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\begin{document}
\centerline{\bf Lecture 4}
\bigskip
In last lecture we discussed how one can solve the Schroedinger equation
to determine the allowed energies of a particle trapped in a spherically
symmetric potential. We will apply these methods to the case of neutrons
and protons bound in the nucleus. First we'll discuss some properties
of nuclei that pertain to our problem.
\bigskip
\noindent {\bf Magic Numbers}
\bigskip
The properties of atoms show patterns. Similar chemical properties are
displayed graphically in the periodic table: elements in the same column
have similar chemical properties. The "magic numbers" for atomic properties
are: 2, 6, 10, 14. These numbers can be understood by solving the
Schroedinger equation with a spherically symmetric mean field potential
derived from the Coulomb interaction. The numbers are equal to
$2(2l+1)$, where $l=0,1,2,3, \cdots$. $l$ is the orbital angular momentum
quantum number.
Nuclei also show patterns, and have their own "magic numbers". This is
demonstrated on the next page. The graph is plot of the energy needed to
remove a neutron versus the number of neutrons in the nucleus. The different
curves correspond to the cases when the number of neutrons, $N$, minus the
number of protons, $Z$, equals $1,2,3, \cdots$. The quantity $N-Z$ is the
number of extra neutrons than protons.
\begin{figure}
\includegraphics[width=14cm]{magicnumbers.png}
\end{figure}
From the figure, one can see that energy needed to remove a neutron drops
after $N$ equals $2, 8, 20, 28, 50, 82, \cdots$. These
numbers are the "magic numbers" for nuclei. There is other experimental
evidence that these numbers to bring on changes in nuclear properties.
Because of the magic number phenomena, one is lead to a shell-model
to describe some nuclei. The data suggest that neutrons and protons
each (separately) have closed shells when their number equals the
{\bf nuclear magic numbers}:
\centerline{\bf $2$ or $8$ or $20$ or $28$ or $50$ or $82$}
\newpage
\noindent {\bf Nuclear Shell Model}
\bigskip
The independent particle model, or shell model, of the nucleus
has been successful in understanding many properties of nuclei.
In this model, the nucleons are treated as independent particles
that move in an average potential (mean field) due to the other nucleons in
the nucleus.
The mean field potential consists of a "strong force" part that both
the neutrons and protons experience. Since the protons possess
charge, they will have an electrostatic Coulomb potential in addition
to the strong force.
A common potential that is used to represent the strong interaction between a nucleon
and the rest of the nucleus is the "Woods-Saxon" potential:
\begin{equation}
V(r) = {{- V_0} \over {1 + e^{(r-c)/a}}}
\end{equation}
\noindent This potential is a smoothed out square well, and the potential
strength $V_0$ is the same for both neutrons and protons.
For your homework exercise, you will calculate the bound state energies
for neutrons and protons in various nuclei. To reduce the number of parameters,
we will take use a simple model for the potential. For the strong potential
{\bf we will use a spherical square well potential} as we did for the $\Lambda$
particle. For the electrostatic potential that the proton experiences,
we will use the potential from a uniformly charged sphere of radius $R$
and total charge $Ze$:
\begin{eqnarray*}
V_{Coulomb}(r) & = & Ze^2 {{3R^2-r^2} \over {2R^3}} \hspace{1cm} if \; r \le R \\
& = & {{Ze^2} \over r} \hspace{1cm} if \; r>R
\end{eqnarray*}
\newpage
\noindent {\bf Energy levels in Nuclei} \\
For your homework assignment, you will solve for the energy levels in a
spherical square well. You will see that as the radius of the well
increases, or the potential depth $V_0$ increases that more states
are allowed in the square well. You might wonder, what is the ordering
of these levels. The lowest level (ground state) will have $l=0$, and the function
$u(r)$ will have one maximum within the well. This is the same
result as the ground state for the hydrogen atom. For hydrogen, the
next $l=0$ and the first $l=1$ state have the same energy. However, the three
dimensional square well will not have this property. The second $l=0$
and the first $l=1$ states will not have the same energy (i.e. not
be degenerate). Only for a perfect $1/r$ potential will the bound state
energies be independent of the orbital angular momentum quantum number $l$.
