In short

Every atomic nucleus is made of two kinds of particles, collectively called nucleons:

  • proton — mass 1.6726×10⁻²⁷ kg, charge +e, spin ½
  • neutron — mass 1.6749×10⁻²⁷ kg, charge 0, spin ½

Two integers identify any nucleus:

  • the atomic number Z (number of protons — determines the chemical element),
  • the mass number A (total number of nucleons, protons + neutrons).

The neutron number is N = A - Z. Nuclides are written ^A_Z X (e.g. ^{14}_6\text{C} is carbon-14: 6 protons, 8 neutrons, 14 nucleons). Two nuclei with the same Z but different A are isotopes — chemically identical, nuclearly different.

The radius of a nucleus scales with the cube root of the number of nucleons:

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

This is the empirical consequence of nuclear matter having approximately constant density, about

\rho_\text{nuc} \approx 2.3\times 10^{17}\ \text{kg/m}^3,

independent of which nucleus you pick. A nucleon claims roughly the same volume (\tfrac{4}{3}\pi R_0^3 \approx 7.2\ \text{fm}^3) whether it lives in a deuteron or in a uranium-238 nucleus.

The force that holds nucleons together is the strong nuclear force: short-range (range \sim 1 fm), charge-independent (acts equally on protons and neutrons), strongly attractive at distances of 1 fm, and strongly repulsive at distances below 0.5 fm. Its competition with the long-range Coulomb repulsion between protons is what sets the stability pattern of the whole periodic table.

Rutherford's 1911 experiment left you staring at a tiny fact. Almost all the mass of an atom sits in a region 100,000 times smaller than the atom itself — a sphere less than 10^{-14} metres across, positively charged, almost infinitely dense compared to anything in everyday life. That is the nucleus. The question his experiment could not answer was: what is it made of?

In 1919, Rutherford himself found the first constituent. He bombarded nitrogen gas with alpha particles and detected a new particle flying out — lighter than an alpha, positively charged, with mass equal to the hydrogen ion. He called it the proton and conjectured that every nucleus was built from them. For hydrogen-1 (a single proton in the nucleus), that worked beautifully. For helium-4, it did not — helium has charge +2e but mass roughly 4 proton masses, which left the obvious question of why the other two mass units carried no charge. For thirteen years physicists guessed that the extra mass came from tightly bound proton-electron pairs stuffed into the nucleus. It was an ugly guess.

The real answer arrived in 1932. James Chadwick, a student of Rutherford, noticed that when alpha particles hit beryllium, a neutral penetrating radiation came out that could not be gamma rays (its penetrating power was wrong). A few careful experiments showed that this radiation was made of particles — neutrons — with mass almost equal to the proton's but no electric charge. Chadwick's paper, three pages long, named the particle that cleaned up the story of every nucleus heavier than hydrogen. Helium-4 is two protons and two neutrons. Oxygen-16 is eight of each. Uranium-238 is 92 protons and 146 neutrons. All nuclei, all the time, are just protons and neutrons.

India's own nuclear physics community got going in precisely this era. Meghnad Saha at Allahabad University had already derived (in 1920) the ionisation equation that bridges atomic physics and astrophysics. Homi Bhabha, arriving at Cambridge in 1927, would go on to calculate electron-positron scattering cross-sections (the Bhabha cross-section, still taught today) and later return to India to found what is now BARC in Trombay. By 1945, nuclear composition was the starting point for reactor design and for understanding the cosmic-ray showers that Bhabha measured on Indian mountain tops. The nucleus became applied physics within a generation of its composition being settled.

This article puts the pieces together: what nucleons are, how Z and A label every nucleus, how big nuclei are (and why the radius goes as A^{1/3}), how dense they are, and what holds them together.

