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 $f=1 / 2$.

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 $1.67 \cdot 10^{-27} \mathrm{~kg}$.

Solution:
The mass density of the nucleus is:
$$
\rho_{m}=f \frac{m_{N}}{V_{N}}=0.52 \cdot \frac{1.67 \cdot 10^{-27}}{4 / 3 \cdot\left(0.85 \cdot 10^{-15}\right)^{3} \pi} \approx 3.40 \cdot 10^{17} \frac{\mathrm{kg}}{\mathrm{m}^{3}}
\tag{4}
$$
Taking into account the approximation that the number of protons and neutrons is approximately equal, for charge density we get:
$$
\rho_{c}=\frac{f}{2} \frac{e}{V_{N}}=\frac{0.52}{2} \cdot \frac{1.6 \cdot 10^{-19}}{4 / 3 \cdot\left(0.85 \cdot 10^{-15}\right)^{3} \pi} \approx 1.63 \cdot 10^{25} \frac{\mathrm{C}}{\mathrm{m}^{3}}
\tag{5}
$$
The number of nucleons in a given nucleus is $A$. The total volume occupied by the nucleus is:
$$
V=\frac{A V_{N}}{f}
\tag{6}
$$
which gives the following relation between radii of nucleus and the number of nucleons:
$$
R=r_{N}\left(\frac{A}{f}\right)^{1 / 3}=\frac{r_{N}}{f^{1 / 3}} A^{1 / 3}=\frac{0.85}{0.52^{1 / 3}} A^{1 / 3}=1.06 \mathrm{fm} \cdot A^{1 / 3}
\tag{7}
$$
The numerical constant $(1.06 \mathrm{fm})$ in the equation above will be denoted as $r_{0}$ in the sequel.

Final answer: $3.40 \cdot 10^{17}$ , $1.63 \cdot 10^{25} $ , $1.06$ $ \frac{\mathrm{kg}}{\mathrm{m}^{3}}$, $\frac{\mathrm{C}}{\mathrm{m}^{3}}$, $ \mathrm{fm} \cdot A^{1 / 3}$