KAHIBARO
Table of Contents
Finite Size of Nuclear Charge
A nucleus is not a point of charge. Its positive charge comes from protons, and those protons are spread through a very small but finite volume. The way this positive charge is distributed inside the nucleus is called the nuclear charge distribution.
If all the nuclear charge were concentrated at one point, the electric field and electric potential near the nucleus would follow the simple point charge formulas everywhere. Real nuclei do not behave that way at extremely small distances. Their finite size changes the electric field inside the nucleus and slightly affects measurable quantities such as electron scattering and atomic energy levels.
The nuclear charge distribution describes how the positive charge of the protons is spread inside the nucleus as a function of distance from the center.
Charge Density
To describe the distribution mathematically, physicists use the charge density, written as $\rho(\mathbf{r})$. This tells us how much charge is contained per unit volume at position $\mathbf{r}$.
If the nucleus is approximately spherical, which is a good first model for many nuclei, the charge density depends mainly on the distance $r$ from the center, so we write
$$
\rho = \rho(r).
$$
The total nuclear charge must equal the charge of all protons in the nucleus:
$$
Q = Ze,
$$
where $Z$ is the atomic number and $e$ is the elementary charge. Therefore the charge density must satisfy
$$
\int \rho(\mathbf{r}) \, dV = Ze.
$$
For spherical symmetry this becomes
$$
\int_0^\infty \rho(r)\, 4\pi r^2 \, dr = Ze.
$$
This condition is called normalization of the charge distribution.
For any physically correct nuclear charge distribution,
$$
\int \rho(\mathbf{r})\, dV = Ze.
$$
The total integrated charge must equal the charge of all protons.
Simple Models of Charge Distribution
A very simple model assumes the charge is uniformly distributed throughout a sphere of radius $R$. Then
$$
\rho(r) =
\begin{cases}
\rho_0, & r \le R \\
0, & r > R
\end{cases}
$$
where $\rho_0$ is a constant. Using the total charge condition,
$$
\rho_0 \cdot \frac{4}{3}\pi R^3 = Ze,
$$
so
$$
\rho_0 = \frac{3Ze}{4\pi R^3}.
$$
This model is easy to use and captures the idea that the nucleus has a finite size. However, real nuclei usually do not have a perfectly sharp edge. Instead, the charge density tends to stay nearly constant in the interior and then fall smoothly near the surface.
For this reason, a more realistic description often uses a smooth distribution. One common example is the Fermi, or Woods-Saxon type, form for charge density:
$$
\rho(r) = \frac{\rho_0}{1 + \exp\left(\frac{r-c}{a}\right)}.
$$
Here, $c$ gives a central size scale, and $a$ describes how gradually the density falls off at the surface. A small $a$ means a sharper edge, while a larger $a$ means a more diffuse surface.
Central Region and Surface Region
The charge distribution of many medium and heavy nuclei has two important features. In the interior, the charge density is roughly constant. Near the outer boundary, it decreases smoothly to zero over a short distance. This outer layer is called the surface region.
This means that a nucleus is not like a hard solid ball with a perfectly sharp boundary. It is better pictured as having a dense core and a fuzzy edge.
The difference between a sharp boundary and a diffuse boundary matters in experiments. For example, scattering measurements are sensitive to how suddenly or gradually the charge density changes.
Mean Square Charge Radius
A useful way to summarize the charge distribution is with the mean square charge radius. This quantity measures how far, on average, the nuclear charge lies from the center.
It is defined as
$$
\langle r^2 \rangle = \frac{1}{Ze} \int r^2 \rho(\mathbf{r})\, dV.
$$
For spherical symmetry,
$$
\langle r^2 \rangle = \frac{1}{Ze} \int_0^\infty r^2 \rho(r)\, 4\pi r^2\, dr.
$$
The square root of this quantity,
$$
r_{\text{rms}} = \sqrt{\langle r^2 \rangle},
$$
is called the root mean square charge radius, or rms charge radius. It is one of the most important measurable quantities related to nuclear charge distribution.
