Download Physics 228 Today: April 22, 2012 Ch. 43 Nuclear

Physics 228 Today: April 22, 2012
Ch. 43 Nuclear Physics
Website: Sakai 01:750:228 or www.physics.rutgers.edu/ugrad/228
Wednesday, April 24, 2013
Nuclear Sizes
Nuclei occupy the center of the atom. We can view them as being
more or less spherical, with a typical radius of a few times 10-15 m
- a few fm (pronounced femtometer or "Fermis").
Studies of nuclei have revealed that the protons and neutrons, are
strongly attracted to each other, with the result that they are
packed densely into a nucleus. It is a semi-reasonable
approximation for nuclei to consider them to be spherical, and
made up of dense hard-packed spheres, protons and neutrons.
Packing A spheres into a nucleus gives a volume proportional to A,
and a radius proportional to A1/3. It has been found that the
radius is about r = R0A1/3 with R0 = 1.2 fm.
Nuclear matter is the densest matter known. For normal nuclei,
ρ ≈ 0.14 nucleons/fm3 ≈ 2.3x1017 kg/m3, vs 103 kg/m3 for water
and about 104 kg/m3 for the densest atoms. Neutron stars are
believed to be about 5 times denser than normal nuclei, based on
theoretical extrapolations. (Note neutron stars exist because they are bound
by gravity, not the nuclear force.)
Wednesday, April 24, 2013
Nuclear Masses
It is a good starting approximation that the mass of a nucleus is
simply proportional to the number of nucleons in the nucleus, A.
But the nucleons are bound into the nucleus, so the mass of the
nucleus is less than the sum of the masses of the constituent
nucleons.
The mass of a nucleus is roughly proportional to the number of
nucleons times 931.5 MeV = 1.66x10-27 kg = 1 u, an atomic mass
unit. You can see from this that nucleons are bound into a nucleus
by ≈ 939 - 931 = 8 MeV, about 1% of their mass. The a.m.u. is
defined so that the mass of the 12C atom is exactly 12.
Note we have the same thing for electrons in atoms, the mass of
the atom is less than the mass of its constituents, but there the
effect is about 10 eV / 511,000 eV, or about 2x10-3%, if we use
the electron mass, or much less using the atomic mass.
Wednesday, April 24, 2013
Binding Energy
In determining nuclear masses, we usually deal with atoms, so
tabulated masses are usually for atoms. The atomic electron masses
and binding energy make a small difference, but usually if we
ignore the inclusion of electrons it is not an issue, since the
number of nucleons and electrons is constant for all processes we
will consider here.
The total nuclear binding energy is the total energy needed to
separate a nucleus into its constituents, Z free hydrogen atoms plus
N free neutrons: EB = (ZmH + Nmn - AM)c2.
One can also refer to the binding energy of the most loosely bound
proton or neutron.
If all the masses are known in a.m.u., the binding energy
calculations shown is straightforward, with the binding energy
typically given in MeV. This uses mp = 1.007825 u, mn = 1.008665 u,
md = 2.014102 u, etc. The deuteron binding energy is 1.007825 u +
1.008665 u - 2.014102 u = 0.002388 u = 2.224 MeV.
Wednesday, April 24, 2013
Mass Excess?
Since the nuclear mass is nearly A u, and since the number of
nucleons is conserved, it is common for practicing nuclear
physicists to use the concept of mass excess. The mass excess Δ is
the difference between the mass and the approximation that the
mass is A u. To have numbers of order 1, units of MeV are used.
The proton mass excess is 7.289 MeV. The neutron mass excess is
8.071 MeV. The deuteron mass excess is 13.136 MeV.
Thus, the binding energy for the deuteron is 7.289+8.071-13.136 =
2.224 MeV. Since the number of nucleons is constant, the two
average nucleon masses would be added and subtracted, and thus
cancel out, if we included them.
Note also to get the deuteron mass using amu, we needed to use
7-digit long numbers. Here we needed 5 digits to get the same
precision answer, so it is more convenient.
