Nuclear radius from alpha-particle closest approach

An alpha particle and a nucleus are both positively charged, so they repel each other. An alpha particle travelling directly towards a stationary nucleus slows down as it approaches: its kinetic energy is transferred to electrostatic potential energy.

In a head-on approach, the particle momentarily stops at its distance of closest approach before moving back along its path. The distance is measured between the centres of the alpha particle and the nucleus.

Calculating the closest approach

Assume that the nucleus gains negligible kinetic energy through recoil and that the alpha particle starts sufficiently far away for its initial electrostatic potential energy to be negligible. At closest approach, all of the alpha particle’s initial kinetic energy has become electrostatic potential energy.

For an alpha particle of charge $$2e$$ approaching a nucleus of charge $$Ze$$, energy conservation gives:

Here, $$E_k$$ is the alpha particle’s initial kinetic energy, $$r$$ is the distance of closest approach, $$Z$$ is the proton number of the nucleus, $$e$$ is the elementary charge and $$\varepsilon_0$$ is the permittivity of free space.

For the same nucleus, a higher initial kinetic energy gives a smaller closest approach: the alpha particle can travel closer before it is brought to rest.

Closest approach and nuclear radius

The calculated closest approach gives an estimate of the maximum possible nuclear radius, assuming electrostatic repulsion remains the relevant interaction up to the turning point. The alpha particle can turn back before reaching the nuclear surface, so the nuclear radius may be smaller than the calculated distance. Closest approach is therefore an upper-bound estimate, not necessarily a measurement of the radius itself.

Nuclear radius from electron diffraction

Electrons can behave as waves. When a beam of electrons is scattered by a nucleus, the wave nature of the electrons produces a diffraction pattern. If the electron wavelength is comparable to the dimensions of the nucleus, the pattern can reveal the nuclear radius.

The electron wavelength is its de Broglie wavelength:

Here, $$\lambda$$ is the wavelength, $$h$$ is the Planck constant and $$p$$ is the electron momentum.

The electron diffraction pattern

Electrons are scattered through different angles as they interact with the nucleus. Waves associated with scattering from different parts of the nucleus combine. At some angles they reinforce one another; at others they cancel strongly. This produces variations in the intensity of scattered electrons.

A graph of scattered intensity against scattering angle shows a central maximum, followed by minima and secondary maxima at larger angles. The minima occur at angles where cancellation is particularly strong.

Determining the nuclear radius

The angles of the diffraction minima depend on both the electron wavelength and the size of the nucleus. To determine the radius, the observed pattern is compared with patterns expected for different nuclear sizes at the known electron wavelength. The radius is inferred from the pattern that matches the observations.

For a fixed wavelength, a larger nucleus produces minima at smaller scattering angles. Thus, if the minima are closer to the central maximum, the inferred nuclear radius is larger.

Nuclear radius and density

Nuclei are extremely small: their radii are typically of the order of $$10^{-15}$$ to $$10^{-14},\text{m}$$. A nucleus containing more nucleons is generally larger, but its radius does not increase in direct proportion to its nucleon number.

Measurements show that the nuclear radius $$R$$ depends approximately on the nucleon number $$A$$ according to:

Here, $$A$$ is the total number of protons and neutrons in the nucleus, and $$R_0 \approx 1.2\times10^{-15},\text{m}$$. The cube-root relationship means, for example, that a nucleus with eight times as many nucleons has approximately twice the radius.

Nuclear volume and nucleon number

Treating the nucleus as a sphere, its volume is $$\frac{4}{3}\pi R^3$$. Substituting the radius relationship gives:

The cube of $$A^{1/3}$$ is $$A$$, so nuclear volume is proportional to nucleon number. A nucleus with twice as many nucleons has approximately twice the volume.

Nuclear density

Density is mass divided by volume. If the mass of one nucleon is approximately $$m_n$$, the mass of a nucleus containing $$A$$ nucleons is approximately $$Am_n$$. Therefore:

The nucleon number $$A$$ cancels because both nuclear mass and nuclear volume are approximately proportional to it. The measured radius relationship is therefore evidence that different nuclei have approximately the same density.

Using $$m_n\approx1.67\times10^{-27},\text{kg}$$ and $$R_0\approx1.2\times10^{-15},\text{m}$$ gives a nuclear density of approximately $$2.3\times10^{17},\text{kg m}^{-3}$$.