Size of nucleus

Every atom is almost entirely empty space wrapped around a speck so small that, for decades after the electron was discovered, nobody knew it was even there. Rutherford's alpha-particle scattering experiment changed that: most alpha particles sailed straight through gold foil as if nothing were in their way, but a rare few bounced almost straight back, as if they had struck something unimaginably dense and unimaginably small. That "something" is the nucleus, and pinning down exactly how small it is — and how its size scales as you add more protons and neutrons — is the question this topic answers. The result, refined by later alpha- and electron-scattering experiments, is a strikingly simple empirical law: nuclear radius grows as the cube root of the mass number, $R = R_0 A^{1/3}$ . Behind that one-line formula sits a beautiful piece of physics. It says nucleons are packed into the nucleus like incompressible marbles into a bag — add more marbles and the bag's volume grows in proportion, so the density of the packing never changes, no matter whether you are looking at helium or uranium. That single fact — constant nuclear density — is the thread that ties nuclear size to nuclear structure, and it is the quiet assumption sitting underneath everything the "Nuclei" chapter builds next: binding energy, the stability curve, fission and fusion. This article works through where the $R = R_0 A^{1/3}$ law comes from, what it predicts about nuclear density and the astonishing emptiness of the atom, and how to use it confidently in board-style and competitive-exam calculations — including the aluminium and uranium numbers you will meet again when the chapter turns to binding energy per nucleon.

Key takeaways

Nuclear radius obeys an empirical cube-root law: $R = R_0 A^{1/3}$ , with $R_0 \approx 1.2\ \text{fm}$ ( $1\ \text{fm} = 10^{-15}\ \text{m}$ ), found from alpha- and electron-scattering experiments.
Volume is proportional to mass number A, not to $R$ directly — because volume goes as $R^3 \propto (A^{1/3})^3 = A$ .
Nuclear density is essentially the same for every nucleus, roughly $2.3\times10^{17}\ \text{kg/m}^3$ , whether the nucleus is light (helium) or heavy (uranium) — the $A$ cancels out of the density formula.
The nucleus is millions of times smaller than the atom: nuclear radii run from about $1$ – $2$ fm to $7$ – $8$ fm, against a typical atomic radius of about $10^{-10}$ m, so the atom is overwhelmingly empty space.
Probing something this small needs matter waves, not light — visible light's wavelength ( $\sim10^{-7}$ m) is far too coarse, which is why high-energy electron or alpha-particle beams are used to measure $R$ .
$R_0$ is an empirical fit, not an exact universal constant — real nuclei deviate slightly, and different experimental techniques (charge radius from electrons vs. matter radius from alphas) give slightly different values clustered near $1.1$ – $1.2$ fm.
Key idea: because $R = R_0 A^{1/3}$ makes nuclear volume directly proportional to mass number A, nuclei of every size pack nucleons at the same constant density, about $2.3\times10^{17}\ \text{kg/m}^3$ , confirming that nuclei behave like closely-packed, incompressible spheres of nuclear matter.

The core insight

Scattering experiments show nuclear radius follows the empirical law $R = R_0 A^{1/3}$ with $R_0 \approx 1.2$ fm, which forces nuclear volume to scale directly with mass number $A$ and therefore makes nuclear density — about $2.3\times10^{17}\ \text{kg/m}^3$ — essentially the same constant for every nucleus in the periodic table.

Where it fits in the syllabus

This topic opens the CBSE Class 12 Physics chapter "Nuclei" (Unit IX, Atoms and Nuclei), coming right after nuclear composition (protons, neutrons, mass number $A$ , atomic number $Z$ ) and just before mass-energy and binding energy. The prescribed 2025-26 scope keeps this squarely as: the empirical size relation $R = R_0 A^{1/3}$ , the consequence of constant nuclear density, and the qualitative picture of a short-range, charge-independent nuclear force holding this dense packing together — no derivation of $R_0$ from scattering cross-sections is expected, and no mention of the nuclear force's meson-exchange mechanism belongs here. One honesty note for anyone comparing against older resources: the 2025-26 rationalised syllabus drops the entire radioactive-decay-law block ( $N = N_0 e^{-\lambda t}$ , half-life, activity) that used to sit in this chapter — none of that appears in this article, and nuclear size stands on its own as background for the binding-energy-per-nucleon curve that follows.

