Nuclear force

Pack ninety-two protons into a space only about fifteen femtometres across, as nature does inside a uranium nucleus, and simple electrostatics predicts disaster. Like charges repel, and repel harder the closer they are squeezed — the Coulomb force between two protons a couple of femtometres apart is already strong enough that a light-weight, unbound object would fly apart in an instant. Yet uranium nuclei sit quietly in rock for billions of years. Something inside the nucleus is holding those protons together against their own electrical fury, and whatever it is must be the strongest force in nature, at least at these vanishingly small distances. That something is the nuclear force — also called the strong nuclear force — and it is one of the strangest forces in physics. It does not care whether a nucleon is a proton or a neutron. It reaches out only a few femtometres and then vanishes almost completely, unlike gravity or electromagnetism, which fade slowly and reach across the universe. And it flips from powerfully attractive to fiercely repulsive if two nucleons are pushed too close together, which is what stops a nucleus from collapsing into a point. This article builds the case for the nuclear force from first principles: why the failure of pure electrostatics to explain nuclear stability forces us to invent it, what its four defining qualitative properties are, and how a little order-of-magnitude arithmetic with Coulomb's law — well within your existing toolkit — lets you estimate just how strong this new force must be, even though the CBSE syllabus asks for no formula of the nuclear force itself. By the end you should be able to explain not just that nuclei are stable, but why heavier nuclei need proportionally more neutrons to stay that way.

Key takeaways

Electrostatics alone cannot hold a nucleus together. Protons packed a few femtometres apart repel each other strongly via the Coulomb force; a separate, stronger force must exist to bind the nucleus.
The nuclear force is the strongest known fundamental interaction at nuclear distances — far stronger than the electromagnetic force it must overcome, and utterly dwarfing gravity.
It is short-range, acting strongly only across about $1$ to $3$ femtometres and dropping to nearly zero beyond about $3$ to $4$ fm, unlike the Coulomb force which extends indefinitely as $1/r$ .
It is charge-independent — proton-proton, neutron-neutron, and proton-neutron pairs feel (very nearly) the same nuclear attraction once the extra Coulomb term for two protons is separated out.
It saturates: because each nucleon only feels its near neighbours, total nuclear binding grows roughly in proportion to the number of nucleons $A$ , while total Coulomb repulsion (which reaches every proton, however far) grows faster, roughly as $Z^2$ — this is why binding energy per nucleon levels off rather than climbing forever.
A quick Coulomb's-law estimate makes the size of the force concrete: two protons at a typical $2$ fm separation repel with about $0.72$ MeV of potential energy, tiny next to the several-MeV-per-nucleon binding the nuclear force actually supplies.
Key idea: the nuclear force is a short-range, charge-independent, saturating attraction between nucleons that must be far stronger than the Coulomb repulsion at femtometre distances to hold any nucleus together, and its short range is precisely why heavier nuclei need extra neutrons to stay stable.

The core insight

Because Coulomb repulsion between protons never switches off with distance while the nuclear force does, the nuclear force must be strongest, charge-independent, and short-range enough to saturate — binding only near neighbours — which is exactly why very heavy nuclei need a growing surplus of neutrons to remain stable.

Where it fits in the syllabus

This is the opening idea of the CBSE Class 12 Physics chapter "Nuclei" (Unit IX, Atoms and Nuclei), introduced right alongside the composition and size of the nucleus and just before binding energy is defined quantitatively. The board syllabus treats the nuclear force qualitatively — you are not asked to derive or quote a mathematical formula for the nuclear force's potential energy (that would require physics beyond this course, such as the Yukawa meson-exchange model, which is explicitly out of scope). What you are expected to do is reason clearly with its four qualitative properties, and to combine them with the plain Coulomb's law you already know from electrostatics to justify, semi-quantitatively, why the nuclear force must dominate at nuclear distances. Radioactive decay, which used to sit in this chapter, has been removed from the 2025-26 CBSE theory list, so this article keeps strictly to force properties, stability, and their link to binding energy.

The physical picture

Two protons separated by a distance $r$ repel each other with an electrostatic potential energy given by Coulomb's law, a formula you already know from electrostatics and can safely reuse here because it needs no re-derivation:

U_{\text{Coulomb}} = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{e^2}{r}. \qquad (1)

where: $U_{\text{Coulomb}}$ is the electrostatic potential energy of the pair (joules, or MeV when convenient for nuclear scales), $e = 1.6\times10^{-19}\ \text{C}$ is the proton charge, $r$ is the separation between the two protons, and $\dfrac{1}{4\pi\varepsilon_0} = 8.99\times10^{9}\ \text{N m}^2/\text{C}^2$ is Coulomb's constant. The physical reading is simple: this repulsion never switches off, it just gets weaker as $r$ grows, falling as $1/r$ . For two protons even close to touching inside a nucleus, $r$ is only a few femtometres ( $1\ \text{fm} = 10^{-15}\ \text{m}$ ), and $U_{\text{Coulomb}}$ is correspondingly large on the atomic energy scale — large enough that, left unopposed, it would fling protons apart. For the nucleus to survive, an attractive force must beat this repulsion at short range. Because no known formula for it is prescribed at this level, the syllabus instead specifies its qualitative signature through four properties established by scattering and binding-energy experiments:

Strongest known force. At nuclear separations, the nuclear force between nucleons is stronger than the electromagnetic repulsion it has to overcome, and enormously stronger than the gravitational attraction between the same two nucleons, which is utterly negligible at this scale.