Protons and neutrons — the nucleons

The two kinds of particles inside a nucleus share the name nucleon because they are nearly interchangeable for most purposes:

Properties of the proton and neutronA two-column comparison table. Column headers: proton and neutron. Rows: mass, charge, spin, magnetic moment, stability. Proton: 1.6726 times ten to the minus twenty seven kilograms, plus e, spin one half, positive 2.79 nuclear magnetons, stable. Neutron: 1.6749 times ten to the minus twenty seven kilograms, zero charge, spin one half, negative 1.91 nuclear magnetons, unstable outside the nucleus with 15 minute half-life.protonneutronmass1.6726×10⁻²⁷ kg (938.27 MeV/c²)1.6749×10⁻²⁷ kg (939.57 MeV/c²)charge+e0spin½ ℏ½ ℏmagnetic moment+2.79 μ_N−1.91 μ_Nstabilitystable (τ > 10³⁴ yr)unstable (half-life 10.2 min)
Properties of the proton and neutron. Their masses differ by only 0.14%, and both carry spin ½ (they are fermions, like electrons). The proton is stable on any practical timescale (no proton decay has been seen despite searches in water tanks for decades — this is what the Super-Kamiokande experiment and India's proposed INO at Theni have been looking for). A *free* neutron is unstable and beta-decays into a proton, electron, and antineutrino with a half-life of about 10 minutes; a *bound* neutron inside most nuclei is stable because the decay is energetically forbidden.

One curious fact: the neutron has a magnetic moment. A truly neutral elementary particle would have no magnetic moment — but the neutron is not elementary. Deep-inelastic scattering experiments at SLAC in the 1960s (and at the IUCAA and TIFR groups that joined the international H1/ZEUS collaborations later) showed that protons and neutrons are made of quarks: the proton is uud (two up quarks, one down), the neutron is udd. The up quark carries charge +2e/3; the down quark carries -e/3. Add them up and you get +e for the proton and 0 for the neutron — but the charge distribution inside the neutron is not spatially uniform, so it has a magnetic moment even though its total charge is zero. For almost all of chemistry, nuclear physics, and JEE-relevant problems, you can treat protons and neutrons as if they were point particles with the masses and charges above. The quark substructure becomes important only at energies thousands of times higher than anything inside a normal nucleus.

Atomic number, mass number, isotopes

Two integers completely identify any nucleus:

Atomic number Z — the number of protons. Determines the chemical element (hydrogen has Z = 1, carbon has Z = 6, iron has Z = 26, uranium has Z = 92).

Mass number A — the total number of nucleons = protons + neutrons. So the neutron number is N = A - Z.

The universally accepted notation places A as a superscript and Z as a subscript before the element symbol:

\mathbf{^{A}_{Z}X}

Since the chemical symbol X already fixes Z, the subscript is often dropped: ^{12}\text{C}, ^{14}\text{N}, ^{235}\text{U}. Any nucleus is fully specified by its element and mass number.

Isotopes of an element are nuclei with the same Z but different A (same protons, different neutron counts). Chemically they are indistinguishable — chemistry only cares about Z, which sets the electron configuration — but they are distinct nuclear species with different masses, different nuclear spins, and sometimes different stability.

Hydrogen isotopes and uranium isotopesSix cartoon nuclei. Top row: hydrogen isotopes — protium (one red dot labelled p), deuterium (one red dot p, one dark dot n), tritium (one p, two n). Bottom row: uranium isotopes — U-234 (blob labelled 92p + 142n), U-235 (blob with 92p + 143n), U-238 (blob with 92p + 146n).^1H (protium)pZ = 1, A = 1^2H (deuterium)pnZ = 1, A = 2^3H (tritium)pnnZ = 1, A = 3^234U92p+142nZ = 92, A = 234^235U92p+143nZ = 92, A = 235^238U92p+146nZ = 92, A = 238
Isotopes in pictures. The three stable and radioactive isotopes of hydrogen differ only in neutron count, but their chemistry is identical — heavy water ($D_2O$, built from deuterium) behaves chemically exactly like ordinary water but has subtly different physical properties (higher boiling point, slightly different reaction rates), which is why it is used as a moderator in the pressurised heavy water reactors built by NPCIL across India. Uranium has 23 known isotopes; the fissile $^{235}U$ is 0.72% of natural uranium (the rest is mostly $^{238}U$). Enriching the $^{235}U$ fraction is physically difficult because the two isotopes are chemically identical — their separation relies on the 1% mass difference alone.