The physical picture

Rutherford-style scattering experiments — alpha particles fired at nuclei, later refined using high-energy electron beams that probe even finer detail via diffraction — consistently show that a nucleus behaves, to a very good approximation, as a sphere of definite radius. That radius grows with the number of nucleons $A$ according to an empirical formula, not a formula derived from first principles:

R = R_0\, A^{1/3}

where: $R$ is the nuclear radius, $A$ is the mass number (total number of protons plus neutrons), and $R_0 \approx 1.2\times10^{-15}\ \text{m} = 1.2\ \text{fm}$ is an empirically fitted constant (one femtometre, $10^{-15}$ m, is also called a fermi in nuclear physics, honouring Enrico Fermi). The physical reading is that radius does not grow in simple proportion to the number of nucleons — it grows only as the cube root, because a sphere's radius scales as the cube root of its volume, and it is volume that ends up proportional to $A$ . That last point is worth making explicit. The volume of the (roughly spherical) nucleus is

V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3\, A

so nuclear volume is directly proportional to $A$ , the total nucleon count — double the nucleons and you double the volume, not the radius. Since the mass of the nucleus is also very nearly proportional to $A$ (every nucleon has almost the same mass, close to $1$ u $= 1.66\times10^{-27}$ kg, whether it is a proton or a neutron), the nuclear density

\rho = \frac{\text{mass}}{\text{volume}} = \frac{A\, m_{\text{nucleon}}}{\frac{4}{3}\pi R_0^3\, A} = \frac{m_{\text{nucleon}}}{\frac{4}{3}\pi R_0^3}

comes out completely independent of $A$ : it cancels cleanly out of both numerator and denominator. Numerically this constant works out to about $2.3\times10^{17}\ \text{kg/m}^3$ — a genuinely enormous number next to water's $10^3\ \text{kg/m}^3$ , and it is the same number whether you compute it for helium-4 or uranium-238. This constant-density behaviour is exactly what you would expect if nucleons pack together like closely-fitted, essentially incompressible spheres of nuclear matter, each nucleon occupying roughly the same little volume of "nuclear stuff" no matter which nucleus it happens to belong to.

Figure

Three circles drawn to scale showing nuclear radius growing with the cube root of mass number for helium four aluminium twenty seven and uranium two thirty eight alongside a much larger circle representing a typical atom to show how empty the atom is

Typical nuclear radii range from about $1$ – $2$ fm for the lightest nuclei (hydrogen, helium) up to about $7$ – $8$ fm for the heaviest natural nuclei such as uranium, compared with a typical atomic radius of order $10^{-10}\ \text{m} = 10^{5}\ \text{fm}$ . The nucleus therefore occupies roughly $(10^{-5})^3 = 10^{-15}$ of the atom's volume — a direct, numerical confirmation of the conclusion Rutherford reached qualitatively: the atom is overwhelmingly empty space, with essentially all its mass concentrated in a central speck a hundred-thousandth of the atom's size. This also explains why the nucleus can never be resolved with ordinary visible light: to image a structure of size $R$ , the probing wave's wavelength must be comparable to or smaller than $R$ . Visible light, with wavelengths near $10^{-7}$ m, is millions of times too coarse; only high-energy electron or alpha-particle beams — whose de Broglie wavelength shrinks as their energy rises — carry a short enough wavelength to probe femtometre-scale structure, which is exactly the technique that measured $R_0$ in the first place.

