KAHIBARO
Table of Contents
Size scale of the nucleus
The atomic nucleus is extremely small compared with the whole atom. While an atom has a size of about $10^{-10}\,\text{m}$, the nucleus typically has a size of about $10^{-15}\,\text{m}$. This nuclear length scale is called the femtometer, written as $\text{fm}$, where
$$
1\,\text{fm} = 10^{-15}\,\text{m}
$$
So the nucleus is about $100{,}000$ times smaller than the atom that contains it. Even though it is so small, it contains almost all of the atom's mass.
What is meant by nuclear radius
When we speak of the nuclear radius, we mean a characteristic distance from the center of the nucleus to its outer edge. Unlike a hard solid ball, a nucleus does not have a perfectly sharp boundary. Its positive charge and its nucleons are spread over a very small region, and the density falls gradually near the surface. Because of this, the nuclear radius is an approximate but very useful quantity.
For many purposes, the nucleus can be treated as nearly spherical, especially for simple estimates. In that case, the radius gives a convenient measure of nuclear size.
Empirical formula for nuclear radius
Experiments show that the radius of a nucleus depends mainly on its mass number $A$, which is the total number of protons and neutrons. The approximate relation is
$$
R = R_0 A^{1/3}
$$
where $R_0$ is a constant of about
$$
R_0 \approx 1.2\,\text{fm}
$$
Sometimes values between $1.1\,\text{fm}$ and $1.3\,\text{fm}$ are used depending on the model and measurement method.
Important nuclear size formula:
$$
R = R_0 A^{1/3}, \qquad R_0 \approx 1.2\,\text{fm}
$$
This means nuclear radius grows with the cube root of the mass number, not directly with $A$.
Why the cube root appears
The relation $R \propto A^{1/3}$ tells us that if we increase the number of nucleons, the radius grows slowly. This happens because nuclei have roughly constant density. If density stays about the same, then the volume must be proportional to the number of nucleons:
$$
V \propto A
$$
For a roughly spherical nucleus,
$$
V = \frac{4}{3}\pi R^3
$$
So if $V \propto A$, then
$$
R^3 \propto A
$$
and therefore
$$
R \propto A^{1/3}
$$
This is one of the clearest signs that nuclear matter is packed with nearly uniform density.
Typical values
Here are some approximate nuclear radii for a few nuclei.
| Nucleus | Mass number $A$ | Approximate radius $R = 1.2A^{1/3}\,\text{fm}$ |
|---|---|---|
| Hydrogen-1 | 1 | $1.2\,\text{fm}$ |
| Helium-4 | 4 | $1.9\,\text{fm}$ |
| Carbon-12 | 12 | $2.7\,\text{fm}$ |
| Oxygen-16 | 16 | $3.0\,\text{fm}$ |
| Iron-56 | 56 | $4.6\,\text{fm}$ |
| Lead-208 | 208 | $7.1\,\text{fm}$ |
These values are only approximate, but they show the general scale very well.
Example calculation
For iron-56, with $A = 56$,
$$
R = 1.2 \times 56^{1/3}\,\text{fm}
$$
Since
$$
56^{1/3} \approx 3.8
$$
we get
$$
R \approx 1.2 \times 3.8 = 4.6\,\text{fm}
$$
So the radius of an iron nucleus is about $4.6\,\text{fm}$.
Consequence for nuclear volume
Since the radius follows $R = R_0 A^{1/3}$, the volume of the nucleus is
$$
V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A
$$
So the volume is directly proportional to $A$. If one nucleus has twice as many nucleons as another, its volume is about twice as large.
Key consequence:
If
$$
R = R_0 A^{1/3}
$$
then
$$
V \propto A
$$
This supports the idea that nuclei have nearly constant density.
Visual picture
A larger nucleus is not enormously bigger in radius than a smaller one. It contains many more nucleons, but because the radius grows only as $A^{1/3}$, the increase in size is moderate.
Comparison with atomic size
It is helpful to compare nuclear size with atomic size.
| Object | Typical radius |
|---|---|
| Atom | $\sim 10^{-10}\,\text{m}$ |
| Nucleus | $\sim 10^{-15}\,\text{m}$ |
This means that most of the atom is empty space. The electrons occupy a much larger region around a tiny central nucleus.
Limits of the simple radius formula
The formula $R = R_0 A^{1/3}$ is very useful, but it is not exact for every nucleus. Real nuclei may not be perfectly spherical, and the edge of the nucleus is not perfectly sharp. Light nuclei can also deviate somewhat from the simple rule. Still, for basic nuclear physics, it gives an excellent first estimate of nuclear size.
The nuclear radius formula is an empirical approximation, not an exact law. It works best as a general estimate of nuclear size.
Main idea to remember
The most important fact about nuclear radius is that nuclear size increases with the cube root of the number of nucleons:
$$
R = R_0 A^{1/3}
$$
This tells us that nuclei remain extremely small, even when they contain many nucleons, and it leads directly to the idea of nearly constant nuclear density.