1
Nuclear Astrophysics
Nuclear energies: The Liquid Drop Model.
Nuclei have internal baryon densities n ' ns = 0.16 fm−3 , corresponding
to mass densities ρs = nsmn ' 2.7 · 1014 g cm−3 . They also have roughly
equal numbers of neutrons and protons: x = Z/A ' 1/2.
E (Z, N ) = Ebulk A + Esurf A2/3 + ECoul A5/3 + · · ·
K
n 2
+ SV (1 − 2x)2 + · · ·
1−
Ebulk ' −16 +
18
ns
K ≈ 240 MeV is incompressibility parameter, SV ≈ 30n/ns MeV is volume
symmetry parameter.
Esurf ' 18 − SS (1 − 2x)2 MeV,
where SS ' 45 MeV is the surface symmetry parameter.
ECoul =
3 x 2 e2
' 0.75x2 MeV,
5 r0
where r0 = (4πns /3)−1/3 ' 1.12 fm.
Consider an infinite nucleus. Saturation density is where energy per
particle for a given composition is minimized:
∂Ebulk P
= 0.
= bulk
∂n x
n2
This occurs where pressure vanishes, or n = ns if x = 1/2. If x 6= 1/2, then
n
∂SV
Pbulk
K
1
−
+
(1 − 2x)2 ,
=
−
2
9ns
ns
∂n
n
9nSV0 (1 − 2x)2 n
' 1−
' 1 − 1.1 (1 − 2x)2 .
ns
K
ns
Optimize energy per particle with respect to composition:
∂Ebulk = − µn − µp = −4SV (1 − 2x) = 0,
∂x n
that is, x = 1/2 for all n. However, we’ve neglected electrons. Optimum
composition determined by:
1/3
2
µn − µp = µe , 4SV (1 − 2x) = h̄c 3π nx
.
2
SV is assumed to be S0 n/ns , so as n → 0, x → 0. But for n → ∞,
2
n = 3π x
2/3 3/2
h̄cns
4S0 (1 − 2x)
where S0 = 30 MeV. Thus x → 1/2 in this limit. When n = ns , x ≈ 0.04.
A finite nucleus has an optimum mass, for a given x:
Esurf
∂E/A 2ECoul
= 0.
= − 4/3 +
∂A x
3A
3A1/3
This becomes Esurf A2/3 = 2ECoul A5/3, the so-called Nuclear Virial Theorem. So
Aopt = Esurf / (2ECoul ) .
(1)
This increases with decreasing x near 1/2 roughly as x−2 . For x = 1/2,
Aopt = 18/.375 ' 48. A nucleus also has an optimum charge, for a given A:
∂E/A −1/3
(1 − 2x) + 2ECoul A2/3/x = 0,
= −4 SV − SS A
∂x A

0.75A2/3
−1

xopt = 2 + −1/3
2 SV − S S A
.
(2)
This path represents the Valley of Beta Stability in the Chart of the Nuclides. If there was no Coulomb energy, or the mass is very small, then
xopt = 1/2. For larger masses, xopt decreases from 1/2. The simultaneous
solution of Eqs. (1) and (2) yields xopt ' 0.432, Aopt ' 61, Zopt ' 26, or
61 Fe. Along the Valley of Beta Stability, the nuclear energy is
E/A = − 16 + 18A−1/3 + 0.375xopt A2/3
−1
16
1
−1/3
= − 16 + 18A
+
.
+
3A2/3 SV − SS A−1/3
This rises steeply with A to the maximum, then decreases relatively slowly
beyond the maximum.
3
The binding energies of 4 He and 56Fe, per baryon, are about 7.1 and 8.9
MeV, respectively, relative to individual neutrons and protons. The nuclear
energy release in H burning is far greater than the energy released in all
subsequent burning stages, which end in Fe formation.
Behavior of Nuclei at High Density
We treated nuclei in isolation. By the end of stellar evolution in massive
stars, ρ ' 107 g cm−3 . The filling factor of nuclei is u ' ρ/ρs ' 3.7 · 10−8 ,
so the distance between nuclei is about 2u−1/3 ≈ 600 nuclear radii. This
is large enough that the effective nuclear Coulomb energy is diminished
because of electron screening. For uniformly distributed electrons, the
Coulomb energy becomes:
3 x 2 e2
ECoul =
5 r0
3 1/3 u
.
1− u
+
2
2
Therefore, at high densities, the average nuclear size increases as
(1 − 32 u1/3 + u/2)−1/3 . However, decreasing proton fractions in nuclei lead
to neutron drip.
4
Neutron Drip
The energies of the last neutron and proton in nuclei (also called the
separation energy) are µn and µp :
Enuc Z ∂ (Enuc /A) ∂Enuc µn =
−
=
∂N Z
A
A ∂ (Z/A) A
∂Enuc Enuc
Z ∂ (Enuc /A) µp =
+ 1−
=
.
∂Z N
A
A
∂ (Z/A) A
So
2/3
−1/3
2
µn = − 16 + 18A
+ SV − S S A
1 − 4x − ECoul A2/3,
(2 − x)
2/3
−1/3
ECoul A2/3.
µp = − 16 + 18A
+ SV − S S A
(1 − x) (2x − 3) +
x
When µn > 0, it is energetically favorable for neutrons to “drip” out of
nuclei. This occurs around the density 4 · 1011 g cm−3 when the nuclear
proton fraction drops below about x = 0.3. For small proton fractions,
the surface energy as given earlier becomes unphysically small. A better
parametrization is
h
i−1
−3
−3
Esurf = α x + (1 − x) + β
2
α = 96Esurf
(x = 1/2) /Ss ,
β = 96Esurf (x = 1/2) /Ss − 16.