Derivation AWhy nuclear density comes out the same for every nucleus

The goal is to show, using nothing beyond the empirical radius law and the definition of density, that the density of nuclear matter must be independent of mass number $A$ — the same constant number for a light nucleus like helium and a heavy one like uranium. This is the single most important consequence of $R = R_0 A^{1/3}$ , and the reasoning uses only algebra: no calculus, no scattering theory, just substitution and cancellation.

Step 1. Write the mass of a nucleus with mass number $A$ . Since every nucleon (proton or neutron) has very nearly the same mass $m_{\text{nucleon}} \approx 1\ \text{u} = 1.66\times10^{-27}\ \text{kg}$ , the total nuclear mass is simply $m = A\, m_{\text{nucleon}}$ — mass is directly proportional to $A$ , the total nucleon count.

Step 2. Write the volume of the nucleus, treating it as a sphere of radius $R$ . Using $R = R_0 A^{1/3}$ from the empirical scattering law, $V = \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi \left(R_0 A^{1/3}\right)^3 = \dfrac{4}{3}\pi R_0^3\, A$ . Cubing the cube root simply returns $A$ , so volume too is directly proportional to $A$ — this is the key algebraic step, and it is exactly why radius grows only as $A^{1/3}$ rather than $A$ : it is volume, not radius, that nature makes proportional to nucleon count.

Step 3. Divide mass by volume to get density. $\rho = \dfrac{m}{V} = \dfrac{A\, m_{\text{nucleon}}}{\frac{4}{3}\pi R_0^3\, A}$ . Both the numerator and the denominator carry a factor of $A$ .

Step 4. Cancel the $A$ . $\rho = \dfrac{m_{\text{nucleon}}}{\frac{4}{3}\pi R_0^3}$ — mass number has disappeared from the expression entirely. Density depends only on the nucleon mass and the fixed constant $R_0$ , neither of which changes from one nucleus to the next.

Step 5. Evaluate the number and read the physics. Substituting $m_{\text{nucleon}} \approx 1.66\times10^{-27}\ \text{kg}$ and $R_0 = 1.2\times10^{-15}\ \text{m}$ gives $\rho \approx 2.3\times10^{17}\ \text{kg/m}^3$ for any $A$ . The physical picture behind the algebra is a bag of marbles: adding more nucleons does not squeeze them tighter or spread them thinner, it simply makes a bigger bag at the same packing density — direct evidence that nucleons behave as closely-packed, essentially incompressible objects held together by a short-range force, rather than as point charges free to spread out or clump arbitrarily.

Worked examples

Board level - radius and relative size of an aluminium nucleus

Question: Estimate the radius of a $^{27}_{13}\text{Al}$ (aluminium) nucleus using $R_0 = 1.2$ fm, and state roughly what fraction of a typical atom's diameter this represents.

Step 1. Identify the mass number: for aluminium, $A = 27$ .

Step 2. Apply the empirical law $R = R_0 A^{1/3}$ : $R = 1.2 \times 27^{1/3}$ fm.

Step 3. Simplify the cube root. Since $27 = 3^3$ , $27^{1/3} = 3$ exactly — a convenient number chosen deliberately for board-level practice.

Step 4. Compute: $R = 1.2 \times 3 = 3.6\ \text{fm} = 3.6\times10^{-15}\ \text{m}$ .

Step 5. Compare with a typical atomic radius of $\sim10^{-10}\ \text{m}$ : the ratio of radii is $3.6\times10^{-15} / 10^{-10} \approx 3.6\times10^{-5}$ , so the aluminium nucleus is roughly thirty thousand times smaller in radius than the atom that contains it — and, since volume scales as the cube of that ratio, about $(3.6\times10^{-5})^3 \approx 4.7\times10^{-14}$ of the atom's volume.

The aluminium nucleus has a radius of about $3.6$ fm, occupying only about $5\times10^{-14}$ of the volume of the atom around it.

