Nuclear Radius Estimation from Scattering

Understanding the size of an atomic nucleus is fundamental to nuclear physics, yet we cannot see it directly. The two primary experimental methods—alpha particle scattering and electron diffraction—provide a powerful one-two punch for estimation. Classical alpha scattering gives us a rough, first-order upper limit, while the quantum wave nature of electrons delivers a far more precise measurement. Together, they reveal a surprising and crucial property of all nuclei: their density is approximately constant.

The Classical Method: Distance of Closest Approach

The first estimates of nuclear size came from the famous Geiger-Marsden experiment (or Rutherford scattering), where alpha particles were fired at a thin gold foil. Most passed through, but some were deflected through large angles, even bouncing straight back. This implied a tiny, massive, positively charged core—the nucleus.

We can estimate its maximum size by considering a head-on collision. In this scenario, an alpha particle approaches the nucleus directly, slows down due to electrostatic repulsion, comes to a momentary stop at the distance of closest approach ($r_0$), and then is repelled back along its original path. At this point of closest approach, the alpha particle's initial kinetic energy ($E_k$) has been entirely converted to electrostatic potential energy.

The electrostatic potential energy between two point charges is given by the formula: $$U=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r}$$ where $q_1$ is the charge of the alpha particle ($2e$), $q_2$ is the charge of the nucleus ($Ze$, where $Z$ is the atomic number), and ${ϵ}_0$ is the permittivity of free space.

Setting the initial kinetic energy equal to the potential energy at $r_0$ gives: $$E_k=\frac{1}{4\pi {ϵ}_0}\frac{(2e)(Ze)}{r_0}$$

Solving for $r_0$ provides our estimate: $$r_0=\frac{1}{4\pi {ϵ}_0}\frac{2Z{e}^2}{E_k}$$

Worked Example: For a 5.0 MeV alpha particle ($E_k=5.0\times 10^6\times 1.602\times 10^{-19}$ J) scattering off a gold nucleus ($Z=79$), the calculation is:

1. Calculate the constant $\frac{1}{4\pi {ϵ}_0}=8.99\times 10^9{\text{ N m}}^2{\text{ C}}^{-2}$.
2. Use $e=1.602\times 10^{-19}$ C.
3. Substitute into the formula:

$$r_0=\frac{(8.99\times 10^9)\times 2\times 79\times (1.602\times 10^{-19}{)}^2}{5.0\times 10^6\times 1.602\times 10^{-19}}\approx 4.55\times 10^{-14}\text{ m}$$

This value, around 45.5 femtometres (fm), is an upper limit for the nuclear radius. The true radius is smaller because for a head-on collision to occur, the alpha particle must actually reach the nuclear surface without any other interactions.

The Quantum Method: Electron Diffraction

While alpha scattering provides an order-of-magnitude check, electron diffraction is the preferred modern technique for accurate nuclear radius measurement. Here, high-energy electrons are used as probes. Electrons are leptons and do not interact via the strong nuclear force; they interact with the nucleus primarily through the electromagnetic force.

The key is the de Broglie wavelength of the electron, given by $\lambda =h/p$, where $h$ is Planck's constant and $p$ is the electron's momentum. To resolve detail as small as a nucleus (on the order of $10^{-15}$ m), the probing wavelength must be comparable or smaller. This requires highly relativistic electrons.

When a beam of these electrons is directed at a thin sample, they are diffracted by the nuclear charge distribution. The resulting diffraction pattern on a screen shows a central maximum and a series of minima. The angle $\theta$ to the first minimum is related to the nuclear radius $R$ by a formula analogous to that for single-slit diffraction: $$\sin \theta \approx \frac{0.61\lambda }{R}$$ A more complete quantum scattering analysis yields a similar relationship: the first minimum in the diffraction pattern occurs when $qR\approx 4.49$, where $q$ is the momentum transfer. By measuring the angle to the first minimum and knowing the electron's de Broglie wavelength, $R$ can be calculated with high precision. For gold, this method gives a radius of about 7.3 fm—much more precise and reliable than the classical estimate from alpha scattering.