The goal of your homework assignment is to see what the energy level spectrum
for the three dimensional square well is.
On the next page, is a diagram of a calculation for the energy spectrum
for a Woods-Saxon potential. You should get similar results using the
3-D square well with the levels at the extreme left.
The number of allowed states for a given orbital angular momentum is
$2(2l+1)$. There are $2l+1$ values for $m_l$, and an additional factor
of $2$ due to the electron having a spin of $1/2$.
\begin{center}
\begin{tabular}{c|c}
$l$ & number of states = 2(2l+1) \\
\hline
0 & 2 \\
1 & 6 \\
2 & 10 \\
3 & 14 \\
\end{tabular}
\end{center}
\begin{figure}
\includegraphics[width=14cm]{shellmod.png}
\end{figure}
\newpage
\noindent {\bf Spin Orbit interaction}\\
Using the ordering of the levels on the left, the numbers are not
magical. Since two neutrons can occupy the same level, closed shells
occur when the neutron number is $2, 8, 28, 34, \cdots$. The agreement
starts out OK, but fails with $34$. To correct the problem, Maria Mayer
and Hans Jenkens suggested that a spin-orbit correction might be strong enough to
produce closed shells which match with experiment. The ordering of the
nuclear levels including the spin-orbit term are shown at the right in
the diagram. As you can see, the addition of the spin-orbit correction
results in the correct magic numbers. In the assignment, your task is to
see if the level ordering without the spin-orbit correction agrees with the
left column.
To check your code, I will give you results for two nuclei: the valance neutron
in $^{13}C$ and the valance proton in $^{13}N$. Using $R=1.28A^{1/3}$ I get
\begin{center}
\begin{tabular}{c|c|c|c|c|c}
A & Z & $V_0$(MeV) & $l$ & $|E|_{calculated}$(MeV) & $|E|_{experiment}$(MeV) \\
\hline
12 & 0 & 32.3 & 1 & 4.97 & 4.95 \\
12 & 6 & 32.3 & 1 & 1.79 & 1.94 \\
\end{tabular}
\end{center}
\noindent which match the experimental data quite well. As expected the energy
levels for the proton are higher than for the neutron. This phenomena explains
why neutrons are stable when they are bound in nuclei. In free space the
neutron will undergo beta decay into a proton via the weak interaction:
\begin{equation}
n \rightarrow p + e + \bar{\nu}_e
\end{equation}
\noindent since the mass of the neutron is greater than the sum of the proton and electron
mass. The lifetime is around 12 minutes. However, the valance neutron in $^{13}C$ cannot
beta decay into a proton, since in the resulting $^{13}N$ nucleus the proton will be
at an energy around $3$ MeV higher than the initial neutron. Neutrons should be grateful
to the Coulomb repulsion of the protons, which gives them their stability when bound in
nuclei.
In the following table, I list results of a Cal Poly Pomona senior project. We used a
Woods-Saxon potential, with $a=0.6 fm$, for the strong interaction and found values of
$V_0$ and $R=r_0A^{1/3}$ that agreed with experiment for valance nucleons.
