Size of the Nucleus

The nucleus is extremely small compared with the size of the atom, but it contains almost the entire mass of the atom. Experimental studies such as alpha-particle scattering and electron scattering show that the nucleus has a finite size and that its radius depends on the mass number A.

The nuclear radius is the effective distance from the centre of the nucleus to its outer boundary. Since the nucleus does not have a perfectly sharp surface, this radius is treated as an approximate effective value in physics.

The experimentally observed relation is: R = R₀A^1/3

where:

• R = radius of the nucleus
• R₀​ = radius constant
• A = mass number of the nucleus.

For nuclei,

• R₀ ≈ 1.2 × 10⁻¹⁵ m.

Rutherford’s scattering idea

Rutherford studied the scattering of alpha-particles by thin metal foils. When an alpha particle moves toward a positively charged nucleus, electrostatic repulsion slows it down and can bring it to a momentary stop before it moves away.

The minimum distance reached during this process is called the distance of closest approach. For a 5.5 MeV alpha-particle, this distance is of the order of 4.0 × 10⁻¹⁴ m, which shows that the nucleus is much smaller than the atom.

Electron scattering is useful

The source material also notes that more accurate measurements of nuclear size come from electron scattering experiments. These experiments help determine the distribution of charge inside the nucleus and give a better estimate of nuclear radius.

Cube-root dependence

If A increases, radius increases as A^1/3, not as A. This means heavy nuclei are larger than light nuclei, but not enormously larger.

Volume of the nucleus

Since the nucleus is approximately spherical, its volume is:

Substituting R = R₀A^1/3, the volume becomes proportional to A. Thus, nuclear volume increases nearly in direct proportion to mass number.

Nearly constant nuclear density

The mass of the nucleus is roughly proportional to A, and the volume is also proportional to A. Therefore, the density of nuclear matter is nearly the same for all nuclei.

The source material gives the approximate density of nuclear matter as:

• Nuclear density ≈ 2.3 × 10¹⁷ kg m⁻³.

This value is extraordinarily large compared with the density of ordinary substances. The material also connects this idea with neutron stars, whose matter has a density of the same order of magnitude.

Density comparison

The source material includes a standard example for an iron nucleus of mass number 56 and mass 55.85 u, showing that the nuclear density is about 2.29 × 10¹⁷ kg m⁻³.

Given

• Mass number, A = 56
• Mass of iron nucleus = 55.85 u
• Radius relation: R = R₀A^1/3, where R₀ = 1.2 × 10⁻¹⁵ m.

Method

1. Find the radius using R = R₀A^1/3.
2. Find the volume using V = \[\frac {4}{3}\]πR³.
3. Convert nuclear mass into SI unit.
4. Use the density formula: ρ = \[\frac {m}{V}\]

Result

• Density of iron nucleus ≈ 2.29 × 10¹⁷ kg m⁻³.

If an atom were expanded to the size of a large stadium, the nucleus would still occupy only a tiny region near the centre, yet almost all the mass would remain concentrated there. This captures the idea that the nucleus is tiny but extremely dense.

Scientific significance

The nearly constant density of nuclei suggests that nuclear matter behaves in a highly compact way. This idea is important in nuclear physics as well as astrophysics, especially in understanding neutron stars.