Isotope nomenclature — examples you will meet often:

Two related but distinct terms:

How big is a nucleus — the A^{1/3} radius formula

The very first thing to measure about a nucleus is its size. Rutherford's alpha-scattering gave an upper bound — alphas scattered off gold at closest approaches down to about 3\times10^{-14} m without visible deviation from Coulomb repulsion, so the gold nucleus was at most that big. Over the following decades, electron-scattering experiments (pioneered by Robert Hofstadter in the 1950s at Stanford) mapped the charge distributions of dozens of nuclei directly. The pattern that emerged is clean:

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

with R_0 \approx 1.2\ \text{fm} = 1.2\times10^{-15}\ \text{m}. This is the nuclear-radius formula. There is nothing mysterious about the A^{1/3} — it is the signature of a nucleus in which each nucleon occupies roughly the same volume, independent of which nucleus you are looking at.

Step 1. Assume nuclear density is constant (an empirical fact, to be verified in Example 1 below). Then the volume of a nucleus is proportional to the number of nucleons:

V \propto A.

Why: a nucleus of mass number A contains A nucleons. If each one claims the same fixed volume, the total volume scales linearly with A. This is exactly the same reasoning that tells you a pile of identical marbles takes up a volume proportional to the number of marbles.

Step 2. Model the nucleus as a sphere of radius R, so its volume is V = \tfrac{4}{3}\pi R^3.

\tfrac{4}{3}\pi R^3 \propto A.

Why: nuclei are roughly spherical (except the very heaviest and those with unclosed shells, which can be slightly prolate like a rugby ball). The spherical approximation is very good for the purposes of this formula.

Step 3. Solve for R:

R^3 \propto A \;\Longrightarrow\; R \propto A^{1/3} \;\Longrightarrow\; R = R_0 A^{1/3}.

Why: the cube root comes straight from volume-to-radius. The constant R_0 is whatever the empirical data fits — about 1.2 fm from electron scattering, sometimes quoted as 1.1 to 1.4 fm depending on exactly how the nuclear edge is defined (rms radius vs half-density radius).

Step 4. Numerical check with a few nuclei. Use R_0 = 1.2 fm.

Uranium-238 is about six times bigger than hydrogen — even though it weighs 238 times more. Put another way: a factor 238 in mass becomes only a factor 238^{1/3} \approx 6.2 in radius because nuclear matter does not compress further.

Interactive: nuclear radius vs mass numberA plot with mass number A on the horizontal axis (0 to 260) and nuclear radius R in femtometres on the vertical axis (0 to 9). A red curve shows R equals 1.2 times A to the one-third power, rising steeply for small A and flattening at large A. A draggable point slides along the mass-number axis and a dashed vertical line drops from the curve to the value of A. Readouts show A, R in femtometres, and nuclear density in kilograms per cubic metre.mass number Anuclear radius R (fm)206010016022002468R = R₀·A^(1/3)drag the red point along A
Drag the red point to change the mass number $A$. The curve $R = 1.2\,A^{1/3}$ grows steeply for small $A$ (hydrogen to oxygen doubles the radius for a sixteen-fold rise in mass) and flattens for heavy nuclei (going from $^{56}\text{Fe}$ to $^{238}\text{U}$ — a fourfold rise in mass — only raises the radius by about 60%). The density (readout) is essentially constant at $2.3 \times 10^{17}$ kg/m³ for all $A$ — that is the defining property of nuclear matter.

Nuclear density — a number so big it is worth stating twice

The radius formula implies constant density. Compute it.

Step 1. Mass of a nucleus of mass number A. To excellent approximation (we are about to do a rough estimate; see the binding energy article for the 0.8% correction):

M \approx A\, m_N,

where m_N \approx 1.66\times 10^{-27} kg is the average nucleon mass (halfway between the proton and neutron masses — this is close to the atomic mass unit u).

Why: A nucleons, each of mass m_N. The actual mass of a nucleus is slightly less than A\, m_N — that mass-defect is where nuclear binding energy comes from — but the defect is under 1% so it is negligible for a density calculation.

Step 2. Volume of a sphere of radius R = R_0 A^{1/3}.

V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi R_0^3 A.