The rms charge radius is
$$
r_{\text{rms}} = \sqrt{\langle r^2 \rangle}.
$$
It gives a single number that characterizes the spatial spread of nuclear charge.
For a uniformly charged sphere of radius $R$, the result is
$$
\langle r^2 \rangle = \frac{3}{5}R^2,
$$
so
$$
r_{\text{rms}} = \sqrt{\frac{3}{5}}\,R.
$$
Electric Field of a Distributed Nuclear Charge
Because the nuclear charge is spread out, the electric field near the center is different from that of a point charge. For a spherically symmetric distribution, Gauss's law shows that only the charge inside radius $r$ contributes to the field at that radius.
If $Q(r)$ is the charge enclosed within radius $r$, then
$$
Q(r) = \int_0^r \rho(r')\, 4\pi r'^2\, dr'.
$$
The electric field is
$$
E(r) = \frac{1}{4\pi \varepsilon_0}\frac{Q(r)}{r^2}.
$$
Outside the nucleus, where all charge is enclosed, this becomes the same as the field of a point charge:
$$
E(r) = \frac{1}{4\pi \varepsilon_0}\frac{Ze}{r^2}, \quad r \text{ outside the nucleus}.
$$
Inside the nucleus, the field is smaller than the point charge value. For a uniformly charged sphere,
$$
Q(r) = Ze \frac{r^3}{R^3}, \quad r \le R,
$$
so
$$
E(r) = \frac{1}{4\pi \varepsilon_0}\frac{Ze}{R^3}r, \quad r \le R.
$$
Thus the electric field inside rises linearly from zero at the center.
Why Charge Distribution Matters
The nuclear charge distribution is not just a geometric detail. It has real physical consequences. Electrons in atoms, especially those very close to the nucleus, feel the finite size of the nuclear charge distribution. This causes small shifts in atomic energy levels. Scattering experiments also depend strongly on how charge is arranged inside the nucleus. If the nucleus were a point, the scattering pattern would be different.
The charge distribution also gives information about nuclear shape. While many nuclei are approximately spherical, some are slightly deformed. In such cases the charge density is not perfectly spherically symmetric. Then the distribution depends on direction as well as distance from the center.
How It Is Measured
One of the main ways to determine nuclear charge distribution is electron scattering. Electrons are especially useful because they interact electromagnetically with the positive nuclear charge. By observing how electrons scatter from nuclei, physicists can infer how the charge is distributed.
Another source of information comes from atomic spectroscopy. Very precise measurements of atomic energy levels can reveal the effect of the finite nuclear size, especially in heavy atoms or in systems where orbiting particles spend significant time near the nucleus.
This chapter focuses only on the idea of the distribution itself. The detailed methods of scattering and spectroscopy belong to other topics.
Comparison of Idealized Descriptions
| Model | Interior density | Surface | Mathematical simplicity | Realism |
|---|---|---|---|---|
| Point charge | Not finite | None | Very high | Poor for nuclei |
| Uniform sphere | Constant | Sharp cutoff | High | Basic approximation |
| Fermi type distribution | Nearly constant | Smooth falloff | Moderate | More realistic |
Key Physical Picture
The nuclear charge distribution tells us that the nucleus has a finite spatial extent and that its positive charge is spread through that volume rather than concentrated at one point. In many nuclei the charge density is nearly uniform in the interior and decreases smoothly at the surface. The total charge is always $Ze$, and the rms charge radius provides a compact way to describe how far the charge extends from the center.
Important facts about nuclear charge distribution:
$$
\int \rho(\mathbf{r})\, dV = Ze
$$
$$
\langle r^2 \rangle = \frac{1}{Ze}\int r^2 \rho(\mathbf{r})\, dV
$$
$$
r_{\text{rms}} = \sqrt{\langle r^2 \rangle}
$$
A real nucleus has a finite size and usually a diffuse surface, not a perfectly sharp edge.