The Nuclear Radius Formula and Constant Density

Data from electron diffraction experiments on many different nuclei revealed a simple empirical relationship between the nuclear radius $R$ and its nucleon number $A$ (the total number of protons and neutrons): $$R=R_0{A}^{1/3}$$ where $R_0$ is a constant empirically determined to be approximately $1.2\times 10^{-15}$ m, or 1.2 fm.

This cubic relationship is profoundly important. It implies that the volume of a nucleus ($V=\frac{4}{3}\pi R^3$) is proportional to its nucleon number $A$: $$V=\frac{4}{3}\pi R^3=\frac{4}{3}\pi (R_0^{3}A)$$

Since the mass of a nucleus is also approximately proportional to $A$ (the mass of a nucleon is roughly constant), this leads to a critical conclusion: Nuclear density is approximately constant. We can confirm this:

1. Nuclear mass: $m\approx A\times u$, where $u$ is the unified atomic mass unit.
2. Nuclear volume: $V=\frac{4}{3}\pi R_0^{3}A$.
3. Nuclear density: $\rho =\frac{m}{V}=\frac{Au}{\frac{4}{3}\pi R_0^{3}A}=\frac{u}{\frac{4}{3}\pi R_0^3}$.

The $A$ cancels out! Substituting $R_0=1.2\text{ fm}$ and $u=1.66\times 10^{-27}\text{ kg}$: $$\rho \approx \frac{1.66\times 10^{-27}}{\frac{4}{3}\pi (1.2\times 10^{-15}{)}^3}\approx 2.3\times 10^{17}{\text{ kg m}}^{-3}$$ This immense, constant density suggests that nucleons are packed together like incompressible droplets, a cornerstone of the liquid drop model of the nucleus.

Common Pitfalls

1. Confusing the radius formulas. A common error is to misuse $R=R_0{A}^{1/3}$ by trying to relate it directly to atomic number $Z$. Remember, $A$ is the mass number (protons + neutrons), not the atomic number (protons alone). The radius depends on the total number of nucleons.
2. Misinterpreting the distance of closest approach. Students often state that $r_0$ is the nuclear radius. It is actually an upper limit or maximum possible radius from that experiment. The true radius is smaller, as proven by the more accurate electron diffraction method.
3. Forgetting the wave condition for diffraction. When discussing electron diffraction, it's crucial to justify why high energy is needed. If the electron's de Broglie wavelength is too large (at low energy), diffraction effects are negligible, and you cannot resolve nuclear-scale details. The probe wavelength must be comparable to or smaller than the object being studied.
4. Incorrect energy conservation setup. In the alpha scattering calculation, ensure you equate the initial kinetic energy (far from the nucleus, where potential energy is zero) to the electrostatic potential energy at the point of closest approach. The kinetic energy at $r_0$ is zero, so the total energy is entirely potential.

Summary

• Alpha particle scattering provides a classical upper limit for the nuclear radius via the distance of closest approach ($r_0$), calculated using conservation of energy: $E_k=\frac{1}{4\pi {ϵ}_0}\frac{2Z{e}^2}{r_0}$.
• Electron diffraction is a precise quantum mechanical method that uses the de Broglie wavelength of high-energy electrons. The diffraction pattern's first minimum is analytically related to the nuclear radius.
• All stable nuclei follow the empirical formula $R=R_0{A}^{1/3}$, where $R_0\approx 1.2\text{ fm}$ and $A$ is the nucleon number.
• The $A^{1/3}$ dependence proves that nuclear volume is proportional to mass, leading to the fundamental conclusion that nuclear density is constant across all nuclei, approximately $2.3\times 10^{17}{\text{ kg m}}^{-3}$.
• These concepts bridge classical and quantum physics, demonstrating how different experimental techniques converge to reveal core properties of matter.