Why: cubing R_0 A^{1/3} gives R_0^3 A. The A dependence comes straight through — the volume scales linearly with the number of nucleons.

Step 3. Density is mass over volume.

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

Why: the A cancels. The density depends only on m_N and R_0 — both are nucleon properties, not nucleus properties. This is exactly the expected behaviour of a substance with a fixed density: how much of it you have does not change the density.

Step 4. Plug in numbers.

\rho = \frac{1.66\times10^{-27}\ \text{kg}}{\tfrac{4}{3}\pi (1.2\times10^{-15}\ \text{m})^3} = \frac{1.66\times10^{-27}}{7.24\times 10^{-45}}\ \text{kg/m}^3.
\boxed{\;\rho \approx 2.3 \times 10^{17}\ \text{kg/m}^3\;}

Why: the denominator evaluates to (4/3)\pi \cdot 1.728\times 10^{-45} = 7.24\times 10^{-45} m³. Ratio: 2.29\times 10^{17}. This is a constant of nature — every nucleus has this density.

Two comparisons to make this number real:

This is also, remarkably, the density of a neutron star: a collapsed stellar core held up against gravity by neutron degeneracy pressure is nuclear matter on a \sim10-km scale. The Giant Metrewave Radio Telescope (GMRT) near Pune has been timing millisecond pulsars — spinning neutron stars — for nearly two decades, constraining the equation of state of nuclear matter from observations of stars made almost entirely out of the stuff we are calculating the density of. Everyday nucleus, distant neutron star: same substance.

The strong nuclear force — what holds the nucleons together

A nucleus is a contradiction. Protons repel each other electrically, with a Coulomb force that, at a separation of 1 fm, amounts to

F_C = \frac{k e^2}{r^2} = \frac{9\times 10^9 \cdot (1.6\times 10^{-19})^2}{(10^{-15})^2} = 230\ \text{N}.

That is a huge force at that scale — about 60 tonnes of force per square micron of cross-section (by which I mean: the pressure this force exerts on a nucleon is about 10^{30} Pa). No electrostatic engineer could ever hold two protons that close together. Yet uranium-238 packs 92 protons into a volume a few femtometres across and stays together for four billion years.

The answer is the strong nuclear force (sometimes the nuclear force), a force that is:

  1. Short-range. Its range is about 1 to 2 fm. Beyond about 2 fm, it drops to essentially zero. At less than 0.5 fm, it flips sign and becomes repulsive (a hard core). Within the attractive-window of 0.5 to 2 fm, it is much stronger than the Coulomb force — at 1 fm, the nuclear force between two nucleons is roughly 10⁴ N, beating the Coulomb 230 N by a factor of 40.

  2. Charge-independent. It acts equally on proton-proton, proton-neutron, and neutron-neutron pairs. The only difference in behaviour is that the electrical Coulomb force is on top of the strong force for pp pairs, absent for nn pairs, and gives a small additional binding for a pn pair in certain states. This isospin symmetry is exact to about 1%.

  3. Spin-dependent. Two nucleons bind more strongly when their spins are parallel. This is why the only stable two-nucleon system is the deuteron (p + n, spins parallel, spin 1) — two protons or two neutrons in a spin-singlet configuration do not bind.

  4. Saturated. Each nucleon only interacts strongly with its nearest few neighbours, not with every other nucleon in the nucleus. This saturation is the micro-reason for the constant-density A^{1/3} pattern — each nucleon settles into roughly the same local environment, regardless of how big the nucleus is.

A useful cartoon is to plot the nucleon-nucleon potential:

Nucleon-nucleon potentialAxes showing nucleon-nucleon separation r on the horizontal axis from 0 to 4 femtometres and potential energy V on the vertical axis. The curve rises to positive infinity at small r (hard core repulsion), crosses zero near r equals 0.5 femtometres, dips to a minimum of about minus 100 MeV near r equals 1 femtometre, and approaches zero for r larger than 2 femtometres. A shaded region indicates the attractive well.separation r (fm)V(r) (MeV)12340−100attractive well: binds nucleonshard core (r < 0.5 fm): repulsiveV → 0 for r > 2 fm
A schematic nucleon-nucleon potential $V(r)$ as a function of the separation $r$ between two nucleons. The hard core at short distances (below 0.5 fm) prevents nucleons from collapsing onto each other — this is what gives nuclear matter its fixed density. The attractive well around 1 fm, with depth about 100 MeV, is what binds nucleons to each other. Beyond 2 fm the force switches off and the nucleons effectively stop seeing each other. The Coulomb repulsion between two protons (long-range, $1/r^2$, never switching off) would lie across the top of this plot as a small additive piece; at nuclear separations it is dominated by the nuclear force.

