Subfield: Modern Physics

3. Simple model of an atomic nucleus 

Introduction

Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below.

Task 1 - Atomic nucleus as closely packed system of nucleons

In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter.

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Context question:
a) Calculate what would be the packing factor $f$ if nucleons were arranged in a "simple cubic" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)].
Context answer:
\boxed{$\frac{\pi}{6}$}


Extra Supplementary Reading Materials:

Important: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$.
Context question:
b) Estimate the average mass density $\rho_{m}$, charge density $\rho_{c}$, and the radius $R$ for a nucleus having $A$ nucleons. The average mass of a nucleon is $1.67 \cdot 10^{-27} \mathrm{~kg}$.
Context answer:
\boxed{$3.40 \cdot 10^{17}$ , $1.63 \cdot 10^{25} $ , $1.06$}


Extra Supplementary Reading Materials:

Task 2 - Binding energy of atomic nuclei - volume and surface terms

Question: Binding energy of a nucleus is the energy required to disassemble it into separate nucleons and it essentially comes from the attractive nuclear force of each nucleon with its neighbors. If a given nucleon is not on the surface of the nucleus, it contributes to the total binding energy with $a_{V}=15.8$ $\mathrm{MeV}\left(1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right)$. The contribution of one surface nucleon to the binding energy is approximately $a_{V} / 2$. Express the binding energy $E_{b}$ of a nucleus with $A$ nucleons in terms of $A, a_{V}$, and $f$, and by including the surface correction.

Solution:
First one needs to estimate the number of surface nucleons. The surface nucleons are in a spherical shell of width $2 r_{N}$ at the surface. The volume of this shell is

$$
\begin{aligned}
V_{\text {surface }} & =\frac{4}{3} R^{3} \pi-\frac{4}{3}\left(R-2 r_{N}\right)^{3} \pi= \\
& =\frac{4}{3} R^{3} \pi-\frac{4}{3} R^{3} \pi+\frac{4}{3} \pi 3 R^{2} 2 r_{N}-\frac{4}{3} \pi 3 R 4 r_{N}^{2}+\frac{4}{3} \pi 8 r_{N}^{3} \\
& =8 \pi R r_{N}\left(R-2 r_{N}\right)+\frac{4}{3} \pi 8 r_{N}^{3}= \\
& =8 \pi\left(R^{2} r_{N}-2 R r_{N}^{2}+\frac{4}{3} r_{N}^{3}\right)
\end{aligned}
\tag{8}
$$
The number of surface nucleons is:
$$
\begin{aligned}
A_{\text {surface }} & =f \frac{V_{\text {surface }}}{V_{N}}=f \frac{8 \pi\left(R^{2} r_{N}-2 R r_{N}^{2}+\frac{4}{3} r_{N}^{3}\right)}{\frac{4}{3} r_{N}^{3} \pi}= \\
& =f 6\left(\left(\frac{R}{r_{N}}\right)^{2}-2\left(\frac{R}{r_{N}}\right)+\frac{4}{3}\right)= \\
& =f 6\left(\left(\frac{A}{f}\right)^{2 / 3}-2\left(\frac{A}{f}\right)^{1 / 3}+\frac{4}{3}\right)= \\
& =6 f^{1 / 3} A^{2 / 3}-12 f^{2 / 3} A^{1 / 3}+8 f= \\
& =6^{2 / 3} \pi^{1 / 3} A^{2 / 3}-2 \cdot 6^{1 / 3} \pi^{2 / 3} A^{1 / 3}+\frac{4}{3} \pi \approx \\
& \approx 4.84 A^{2 / 3}-7.80 A^{1 / 3}+4.19 .
\end{aligned}
\tag{9}
$$
The binding energy is now:
$$
\begin{aligned}
& E_{b}=\left(A-A_{\text {surface }}\right) a_{V}+A_{\text {surface }} \frac{a_{V}}{2}= \\
& =A a_{V}-A_{\text {surface }} \frac{a_{V}}{2}= \\
& =A a_{V}-\left(3 f^{1 / 3} A^{2 / 3}-6 f^{2 / 3} A^{1 / 3}+4 f\right) a_{V}= \\
& =A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}= \\
& =\left(15.8 A-38.20 A^{2 / 3}+61.58 A^{1 / 3}-33.09\right) \mathrm{MeV}
\end{aligned}
\tag{10}
$$

Final answer: $E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}$