Size of the Nucleus

The Size of the Nucleus

The story of the nucleus begins with Rutherford. After his famous gold-foil experiment, he knew that the atom’s positive charge and nearly all its mass were concentrated in a tiny central region — the nucleus. But how small is “tiny”? The experiment gave the first real clue.

Geiger and Marsden, working at Rutherford’s suggestion, fired alpha particles (helium nuclei, with charge $+2e$) at a thin gold foil. Most particles passed straight through, but a few were scattered backwards. Rutherford showed that this backscattering could only happen if the positive charge of the atom were concentrated in a volume far smaller than the atom itself. The key quantity that emerged from the experiment was the distance of closest approach — the minimum distance an alpha particle can get to the nucleus before its kinetic energy is completely converted into electrostatic potential energy and it stops, then turns back.

For an alpha particle of kinetic energy $K = 5.5\ \text{MeV}$ scattering off a gold nucleus ($Z = 79$), the distance of closest approach $r_0$ is given by equating the initial kinetic energy to the Coulomb potential energy at the turning point:

Solving for $r_0$:

Plugging in the numbers gives $r_0 \approx 4.0 \times 10^{-14}\ \text{m}$. This is the distance at which the alpha particle just touches the “edge” of the nuclear charge distribution — but only if the only force acting is the Coulomb repulsion. Since the scattering matched Rutherford’s pure-Coulomb formula perfectly at this energy, the actual size of the gold nucleus must be smaller than $4.0 \times 10^{-14}\ \text{m}$. If the nucleus were larger, the alpha particle would have hit it before reaching that distance, and the scattering pattern would have deviated from Rutherford’s prediction.

Determining Nuclear Radii

If we use alpha particles of higher energy, the distance of closest approach becomes smaller. At some point, the alpha particle gets so close that it feels the strong nuclear force — a short-range attraction that overpowers the Coulomb repulsion at very small distances. When this happens, the scattering pattern no longer matches Rutherford’s pure-Coulomb formula. The distance at which the deviation first appears gives a direct measure of the nuclear radius.

However, the most precise measurements of nuclear sizes come from scattering experiments using fast electrons (rather than alpha particles) as projectiles. Electrons are point-like and feel only the electromagnetic force (no strong nuclear force), so they probe the charge distribution of the nucleus cleanly. By bombarding targets made of various elements with high-energy electrons and measuring how they scatter, physicists have determined the radii of many nuclei with great accuracy.

The result is remarkably simple: the radius $R$ of a nucleus depends only on its mass number $A$ (the total number of protons and neutrons), and follows a universal power law:

where $R_0 = 1.2 \times 10^{-15}\ \text{m} = 1.2\ \text{fm}$ (1 femtometre = $10^{-15}\ \text{m}$).

This formula tells us something profound about nuclear matter.

Volume and Density: The Nucleus as a Liquid Drop

Since the volume of a sphere is $V = \frac{4}{3}\pi R^3$, substituting $R = R_0 A^{1/3}$ gives:

The volume is directly proportional to $A$. The mass of the nucleus is also approximately proportional to $A$ (since each nucleon has nearly the same mass, about $1.67 \times 10^{-27}\ \text{kg}$). Therefore, the density of nuclear matter is:

The factor $A$ cancels out completely. This means the density of nuclear matter is the same for all nuclei, regardless of their size. It is a constant, approximately $2.3 \times 10^{17}\ \text{kg m}^{-3}$.

To appreciate how enormous this density is: ordinary water has a density of $10^3\ \text{kg m}^{-3}$. Nuclear matter is about $10^{14}$ times denser. This is because an atom is mostly empty space — the nucleus occupies only about $10^{-14}$ of the atom’s volume, but contains nearly all its mass.

Worked Example: Density of the Iron Nucleus

Example 13.1 from the NCERT text: Given the mass of an iron nucleus as $55.85\ \text{u}$ and $A = 56$, find the nuclear density.

Solution:

First, convert the mass from atomic mass units to kilograms. $1\ \text{u} = 1.660539 \times 10^{-27}\ \text{kg}$, so:

The nuclear radius for $A = 56$ is:

Now, $56^{1/3} \approx 3.826$. So:

The volume is:

$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (4.59 \times 10^{-15})^3$ …