The modern understanding, from quantum chromodynamics (QCD), is that the nuclear force is a residual effect of the far stronger force that binds quarks into nucleons — somewhat analogous to the way the van der Waals force between two neutral atoms is a residue of the far stronger electromagnetic force inside each atom. The mediator of the nuclear force at nuclear-physics energies is primarily the pion, a meson with mass about 140 MeV/c² and range \lambda = \hbar/(m_\pi c) \approx 1.4 fm — which matches the observed range of the force and was Hideki Yukawa's 1934 prediction (confirmed in 1947 by the discovery of the pion in cosmic rays, including in cloud-chamber work done by the physicist Bibha Chowdhuri in Kolkata). The pion-exchange picture is one of the great case studies of how a phenomenological force constant comes out of a deeper particle physics.

Worked examples

Example 1: Radius and density of $^{56}$Fe — the most stable nucleus

Iron-56 is the nucleus at the peak of the binding-energy curve — the endpoint of stellar nucleosynthesis and the most stable of all nuclides (see the mass defect article). Compute (a) its radius and (b) its density, and compare to the density of solid iron metal.

Iron-56 nucleus to scale with a Bohr atomA large circle of radius 100 pixels on the left represents the electron cloud of an iron atom to some scale. A tiny filled dot at the centre represents the Fe-56 nucleus. The nucleus is labelled 4.6 femtometres while the electron cloud is labelled 1.4 Angstrom. Scale mismatch is noted as the electron cloud being thirty thousand times bigger than the nucleus.nucleus: 4.6 fmelectron cloud: 1.4 Å = 140,000 fmscale:cloud / nucleus = 30,000if cloud were a football stadium,the nucleus would be a grain of rice
Iron-56 drawn roughly to scale with the electron cloud of an Fe atom. The nucleus is more than 30,000 times smaller than the atom. If the Fe atom were the size of a football stadium, the nucleus would be a grain of rice at the centre — and it would contain more than 99.9% of the atom's mass.

Step 1. Compute the radius from R = R_0 A^{1/3}.

R = 1.2\ \text{fm}\cdot 56^{1/3}.

Compute 56^{1/3}: 3^3 = 27, 4^3 = 64, so 56^{1/3} is between 3 and 4. Try 3.83: 3.83^3 = 56.18. Close. So 56^{1/3} \approx 3.826.

R = 1.2 \cdot 3.826 = 4.59\ \text{fm} \approx 4.6\ \text{fm}.

Why: the Fe-56 nucleus is just under 5 fm across — a few times bigger than hydrogen. The cube root grows slowly, so even a 56-fold increase in mass only multiplies the radius by about 3.8.

Step 2. Volume of the nucleus.

V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi (4.59\times 10^{-15})^3 = \tfrac{4}{3}\pi \cdot 9.68\times 10^{-44} = 4.05\times 10^{-43}\ \text{m}^3.

Step 3. Mass — A = 56 nucleons, each \sim m_N = 1.66\times10^{-27} kg.

M \approx 56\cdot 1.66\times 10^{-27} = 9.29\times 10^{-26}\ \text{kg}.

Step 4. Density.

\rho = \frac{M}{V} = \frac{9.29\times 10^{-26}}{4.05\times 10^{-43}} = 2.29\times 10^{17}\ \text{kg/m}^3.

Why: this matches the universal nuclear density derived above — as it must. Any nucleus, plugged into \rho = M/V with the A^{1/3} radius formula, returns the same number.

Step 5. Compare to solid iron metal. Iron has density 7874 kg/m³.