Wednesday, April 24, 2013
Binding Energy / Nucleon
The total binding energy divided by the number of nucleons is the
binding energy per nucleon. It is maximum for medium mass nuclei.
Thus, if two light nuclei fuse into a heavier nucleus, energy is
released. Or if a heavy nucleus splits into two light ones, energy is
released.
This pattern
provides the
underlying reason
why the sun
generates light, and
why nuclear
reactors and
weapons work.
Wednesday, April 24, 2013
The jagged pattern is
due to QM shell effects.
Saturation
The pattern of the binding energy per nucleon being about constant
is an indication of the saturation of the nuclear force. The force is
short ranged, about 1 - 2 fm, so each nucleon largely interacts with
its nearest neighbors rather than with the nucleus as a whole.
As a result, nucleons
in large nuclei all
have about the
same number of
neighbors and the
binding energy per
nucleon is about
constant.
Wednesday, April 24, 2013
The jagged pattern is
due to QM shell effects.
The Nuclear Force
The textbook, on the nuclear force: "physicists have yet to
determine its dependence on the separation r."
This is somewhat misleading.
In old fashioned mid-1900s quantum mechanics the force between
nucleons arises from nucleons exchanging pions. This has a definite
spatial dependence: UYukawa = -C e-mr / r. The exponential converts a
Coulomb-type potential into a short range "screened" Coulomb
potential. The mass of the pion appears as the factor of m in the
exponential.
This idea arises nicely from Heisenberg's uncertainty principle.
A "virtual" photon exchanged in the Coulomb interaction can have m
= ΔE ≈ 0, so Δt can be ≈ ∞, you can have a long range force.
The pion mass is 140 MeV, so a "virtual" pion can live only a short
time and travel a short range. Crudely...
Δx = cΔt = ħc/2ΔE = 200 MeV.fm / 300 MeV ≈ 1 fm.
Wednesday, April 24, 2013
The Nuclear Force
Now it is also understood that subatomic forces cannot be
expressed only as functions of r.
They have terms that depend on r (central force), terms that
depend on momentum (derivatives of positions, if you think back to
the Schrödinger Eq.) (nonlocal force), and terms that depend on the
alignment of the spins of the two nucleons (tensor force), and terms
that depend on the alignment of the spin and orbital angular
momentum of the two nucleons (spin-orbit force).
You might recall things of this sort from our discussions of the EM
force and atomic structure.
Wednesday, April 24, 2013
iClicker
By what factor is the radius of the
nucleus 160Dy bigger than that of 20Ne?
a) <1 - because of the strength of the
nuclear force, nuclei get smaller as you
add more nucleons
b) 1 - all nuclei are about the same
size
c) 2
d) 4
e) 8
Wednesday, April 24, 2013
Nucleon Magnetic Moments
Protons, neutrons, and nuclei also have magnetic moments.
Protons and neutrons are spin-1/2, the same as the electron.
For the proton, similar to the electron, we would expect if it is a
point-like particle for its magnetic moment to be one nuclear
magneton, defined as μn = eħ/2mp = 3.152x10-8 eV/T. For the
neutron, since it is charge 0, we would expect it to have no
magnetic moment.
But actually protons and neutrons have a complex internal
structure, as do nuclei and atoms, so their magnetic moments are
not so simply related to their spins.
For the proton, μ = 2.793 μn, while for the neutron, μ = -1.913 μn.
The "-" sign for the neutron indicates that its magnetic moment is
directed opposite to its spin, unlike the proton.
Wednesday, April 24, 2013
Nuclear Magnetic Moments
For the proton, μ = 2.793 μn, while for the neutron, μ = -1.913 μn.
The "-" sign for the neutron indicates that its magnetic moment is
directed opposite to its spin, unlike the proton and electron.
Nuclear magnetic moments arise from the proton and neutron
moments and their orbital motions.