Competitive level - density and packing fraction of a uranium-238 nucleus

Question: Estimate the radius and density of a $^{238}_{92}\text{U}$ (uranium) nucleus using $R_0 = 1.2$ fm, then find what fraction of a sphere of atomic radius $1.0\times10^{-10}\ \text{m}$ is occupied by this nucleus.

Step 1. Identify $A = 238$ for uranium-238, and take its mass as $238\ \text{u} = 238 \times 1.66\times10^{-27}\ \text{kg} = 3.95\times10^{-25}\ \text{kg}$ .

Step 2. Find $A^{1/3}$ . Since $6.2^3 = 238.3$ , a very close match, $238^{1/3} \approx 6.20$ .

Step 3. Apply $R = R_0 A^{1/3} = 1.2 \times 6.20 = 7.44\ \text{fm} = 7.44\times10^{-15}\ \text{m}$ — this is the uranium radius used again later, when the chapter computes uranium's binding energy per nucleon.

Step 4. Compute the nuclear volume: $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (7.44\times10^{-15})^3$ . Cubing, $(7.44\times10^{-15})^3 \approx 4.12\times10^{-43}\ \text{m}^3$ , so $V \approx 4.18879 \times 4.12\times10^{-43} \approx 1.72\times10^{-42}\ \text{m}^3$ .

Step 5. Compute the density: $\rho = \dfrac{m}{V} = \dfrac{3.95\times10^{-25}}{1.72\times10^{-42}} \approx 2.3\times10^{17}\ \text{kg/m}^3$ — matching the aluminium-derived value from the derivation above to two significant figures, exactly as the constant-density argument predicts.

Step 6. Compare nuclear volume to an atomic-scale sphere of radius $1.0\times10^{-10}$ m: that atomic volume is $\frac{4}{3}\pi (10^{-10})^3 \approx 4.19\times10^{-30}\ \text{m}^3$ . The ratio is $\dfrac{1.72\times10^{-42}}{4.19\times10^{-30}} \approx 4.1\times10^{-13}$ .

The uranium-238 nucleus has radius $7.44$ fm, density $\approx 2.3\times10^{17}\ \text{kg/m}^3$ (matching the light-nucleus value), and occupies only about $4\times10^{-13}$ of a comparably sized atomic sphere — a larger fraction than aluminium's, precisely because a heavier nucleus, while denser-looking in absolute size, still occupies a fixed tiny sliver of a roughly fixed-size atom. Timing note: in a competitive exam, do not compute $(7.44)^3$ digit by digit under time pressure — approximate $7.44 \approx 7.5$ mentally, note $7.5^3 = 421.9$ , and carry that rounded value through; the density still lands within a few per cent of the exact answer, which is well inside the tolerance most objective-type questions accept, and it saves precious seconds for the next question.

Common traps

Thinking heavier nuclei are denser than lighter ones. Because $R \propto A^{1/3}$ makes volume proportional to $A$ , density comes out essentially the same for every nucleus — heavier nuclei are simply bigger in radius and volume, not more tightly packed.
Forgetting to cube the ratio when relating radius to volume. $R = R_0 A^{1/3}$ , but $V \propto R^3 \propto A$ , not $V \propto A^{1/3}$ — a very common algebra slip under exam pressure.
Confusing femtometres (nuclear scale, $10^{-15}$ m) with picometres or angstroms (atomic scale, $10^{-10}$ m) — an order-of-magnitude slip of five whole powers of ten that silently wrecks any comparison between nuclear and atomic size.
Treating $R_0$ as an exact universal constant rather than an empirical fit. Measured nuclear radii cluster around $R_0 \approx 1.1$ – $1.2$ fm depending on the experimental technique (electron scattering measures charge radius; alpha scattering measures the range of the nuclear force) — the formula is a good approximation, not an exact law, and small deviations from it are expected, especially near closed-shell "magic number" nuclei.
Mixing up mass number $A$ (protons plus neutrons, what goes into $R = R_0 A^{1/3}$ ) with atomic number $Z$ (protons only) — nuclear size depends on the total nucleon count, not on how many of those nucleons are charged.