\frac{\rho_\text{nuclear}}{\rho_\text{iron}} = \frac{2.29\times 10^{17}}{7.87\times 10^3} = 2.9\times 10^{13}.

Why: a cubic centimetre of solid iron weighs about 8 grams, of which 99.95% is locked in nuclei that collectively occupy only 1 part in 3\times 10^{13} of the volume. The rest is empty space through which electrons move. All solid matter, including this page and the glass of your phone, is mostly empty; if you compressed the atoms until all the nuclei touched, a human body would fit in a dust speck.

Result. R_{Fe-56} = 4.6 fm, \rho_\text{Fe-56} = 2.3\times 10^{17} kg/m³, roughly 3\times 10^{13} times the density of solid iron.

What this shows. The universal nuclear density comes out the same number for every nucleus, and it is enormous. Nuclear matter is not special to exotic nuclei — it sits inside every atom of iron in the rebar holding up a building.

Example 2: Coulomb barrier for two $^{4}$He nuclei — why fusion needs a million kelvin

Two alpha particles (^4He nuclei, Z = 2, A = 4) approach each other head-on. Compute the electrostatic potential energy when their surfaces just touch. This is the Coulomb barrier — the energy hill an incoming alpha must surmount before the strong nuclear force can take over. Express your answer in MeV.

Two helium-4 nuclei at closest approachTwo spheres representing helium-4 nuclei touching at their surfaces. Each is labelled with charge 2e and radius about 1.9 femtometres. The separation between centres is labelled r equals 2R equals 3.8 femtometres. An arrow shows the direction of approach.⁴Hecharge = +2eR ≈ 1.9 fm⁴Hecharge = +2eR ≈ 1.9 fmr = 2R ≈ 3.8 fm
Two $^4$He nuclei at the instant their surfaces touch. The separation between centres is $r = 2R$, where $R = R_0 A^{1/3} \approx 1.9$ fm for helium. Below this separation the strong nuclear force takes over and fusion can occur; above this separation the Coulomb repulsion dominates.

Step 1. Nuclear radius of ^4He.

R = R_0 A^{1/3} = 1.2\cdot 4^{1/3}\ \text{fm} = 1.2\cdot 1.587 = 1.90\ \text{fm}.

Why: A = 4, so A^{1/3} = 4^{1/3}. Cube root of 4 is \approx 1.587 (since 1.587^3 \approx 3.99).

Step 2. Separation between centres at contact:

r_\text{touch} = 2R = 3.80\ \text{fm} = 3.80\times 10^{-15}\ \text{m}.

Step 3. Coulomb potential energy (each nucleus has charge Ze = 2e):

U = \frac{k(Z_1 e)(Z_2 e)}{r} = \frac{9\times 10^9\cdot (2\cdot 1.602\times 10^{-19})^2}{3.80\times 10^{-15}}.

Numerator: 9\times 10^9 \cdot (3.204\times 10^{-19})^2 = 9\times 10^9 \cdot 1.026\times 10^{-37} = 9.24\times 10^{-28} J·m.

Divide by 3.80\times 10^{-15} m:

U = 2.43\times 10^{-13}\ \text{J}.

Why: this is the work required to push two charge-2e objects from infinity to a separation of 3.8 fm, overcoming their mutual Coulomb repulsion. Keep the units careful — k in SI, charges in coulombs, distance in metres — and the answer pops out in joules.

Step 4. Convert to MeV. One MeV = 1.602\times 10^{-13} J.

U = \frac{2.43\times 10^{-13}\ \text{J}}{1.602\times 10^{-13}\ \text{J/MeV}} = 1.52\ \text{MeV}.

Why: the 10^{-13} scale matches between joules and the MeV cancel, so the answer is a small number on the MeV scale. Nuclear physics favours MeV as its natural energy unit because almost everything in a nucleus — binding energies, reaction energies, excitation energies — falls in the range 0.1 to 10 MeV per nucleon.

Step 5. What temperature would a gas of ^4He nuclei need to have enough thermal kinetic energy to cross this barrier? Thermal energy per particle is \tfrac{3}{2}k_B T.

