Nuclear Physics

1. The Nucleus

The nucleus was discovered by Lord Rutherford from his alpha-particle scattering experiment. An atom consists of a positively charged, extremely dense central core called the nucleus, surrounded by orbiting electrons. The diameter of a nucleus is of the order of $10^{-15}\ \text{m}$ (a few femtometres). The nucleus contains protons and neutrons.

• A proton is a positively charged particle whose charge is equal in magnitude to that of an electron, $+e = 1.6\times10^{-19}\ \text{C}$. Its mass is about $1.67\times10^{-27}\ \text{kg}$.
• A neutron is a chargeless particle whose mass is nearly equal to that of a proton. Because it carries no charge, it has high penetrating power, and a slow neutron can trigger nuclear fission.

Protons and neutrons together are called nucleons.

Atomic number, mass number and notation

• Atomic number $Z$: the number of protons in the nucleus.
• Mass number $A$: the total number of nucleons (protons + neutrons):

$$ \begin{aligned} A &= Z + N \qquad\Rightarrow\qquad N \ &= A - Z \end{aligned} $$

where $N$ is the number of neutrons. A nuclide of element $X$ is written as $^{A}_{Z}X$, with $Z$ the atomic number and $A$ the mass number.

The net charge of a nucleus is $q = +Ze$, since it holds $Z$ protons each of charge $+e$.

---

2. Properties of the Nucleus

(i) Nuclear size

Experiment shows that the radius $r$ of a nucleus is directly proportional to the one-third power of its mass number $A$:

$$ \begin{aligned} r &= r_0 A^{1/3} \ \qquad r_0 &= 1.2\times10^{-15}\ \text{m} \end{aligned} $$

Example: Nuclear radius of uranium $^{238}U$ ($A = 238$):

$$ \begin{aligned} r &= 1.2\times10^{-15}\times(238)^{1/3} \ &= 1.2\times10^{-15}\times 6.197 \ &= 7.44\times10^{-15}\ \text{m} \end{aligned} $$

Worked problem. The radius of a nucleus of mass number $16$ is $3\ \text{fm}$. Find the radius of a nucleus of mass number $128$.

Since $r_1 = r_0 A_1^{1/3}$ and $r_2 = r_0 A_2^{1/3}$, dividing gives:

$$ \begin{aligned} \frac{r_2}{r_1} &= \left(\frac{A_2}{A_1}\right)^{1/3} \ &= \left(\frac{128}{16}\right)^{1/3} \ &= 8^{1/3} \ &= 2 \ r_2 &= 2\times r_1 \ &= 2\times 3 \ &= 6\ \text{fm} \end{aligned} $$

(ii) Nuclear charge

A nucleus of atomic number $Z$ carries charge:

$$ \begin{aligned} q &= +Ze \ \qquad e &= 1.6\times10^{-19}\ \text{C} \end{aligned} $$

(iii) Nuclear mass

The nuclear mass $M$ is the sum of the masses of its protons and neutrons:

$$M = Z m_p + (A - Z)m_n$$

Since $m_p \approx m_n$, this is approximately $M \approx A m_p$.

(iv) Nuclear density

Nuclear density is the nuclear mass per unit nuclear volume. Taking $M \approx A m_p$ and volume $V = \tfrac{4}{3}\pi r^3$ with $r = r_0 A^{1/3}$:

$$ \begin{aligned} \rho &= \frac{M}{V} \ &= \frac{A m_p}{\tfrac{4}{3}\pi (r_0 A^{1/3})^3} \ &= \frac{A m_p}{\tfrac{4}{3}\pi r_0^3 A} \ &= \frac{3 m_p}{4\pi r_0^3} \end{aligned} $$

The mass number $A$ cancels, so nuclear density is independent of $A$. Substituting values:

$$ \begin{aligned} \rho &= \frac{3\times 1.67\times10^{-27}}{4\pi (1.2\times10^{-15})^3} \ &= \frac{5.01\times10^{-27}}{2.17\times10^{-44}} \ &= 2.3\times10^{17}\ \text{kg m}^{-3} \end{aligned} $$

This enormous, mass-number-independent density shows that nuclear matter is packed uniformly and extremely tightly.

---