You should be able to…

State the empirical formula $R = R_0 A^{1/3}$ together with the standard value of $R_0 \approx 1.2$ fm, and explain that it is an experimental fit, not a derived law.
Explain, using the algebra of the derivation above, why nuclear volume is proportional to mass number $A$ while radius grows only as $A^{1/3}$ .
Explain why nuclear density works out to be (approximately) the same constant, about $2.3\times10^{17}\ \text{kg/m}^3$ , for every nucleus, and reproduce that cancellation from first principles.
Calculate the radius, volume, or density of a given nuclide from its mass number, including cube roots that are not perfect cubes.
Compare the relative sizes of nucleus and atom quantitatively, and explain in one sentence why visible light cannot resolve nuclear structure.

Ask the tutor: want to see how this same $R = R_0 A^{1/3}$ calculation feeds directly into computing binding energy per nucleon for uranium-238 in the next topic — ask the AI tutor to carry the numbers forward.

Why does the radius formula use A to the power one third instead of just A?

Because it is the nuclear volume, not the radius, that nature makes proportional to the number of nucleons — each nucleon simply adds its own fixed little volume to the total, like adding one more marble to a bag. Since volume of a sphere goes as $R^3$ , and volume is proportional to $A$ , you need $R^3 \propto A$ , which means $R \propto A^{1/3}$ . If radius itself were directly proportional to $A$ , volume would explode as $A^3$ , and nuclear density would rocket down for heavy nuclei — which scattering experiments flatly do not observe. The cube root is exactly what constant density demands.

Is R0 = 1.2 fm exact, and where does that number actually come from?

It is an empirical best-fit value, not a fundamental constant derived from theory. Physicists fire beams of alpha particles or high-energy electrons at nuclei and measure how the scattering pattern depends on nuclear size (electron scattering probes the charge distribution; alpha scattering probes how close an alpha can approach before the strong nuclear force takes over from electrostatic repulsion). Fitting $R = R_0 A^{1/3}$ to that scattering data across many nuclei gives $R_0$ somewhere around $1.1$ to $1.2$ fm depending on the exact method — CBSE uses $1.2$ fm as the standard working value. Small real deviations from the pure $A^{1/3}$ law do exist, especially for nuclei with "magic" proton or neutron numbers that are unusually stable and slightly more compact, but those refinements sit beyond board-level scope.

If the nucleus is so tiny compared to the atom, why does almost all the mass sit there?

Because electrons are roughly two thousand times lighter than a proton or neutron, so even though electrons occupy essentially the entire volume of the atom, their total mass contribution is negligible next to the nucleus. Picture the atom as a football stadium with the nucleus as a pea at the centre: the pea holds essentially all the stadium's mass, while the enormous stadium volume is filled almost entirely by the ghost-light electron cloud (occupying it in the quantum-mechanical sense of probability, not literally filling space with stuff). This is exactly the picture Rutherford's scattering result forced onto physics in 1911, and the $R = R_0 A^{1/3}$ law is simply the precise, quantitative version of "the pea is small" that this chapter builds on.

Does a bigger nucleus mean a bigger nuclear force holding it together?

Not in the sense of reaching farther — the nuclear (strong) force is famously short-range, acting only over about $1$ – $2$ fm, roughly the spacing between neighbouring nucleons, regardless of how large the whole nucleus is. That is precisely why each nucleon only "feels" its nearest neighbours rather than every other nucleon in the nucleus, which is what makes binding energy per nucleon roughly constant in the middle of the periodic table rather than growing without bound as $A$ increases — a point the very next topic in this chapter, binding energy, builds directly on the constant-density picture established here.