\tfrac{3}{2}k_B T = 1.52\ \text{MeV} \Longrightarrow T = \frac{2 \cdot 1.52 \times 10^6 \cdot 1.602\times 10^{-19}}{3\cdot 1.381\times 10^{-23}} = 1.18\times 10^{10}\ \text{K}.

Why: equate \tfrac{3}{2}k_B T to the barrier energy and solve for T. Eleven billion kelvin is required for a direct thermal crossing of the barrier. The interior of the Sun is only 1.5\times 10^7 K — a thousand times too cold. Fusion in the Sun nevertheless happens because quantum tunnelling lets a tiny fraction of the thermal distribution sneak through the barrier without climbing over it. Without that quantum effect, the Sun would not shine.

Result. Coulomb barrier for ^4He + ^4He fusion is \approx 1.5 MeV, corresponding to a classical ignition temperature of \sim 10^{10} K.

What this shows. The same arithmetic applied to proton-proton fusion (the first step of stellar hydrogen burning) gives \approx 0.7 MeV and \sim 5\times 10^9 K — still a thousand times the Sun's actual core temperature. Quantum tunnelling is not a correction to the classical picture; it is the whole reason stars exist. The Gamow peak (the energy at which the tunnelling probability multiplied by the Maxwell-Boltzmann distribution peaks) gives the real rate. Fusion is always a quantum effect.

Common confusions

If you need the JEE-level story — proton, neutron, Z and A, R = R_0 A^{1/3}, density \sim 2.3 \times 10^{17} kg/m³, strong force — you have it. What follows is for readers who want the electron-scattering derivation of the radius, the semi-empirical mass formula foreshadow, a sketch of the nuclear shell model, and a quick tour of exotic nuclei.

How electron scattering measures nuclear size

The A^{1/3} law is empirical. The cleanest way to measure it is to fire high-energy electrons (say, 500 MeV) at a nuclear target and look at the angular distribution of scattered electrons. The point-like Rutherford cross-section is

\frac{d\sigma}{d\Omega}\bigg|_\text{point} = \left(\frac{Ze^2}{4E\sin^2(\theta/2)}\right)^2.

A real nucleus has a finite charge distribution, so the measured cross-section is reduced from this by a factor called the form factor F(q), where q = 2p\sin(\theta/2)/\hbar is the momentum transfer. The form factor is the Fourier transform of the charge density:

F(q) = \int \rho(\mathbf{r})\, e^{i\mathbf{q}\cdot\mathbf{r}}\, d^3r.

For a uniform sphere of radius R, this works out to F(q) = 3[\sin(qR) - qR\cos(qR)]/(qR)^3 — a function with characteristic zeros at specific qR values. Hofstadter's groundbreaking measurements in the 1950s fit these zeros to extract R for nuclei from ^{12}C up to ^{208}Pb, and the A^{1/3} law fell out as a fit across all of them.

Isospin and the symmetry between p and n

The proton and neutron are so similar that Werner Heisenberg in 1932 proposed treating them as two states of a single particle — a doublet in a new internal symmetry he called isospin. The proton is "isospin up," the neutron is "isospin down," and the strong nuclear force is invariant under isospin rotations to a very good approximation. This symmetry is why proton-proton, neutron-neutron, and proton-neutron interactions are nearly identical (after subtracting the purely electromagnetic Coulomb piece). Isospin is a small mental shift but it underlies huge amounts of nuclear physics, from the near-equality of mirror-nucleus properties (^{11}B vs ^{11}C, ^{15}N vs ^{15}O) to the classification of nuclear excited states.

The semi-empirical mass formula — a preview

The observation that nuclear density is constant, combined with the charge-independence of the nuclear force, allows a remarkable closed-form formula for the binding energy of any nucleus, due to C. F. von Weizsäcker (1935):

B(A, Z) = a_V A - a_S A^{2/3} - a_C \frac{Z(Z-1)}{A^{1/3}} - a_A \frac{(A - 2Z)^2}{A} + \delta(A, Z).