As with the case of atomic electrons, most nucleons form pairs
whose total angular momentum, and total magnetic moment, add to
0. As a result, nuclear magnetic moments often result from only a
few nucleons, and are typically of size a few nuclear magnetons.
In addition, particularly for heavier nuclei, the "entire" nucleus
can rotate leading to orbital angular momenta, like for molecules.
Wednesday, April 24, 2013
MRI and Nuclear Magnetic Moments
Wednesday, April 24, 2013
The Liquid Drop Model
An old idea is that nuclei act like quantum liquids (George Gamow,
1928). This leads to the "semi-empirical mass formula":
2
Z(Z
−
1)
(A
−
2Z)
2/3
−4/3
EB = C1 A − C2 A − C3
−
C
±
C
A
4
5
A
A1/3
C1: 16 MeV - the binding of a nucleon to a complete set of nearest neighbors
C2: 18 MeV - nucleons on the surface do not have neighbors outside to bind to
C3: 0.7 MeV - the Coulomb potential energy of protons
C4: 24 MeV - the symmetry energy... because there are different states for
protons and for neutrons, nuclei with N ≈ Z are more bound (neglecting
Coulomb)
C5: 39 MeV - protons bound in pairs and neutrons bound in pairs are more
bound in nuclei (the two of a pair are in the same quantum state); this term is
positive for even numbers of both, negative for odd numbers of both, and 0 for
one type of nucleon odd, the other even
Wednesday, April 24, 2013
Odd and Even Nuclei
2
Z(Z
−
1)
(A
−
2Z)
2/3
−4/3
EB = C1 A − C2 A − C3
−
C
±
C
A
4
5
A
A1/3
Nuclear pairing is such an important feature that most nuclei in
nature have even numbers of protons and neutrons.
A significant fraction are odd-even, with an even number of one of
the nucleon types, and an odd number of the other.
But there are only a few stable odd-odd nuclei. These are 2H, 6Li,
10B, and 14N, all light nuclei which we can better understand from
ideas of finite quantum mechanical potential wells than from the
idea of a liquid drop.
Wednesday, April 24, 2013
Vibrations and Rotations
Thinking of nuclei as liquid drops, you can imagine classically how
the liquid can have various vibrations and rotations. There are in
fact quantum vibrational, "(n+1/2)ħω", and rotational, l(l+1)ħ2/2I
spectra, particularly for heavier nuclei.
Wednesday, April 24, 2013
The Shell Model
The main "central" nuclear potential is
shaped something like a finite square well.
Because of the curvature it is often
thought of as a finite harmonic oscillator
well.
This leads to the idea of a set of nuclear
states just like the states of a finite well.
For the atom, we found that for particular
numbers of electrons, major atomic shells
(K, L, M, ...) were filled and the atoms were
particularly stable and non-reacting.
For nuclei, a similar phenomena is
observed. This leads to the "shell" model getting the spacing about right requires
using mainly the central force and a spinorbit force, that lowers the energy of
orbits where S || L, not anti-parallel.
Wednesday, April 24, 2013
Magic Numbers
For atoms, the "magic numbers" were Z=2, 10, 18, 36
corresponding to filling 1s, 2s+2p, 3s+3p, 4s+4p+3d, ... shells.
For nuclei, the typical naming of orbits is different, the order
in which they fill is different, and the doubly "magic nuclei"
are 4He with 1s1/2 shell filled for p+n, 16O with 1s1/2 + 1p3/2
+ 1p1/2 shell filled for p+n, 40Ca with ... + 1d5/2 + 2s1/2 +
1d3/2 orbits filled for p+n, ... 48Ca with ... + 1f7/2 filled with
n, ...
These nuclei have higher binding energies than their
neighbors, and also the excited states are further away
from the ground states than non-magic nuclei.
Wednesday, April 24, 2013
iClicker
Which of the following nuclei likely has
the largest binding energy per nucleon?
a)
15
b)
14 N
7
c)
12
d)
13
e)
16
7N
6C
6C
8O
Wednesday, April 24, 2013