such an analogy is an oversimplification,
and the nucleus has many properties that can be explained more simply in terms
of independent-particle behavior rather than in terms of the strong-interaction picture
implied by the liquid drop model. The most primitive independent-particle
model is obtained if the nucleus is treated as a degenerate Fermi gas of nucleons.
The nucleons are assumed to move freely, except for effects of the exclusion principle,
throughout a sphere of radius R = R0A 1/3 ,R0 ≈ 1.2 fm. The situation is
represented in Fig. 16.4 by two wells, one for neutrons and one for protons.
Figure 16.4: Nuclear square wells for neutrons and protons.
The well parameters are adjusted to give the observed binding
energy B ′ .
Free neutrons and free protons,
far away from the wells,
have the same energy, and
the zero levels for the two
wells are the same. The two
wells however, have different
shapes and different depths
because of the Coulomb energy,
Eq. (8.37): The bottom
of the proton well is higher
than the bottom of the neutron
well by an amount Ec,
and the proton potential has
a Coulomb!barrier.
Protons that try to enter the nucleus from the outside are repelled by the positive
charge of the nucleus; they must either “tunnel” through the barrier or have enough
energy to pass over it.
The wells contain a finite number of levels. Each level can be occupied by two
nucleons, one with spin up and one with spin down. It is assumed that, under
normal conditions, the nuclear temperature is so low that the nucleons occupy the
lowest states available to them. Such a situation is described by the term degenerate
Fermi gas. The nucleons populate all states up to a maximum kinetic energy equal
to the Fermi energy EF . The total number, n, of states with momenta up to pmax
follows from Eq. (10.25), after integration over d 3 p,as
n = Vp3max 6π2 . (16.11)
�3 Each momentum state can accept two nucleons so that the total number of one
species of nucleons with momenta up to pmax is 2n. If neutrons are considered,
then 2n = N, the number of neutrons, and N is given by