\newpage
\noindent{\bf Table III.}
Experimental and calculated values for the valence nucleon. The binding
energies are given in units of MeV.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|} \hline
Core & & Binding Energy of & Binding Energy of &
& \\
Nucleus & l(jP) & Valence Neutron & Valence Proton & $r_0$(fm) &
$V_0$(MeV) \\
\hline
He(4) & ----- & ----- & ----- & ----- & ----- \\
Li(6) & 1(3/2-) & 7.250 & 5.606 & 1.52 & 43.11 \\
Be(8) & 1(3/2-) & 1.665 & ----- & ---- & ----- \\
B(10) & 1(3/2-) & 11.454 & 8.689 & 1.34 & 47.41 \\
\hline
C(12) & 1(1/2-) & 4.946 & 1.943 & 1.29 & 35.11 \\
N(14) & 1(1/2-) & 10.833 & 7.297 & 1.33 & 40.11 \\
\hline
O(16) & 2(5/2+) & 4.144 & .601 & 1.34 & 47.71 \\
F(18) & 2(1/2+) & 10.432 & 6.412 & 1.36 & 52.64 \\
Ne(20) & 2(3/2+) & 6.761 & 2.431 & 1.32 & 46.80 \\
Na(22) & 2(3/2+) & 12.419 & 7.580 & 1.32 & 53.09 \\
Mg(24) & 2(5/2+) & 7.331 & 2.271 & 1.29 & 44.91 \\
Al(26) & 2(5/2+) & 13.058 & 7.463 & 1.28 & 51.81 \\
\hline
Si(28) & 0(1/2+) & 8.473 & 2.748 & 1.28 & 46.54 \\
P(30) & 0(1/2+) & 12.312 & 6.133 & 1.30 & 49.91 \\
\hline
S(32) & 2(3/2+) & 8.641 & 2.276 & 1.27 & 41.42 \\
Cl(34) & 2(3/2+) & 12.644 & 5.896 & 1.29 & 45.05 \\
Ar(36) & 2(3/2+) & 8.789 & 1.857 & 1.27 & 39.29 \\
K(38) & 2(3/2+) & 13.077 & 5.764 & 1.29 & 43.39 \\
\hline
Ca(40) & 3(7/2-) & 8.363 & 1.085 & 1.29 & 51.62 \\
Sc(42) & 3(7/2-) & 12.138 & 4.488 & 1.29 & 55.22 \\
Ti(44) & 3(7/2-) & 9.520 & 1.614 & 1.28 & 51.49 \\
V(46) & 1(3/2-) & 13.001 & 4.767 & 1.31 & 58.03 \\
Cr(48) & 3(5/2-) & 10.582 & 2.085 & 1.27 & 51.14 \\
Mn(50) & 3(5/2-) & 13.687 & 4.885 & 1.27 & 53.66 \\
Fe(52) & 3(7/2-) & 10.683 & 1.599 & 1.25 & 50.11 \\
Co(54) & 3(7/2-) & 14.090 & 4.614 & 1.25 & 53.82 \\
\hline
Ni(56) & 1(3/2-) & 10.247 & 0.694 & 1.27 & 51.34 \\
Cu(58) & 1(3/2-) & 12.763 & 2.887 & 1.28 & 53.16 \\
Zn(60) & 1(3/2-) & 10.231 & 0.454 & 1.31 & 47.21 \\
Ga(62) & 1( ? ) & 12.765 & 2.203 & 1.25 & 53.26 \\
\hline
\end{tabular}
\end{center}
\newpage
\begin{center}
{\bf Electrostatic Potential for a \\
uniformly charged sphere}
\end{center}
Let's remind ourselves how we can obtain the electrostatic potential for
a uniformly charged sphere of charge. Let the total charge be $Q$, and
the radius of the sphere be $R$. Let's first find the electric field
everywhere, then integrate to find the potential.
We can determine the electric field for $r>R$ by
applying Gauss's Law to a spherical surface of radius $r>R$:
\begin{eqnarray*}
\int \vec{E} \cdot \vec{dS} & = & {{Q_{inside}} \over \epsilon_0} \\
4 \pi r^2 |E| & = & {Q \over {\epsilon_0}} \\
|E| & = & {{kQ} \over r^2}
\end{eqnarray*}
\noindent where $k = 1/(4 \pi \epsilon_0)$.
Using a surface within the charged sphere, where $rR$:
\begin{eqnarray*}
V(r) - V(\infty) & = & \int_r^\infty {{kQ} \over r^2} dr \\
V(r) - 0 & = & - {{kQ} \over r} \\
V(r) & = & {{kQ} \over r}
\end{eqnarray*}
\noindent for $r>R$. Now for $r