Size of the Nucleus

How the Size Was Measured

The existence of the nucleus was established by Rutherford through the alpha-scattering experiment that Geiger and Marsden performed on thin gold foils. That same experiment also delivered the first quantitative bound on nuclear size. Geiger and Marsden found that an alpha particle of kinetic energy $5.5\ \text{MeV}$ approached a gold nucleus to a distance of closest approach of about $4.0 \times 10^{-14}\ \text{m}$ before Coulomb repulsion turned it back.

Rutherford understood this scattering by assuming the Coulomb repulsive force alone was responsible. Because the positive charge is confined to the nucleus, the actual nuclear radius must be less than $4.0 \times 10^{-14}\ \text{m}$ — the alpha never reached the nuclear surface. If alpha particles of energy higher than $5.5\ \text{MeV}$ are used, the distance of closest approach shrinks; at some point the projectile is close enough that the short-range nuclear force begins to act, and the scattering departs from Rutherford's pure-Coulomb prediction. The energy at which this deviation sets in lets one infer the nuclear size.

A sharper tool replaced the alpha probe. By performing scattering experiments in which fast electrons bombard targets made of various elements, the nuclear radii of many elements were accurately measured. Electrons interact only electromagnetically and not through the strong force, so they map the nucleus cleanly. These experiments yielded the empirical radius law that anchors this entire topic.

The Radius Law R = R0 A^(1/3)

Treating the nucleus as a sphere, electron-scattering data show that a nucleus of mass number $A$ has a radius

$$R = R_0\,A^{1/3}, \qquad R_0 = 1.2 \times 10^{-15}\ \text{m} = 1.2\ \text{fm}$$

where the femtometre, $1\ \text{fm} = 10^{-15}\ \text{m}$, is the natural unit of length here (NIOS notes that it honours Enrico Fermi). The single constant $R_0 = 1.2\ \text{fm}$ fixes the size of every nucleus. The radius depends only on the mass number $A$ — the total count of nucleons — and not on the atomic number $Z$ or on the neutron-to-proton mix. Two isobars of the same $A$ but different $Z$ therefore have the same radius.

Two reference points fix the scale. For hydrogen, $A = 1$, so $R = 1.2\ \text{fm}$ — the smallest nucleus. For uranium, $A = 238$, so $R = 1.2 \times 238^{1/3} \approx 7.5\ \text{fm}$ (the value NIOS quotes). Across the entire periodic table the heaviest nucleus is only about six times the radius of the lightest, because the cube root crushes a 238-fold range in $A$ into a 6-fold range in $R$. This gentle growth is the signature of the $A^{1/3}$ dependence and is worth committing to memory.

Volume Proportional to A

Modelling the nucleus as a sphere of radius $R$, its volume is $V = \tfrac{4}{3}\pi R^{3}$. Substituting the radius law,

$$V = \frac{4}{3}\pi R^{3} = \frac{4}{3}\pi \left(R_0 A^{1/3}\right)^{3} = \frac{4}{3}\pi R_0^{3}\, A.$$

Because $R_0$ is a constant, the volume of the nucleus is directly proportional to the mass number $A$. The factor $\tfrac{4}{3}\pi R_0^{3}$ is the fixed volume occupied by a single nucleon. Each proton or neutron added to a nucleus contributes the same parcel of volume, so doubling the number of nucleons doubles the nuclear volume. This is precisely the behaviour of a liquid drop: NCERT remarks that different nuclei are "like a drop of liquid of constant density," and this picture underpins the liquid-drop model used elsewhere in nuclear physics.

Density Independent of A

The single most examined consequence of the radius law is the constancy of nuclear density. The mass of a nucleus is approximately the number of nucleons times the mass $m$ of one nucleon, $M \approx A\,m$. Its volume is $V = \tfrac{4}{3}\pi R_0^{3} A$. The density is the ratio:

$$\rho = \frac{M}{V} = \frac{A\,m}{\tfrac{4}{3}\pi R_0^{3}\,A} = \frac{m}{\tfrac{4}{3}\pi R_0^{3}}.$$

The mass number $A$ cancels exactly. The density depends only on the nucleon mass and the constant $R_0$, so it is the same for every nucleus — light or heavy, the matter is packed identically tightly. Numerically NCERT gives $\rho \approx 2.3 \times 10^{17}\ \text{kg m}^{-3}$, and NIOS confirms the same order for both hydrogen and oxygen. NCERT Example 13.1 reaches $2.29 \times 10^{17}\ \text{kg m}^{-3}$ from the iron nucleus. This is the algebraic content of NCERT Exercise 13.10, which asks you to show density is independent of $A$ from the radius relation.

This density is staggering. Water has $\rho = 10^{3}\ \text{kg m}^{-3}$, so nuclear matter is some $10^{14}$ times denser than water. NCERT explains the contrast: ordinary matter is mostly empty space because the atom is overwhelmingly empty, while the nucleus is matter at full packing. NIOS dramatises the figure — if the Earth were compressed to nuclear density it would be a sphere of radius about $184\ \text{m}$. NCERT adds that neutron stars have densities comparable to this, so a neutron star behaves like one gigantic nucleus.

Nucleus Versus Atom

The nucleus is dwarfed by the atom that surrounds it. Alpha-scattering experiments show the nuclear radius is smaller than the atomic radius by a factor of about $10^{4}$. Because volume scales as the cube of the radius, the nuclear volume is roughly $10^{-12}$ times the atomic volume — the atom is almost entirely empty. NCERT's image is exact: enlarge an atom to the size of a classroom and the nucleus is a pinhead. NIOS offers a parallel scale comparison — a nucleus to its atom is like a bucket of water to the water held in a large dam.

Yet that pinhead carries more than 99.9% of the atom's mass. The combination of a tiny volume and almost all the mass is exactly what produces the colossal nuclear density derived above. The two ideas — "atom mostly empty" and "nuclear matter incredibly dense" — are two faces of the same fact and are frequently tested together.

Worked Examples