3.8.1.5 - Nuclear Radius

Closest approach sets an upper limit

A positive alpha particle slows as it approaches a positive nucleus because electrostatic repulsion transfers kinetic energy to electric potential energy. For a head-on approach to a much heavier nucleus, the alpha particle momentarily stops at the turning point. This is a centre-to-centre separation, not the radius of either particle.

Take the electric potential energy to be zero at a very large separation. Neglect recoil of the target and assume the interaction is purely Coulombic outside the nucleus. Conservation of energy then gives

Here $Z$ is the target proton number, $e$ is the elementary charge and $d$ is the minimum centre-to-centre separation. Use joules with SI charges and distances. This energy equation contains $1/d$, whereas the Coulomb force equation contains $1/d^2$.

[DIAGRAM: asset_name: 8.1.5 - Nuclear Radius - Diagram 1; asset_slug: 8.1.5 - Nuclear Radius - Diagram 1; caption: An alpha particle stops outside the nucleus at the head-on turning point; description: An alpha particle stops outside the nucleus at the head-on turning point.]

For an alpha particle with $E_k=6.0\,\mathrm{MeV}$ approaching a nucleus with $Z=50$, use $k=8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}}$, $e=1.60\times10^{-19}\,\mathrm C$ and $1\,\mathrm{MeV}=1.60\times10^{-13}\,\mathrm J$. The energy is $9.60\times10^{-13}\,\mathrm J$, so $d=2.40\times10^{-14}\,\mathrm m=24.0\,\mathrm{fm}$.

The alpha particle has turned around outside the nuclear surface: the result is an upper limit on nuclear radius. Doubling the incident energy halves this calculated distance, although the nucleus has not changed size. Recoil and the finite alpha-particle size also limit the estimate. At sufficiently close approach strong-interaction effects invalidate a purely Coulombic model; increasing energy indefinitely is not a way to measure the radius with this equation.

Electron diffraction probes nuclear charge

High-energy electrons have a de Broglie wavelength $\lambda=h/p$ small enough to resolve a nucleus. A wavelength comparable with a few femtometres requires very large momentum. At hundreds of MeV the electron is highly relativistic, so $E\approx pc$ and $\lambda\approx hc/E$; the non-relativistic expression $p=mv$ is unsuitable at these energies.

Electrons scatter electromagnetically from the nuclear charge distribution. They are leptons and do not experience the strong nuclear interaction. A detector measures scattered intensity at different angles to the incident beam. The finite size of the charge distribution produces a diffraction pattern: a large forward maximum, a first minimum at a nonzero angle, then weaker maxima and minima.

[DIAGRAM: asset_name: 8.1.5 - Nuclear Radius - Diagram 2; asset_slug: 8.1.5 - Nuclear Radius - Diagram 2; caption: Electron scattering has a forward maximum and a first diffraction minimum above zero; description: Electron scattering has a forward maximum and a first diffraction minimum above zero.]