Each term has a physical meaning:

  • a_V A — volume term. Each nucleon contributes a fixed binding energy. This is the constant-density result: if each nucleon feels a fixed-depth local potential, total binding goes as A.
  • -a_S A^{2/3} — surface term. Nucleons on the surface have fewer neighbours, so they are less bound. Surface area of a sphere goes as R^2 \propto A^{2/3}.
  • -a_C Z(Z-1)/A^{1/3} — Coulomb term. Each pair of protons repels. Number of pairs is Z(Z-1)/2, and the typical separation goes as R \propto A^{1/3}.
  • -a_A (A - 2Z)^2/A — asymmetry term. Nuclei prefer N \approx Z. Pauli exclusion forces excess neutrons (or excess protons) into higher energy states, costing energy quadratically in the imbalance.
  • \delta(A, Z) — pairing term. Paired-spin nucleons bind slightly more than unpaired ones.

The numerical constants (in MeV) are roughly a_V \approx 16, a_S \approx 18, a_C \approx 0.7, a_A \approx 23, \delta \approx \pm 12/A^{1/2}. This formula is the foundation of the full mass defect and binding energy article, where the peak at iron emerges from the competition between the surface and Coulomb terms.

The nuclear shell model — magic numbers

Experimentally, nuclei with Z or N equal to 2, 8, 20, 28, 50, 82, or 126 are much more stable than their neighbours. These are the magic numbers. Maria Goeppert Mayer and J. Hans D. Jensen (independently, 1949) explained them with a shell model — nucleons fill quantum mechanical orbitals in a nuclear potential just as electrons fill atomic orbitals, with the magic numbers being "closed shells." The nuclear potential needs a strong spin-orbit coupling for the observed magic numbers to come out; in atomic physics spin-orbit is a small correction, but in nuclear physics it is enormous.

For JEE, all you need is the fact that doubly-magic nuclei (^{4}He: Z = N = 2; ^{16}O: Z = N = 8; ^{40}Ca: Z = N = 20; ^{208}Pb: Z = 82, N = 126) are especially stable. Alpha particles are emitted in decay partly because ^4He is doubly magic and thus the most stable few-nucleon cluster — better than, say, emitting a ^3He plus a neutron.

Exotic nuclei — halos, skins, and islands of stability

Modern experimental nuclear physics (at facilities like GANIL in France, RIBF in Japan, and the proposed Accelerator Driven System under development at BARC Mumbai) can produce nuclei far from stability. Some surprises:

  • Neutron halos. ^{11}Li (3 protons, 8 neutrons) has two loosely bound valence neutrons that extend far beyond the A^{1/3} radius, forming a "halo." ^{11}Li has effectively the same size as ^{48}Ca despite being four times lighter.
  • Neutron skins. Heavy neutron-rich nuclei like ^{208}Pb have a thin outer layer enriched in neutrons. Measuring the neutron-skin thickness (about 0.3 fm for Pb) constrains the equation of state of neutron-star matter — and was the motivation behind the PREX experiment at Jefferson Lab.
  • Island of stability. Theoretical calculations predict extra stability around Z = 114–120 and N = 184, the next magic shell beyond what has been observed. Synthesised superheavy elements around Z = 118 have lifetimes of milliseconds; elements closer to the predicted island might last minutes or longer.

Neutron stars — the same density, 20 km across

A neutron star is a stellar remnant of mass \sim 1.4 M_\odot compressed to a radius of \sim 12 km. Its density is:

\rho = \frac{1.4\cdot 2\times 10^{30}\ \text{kg}}{\tfrac{4}{3}\pi (12\times 10^3\ \text{m})^3} \approx 4\times 10^{17}\ \text{kg/m}^3.

Within a factor of two of ordinary nuclear matter. Physically, a neutron star is a nucleus — of mass number A \sim 10^{57}. The microscopic physics that holds a ^{56}Fe nucleus together is what holds up a neutron star against gravity, scaled up by \sim55 orders of magnitude in nucleon count. This is why pulsar timing (the precision measurement of spinning neutron stars) is a probe of nuclear-matter physics. The Giant Metrewave Radio Telescope (GMRT) near Pune, operated by the National Centre for Radio Astrophysics of TIFR, contributes to timing of millisecond pulsars whose arrival times let theorists constrain the equation of state of dense nuclear matter.

Where this leads next