A-Level Physics: Nuclear Radius and Scattering

To understand the structure of matter itself, you must probe the atom's heart: the nucleus. Before the 20th century, the atom was imagined as a uniform, plum-pudding-like sphere. Revolutionary experiments shattered this model, revealing a tiny, dense core and establishing the field of nuclear physics. The key experimental methods—Rutherford scattering and electron diffraction—allow us to measure the nuclear radius and deduce the profound implications of a nearly constant nuclear density.

Rutherford's Alpha Scattering Experiment and Nuclear Discovery

The Rutherford alpha particle scattering experiment was pivotal in revealing the nuclear atom. Ernest Rutherford, along with Geiger and Marsden, directed a beam of alpha particles (helium nuclei, $H{e}^{2+}$) at a thin gold foil. According to the then-prevailing plum pudding model, these positively charged, high-energy particles should have passed through with minimal deflection. The stunning result was that while most did, a very small fraction—about 1 in 8000—were deflected through angles greater than 90°, some even bouncing straight back.

This observation was only possible if the atom's positive charge and most of its mass were concentrated in an incredibly small, dense region. Rutherford called this the nucleus. The large-angle scattering occurred due to the electrostatic repulsion (Coulomb force) between the positively charged alpha particle and the positively charged nucleus. The fact that most alpha particles passed through undeflected proved the atom is mostly empty space. Rutherford derived a formula for the scattering, showing the probability of deflection is proportional to $1/si{n}^4(\theta /2)$, where $\theta$ is the scattering angle.

Estimating Nuclear Radius from Scattering Data

The Rutherford scattering formula holds true as long as the alpha particle does not come close enough to the nucleus for the strong nuclear force to act. This provides a method to estimate the nuclear radius. At the point of closest approach, the alpha particle's initial kinetic energy is entirely converted into electrostatic potential energy. For a head-on collision where the alpha particle is turned back (180° scattering), you can equate these energies.

The formula is derived as follows: The initial kinetic energy $E_k$ of the alpha particle equals the potential energy at the distance of closest approach, $r_{min}$. $$E_k=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r_{min}}$$ 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 e is the elementary charge. Rearranging gives: $$r_{min}=\frac{1}{4\pi {ϵ}_0}\frac{(2e)(Ze)}{E_k}$$

For a gold nucleus (Z=79) and a typical alpha energy of 5 MeV, $r_{min}$ calculates to about $4.5\times 10^{-14}$ m. This is an upper estimate for the nuclear radius, as it assumes a purely Coulombic interaction. If alpha particles are fired with higher energies, they can overcome the electrostatic barrier and get close enough for the strong, short-range nuclear force to attract them. The deviation from Rutherford's scattering law at these high energies gives a more direct measure, suggesting nuclear radii are on the order of $10^{-15}$ m, or 1 femtometre (fm).

Electron Diffraction: A More Precise Probe

While alpha scattering discovered the nucleus, electron diffraction provides a more precise method for measuring nuclear size. This technique uses electrons because, as leptons, they do not experience the strong nuclear force; their interaction is purely electromagnetic. When high-energy electrons (typically hundreds of MeV) are fired at a thin sample, they behave as waves according to the de Broglie wavelength.

The de Broglie wavelength $\lambda$ is given by $\lambda =h/p$, where h is Planck's constant and p is the electron's momentum. For high energies, relativistic calculations are often needed. When this wavelength is comparable to the nuclear diameter, the electrons are diffracted. The resulting diffraction pattern on a detector consists of a central bright maximum and a series of concentric minima and maxima.

The first minimum in the diffraction pattern is crucial. The angle $\theta$ of this first minimum is related to the nuclear radius R by the formula for diffraction at a circular obstacle: $$\sin \theta \approx \theta \approx \frac{0.61\lambda }{R}$$ By measuring $\theta$ and knowing $\lambda$ from the electron energy, you can solve for R. This method confirms nuclear radii are of order $10^{-15}$ m and provides the data for the fundamental radius-mass relationship.

The Nuclear Radius-Mass Relationship and Constant Density

Measurements from electron diffraction and other techniques reveal a simple but profound relationship: the radius R of a nucleus is proportional to the cube root of its nucleon number A (the total number of protons and neutrons). This is expressed by the formula: $$R=R_0{A}^{1/3}$$ Where $R_0$ is a constant, experimentally determined to be approximately $1.2\times 10^{-15}$ m or 1.2 fm.

This relationship has a critical implication. If you model the nucleus as a sphere, its volume V is proportional to $R^3$: $$V=\frac{4}{3}\pi R^3=\frac{4}{3}\pi (R_0^{3}A)$$ Since volume is proportional to A, and mass M is also essentially proportional to A (as the mass of a nucleon is roughly constant), the density $\rho$ is independent of A: $$\rho =\frac{M}{V}\propto \frac{A}{A}=\text{constant}$$ This means nuclear density is approximately constant across all nuclei, a key piece of evidence that nuclear matter is incompressible and that the strong nuclear force is saturated—a nucleon only interacts with its immediate neighbours, not every nucleon in the nucleus. This constant density is enormous, on the order of $10^{17}$ kg m${}^{-3}$.

Common Pitfalls

1. Confusing the distance of closest approach with the nuclear radius. The $r_{min}$ calculated from energy conservation in Rutherford scattering is an upper limit for the radius. The true radius is smaller, as the calculation assumes no nuclear forces. The deviation from Rutherford scattering at high energies gives the better measure.
2. Misapplying the de Broglie wavelength formula. For the high-energy electrons used in diffraction, their kinetic energy is often comparable to or greater than their rest mass energy. You must use the relativistic formula for momentum $p=\frac{E_{total}}{c}$, not the classical $p=mv$, to calculate $\lambda$ correctly.
3. Forgetting the cube in the volume calculation. A common algebraic error is to state that if $R\propto A^{1/3}$, then $V\propto A^{1/3}$. Remember, volume depends on $R^3$, so $V\propto (A^{1/3}{)}^3=A$. This is the crucial step that leads to the conclusion of constant density.
4. Assuming all nuclei are perfectly spherical. The $R=R_0{A}^{1/3}$ formula describes an average, spherical approximation. Some nuclei, especially those with certain proton/neutron numbers, have non-spherical "deformed" shapes (like rugby balls), but the general density rule still holds.

Summary

• Rutherford's alpha scattering experiment revealed the nuclear atom by showing that a tiny, dense, positively charged nucleus causes large-angle deflections, while most of the atom is empty space.
• Electron diffraction provides a precise method for measuring nuclear radius, utilizing the wave nature of particles (de Broglie wavelength) and the diffraction pattern from a circular object.
• The nuclear radius $R$ is related to the nucleon number $A$ by $R=R_0{A}^{1/3}$, where $R_0\approx 1.2$ fm.
• This relationship leads directly to the conclusion that nuclear density is approximately constant across all nuclei (～$10^{17}$ kg m${}^{-3}$), implying nuclear matter is incompressible and the strong force is short-range.
• Understanding these experiments and relationships is fundamental to nuclear and particle physics, bridging the gap between atomic structure and the forces that bind the universe's core.