Short-range. The nuclear force acts powerfully only across roughly $1$ to $3$ fm, and effectively vanishes beyond about $3$ to $4$ fm. This is the single most important qualitative fact about it, and it is what makes the force behave completely differently from gravity or the Coulomb force, both of which reach to infinity as $1/r^2$ forces (with potential energy falling as $1/r$ ).

Charge-independent. Once you separately account for the extra Coulomb repulsion that exists only between two protons, the underlying nuclear attraction between proton-proton, neutron-neutron, and proton-neutron pairs is (very nearly) identical in strength. The nuclear force does not "know" or care about electric charge; it treats every nucleon alike.

Attractive then sharply repulsive at very short range. Beyond about $0.7$ fm the nuclear force pulls nucleons together strongly; but push two nucleons closer than that and the force turns strongly repulsive, preventing the nucleus from collapsing to a point. Nucleons behave like tightly packed, slightly compressible spheres rather than point charges falling into each other.

Figure

Two overlaid curves of potential energy against separation distance, one a long slowly falling Coulomb repulsion curve extending to large distance, the other a nuclear force curve that plunges into a deep attractive well between about one and three femtometres and then shoots sharply upward at very short distance, illustrating that the nuclear force is short range, deep, and repulsive at very close approach while Coulomb repulsion persists at all distances

Derivation AWhy the nuclear force must overwhelm Coulomb repulsion at nuclear separations

The goal here is to turn the qualitative claim "the nuclear force is the strongest force at nuclear distances" into a number you can actually check with Coulomb's law and the known binding energy scale of nuclei, using nothing beyond algebra, so that the four qualitative properties above stop being assertions to memorise and become the direct consequence of a repulsion the nuclear force is visibly strong enough to beat, and so that the saturation property (attraction growing with $A$ , repulsion growing faster with $Z$ ) follows naturally from counting nucleon pairs rather than from a formula for the nuclear force that the syllabus never asks you to know.

Step 1. Write the Coulomb repulsion between two protons at a typical nuclear separation using equation (1). Take $r = 2\ \text{fm} = 2\times10^{-15}\ \text{m}$ , a reasonable spacing between neighbouring nucleons. Then $U_{\text{Coulomb}} = (8.99\times10^{9})\dfrac{(1.6\times10^{-19})^2}{2\times10^{-15}} = 1.15\times10^{-13}\ \text{J}$ .

Step 2. Convert to the natural energy unit for nuclear physics, the mega-electron-volt, using $1\ \text{MeV} = 1.6\times10^{-13}\ \text{J}$ . This gives $U_{\text{Coulomb}} \approx 0.72\ \text{MeV}$ for a single proton-proton pair at $2$ fm — a small but not negligible amount of repulsive energy, purely electrostatic, with nothing nuclear about it yet.

Step 3. Compare this to the scale of binding the nuclear force actually supplies. From the binding-energy-per-nucleon curve (built in the companion topic on mass defect and binding energy), a typical medium-mass nucleus is bound by roughly $7$ to $8.8$ MeV per nucleon, and even the small alpha particle ( $^4_2\text{He}$ ) has a total binding energy of about $28.3$ MeV. For the nucleus to hold together at all, the net attractive contribution the nuclear force supplies per nucleon must clearly exceed the roughly $0.7$ to $1.4$ MeV of Coulomb repulsion any one proton pair contributes at these separations — by a factor of five to ten. This numerically justifies calling the nuclear force the strongest known force at this range: it is not just bigger, it has to be several times bigger to leave a nucleus net bound rather than net repelled.

Step 4. Now scale this up by counting pairs, which is where saturation comes from. The number of proton-proton pairs in a nucleus with $Z$ protons is $\binom{Z}{2} = \dfrac{Z(Z-1)}{2}$ , growing roughly as $Z^2$ for large $Z$ — every proton repels every other proton, however far apart, because Coulomb's law in equation (1) never actually reaches zero. But the nuclear force, being short-range, only lets each nucleon feel its handful of nearest neighbours; adding one more nucleon to a large nucleus adds only a fixed, roughly constant amount of extra nuclear binding, so total nuclear attraction grows only in proportion to $A$ , not $A^2$ . Total repulsion climbing as $Z^2$ while total attraction climbs only as $A$ is precisely why heavier nuclei need $N$ to grow faster than $Z$ to stay stable, and why the binding-energy-per-nucleon curve flattens and then gently droops for the heaviest nuclei rather than climbing forever.

Worked examples

Board level - range comparison with a Coulomb energy estimate

Question: Two protons inside a light nucleus are separated by $1$ fm. If they could be pulled apart to $10$ fm (well outside any nucleus), (a) state which force dominates their interaction at each separation and why, and (b) find the ratio of their Coulomb potential energies at $1$ fm and at $10$ fm.

Step 1. At $r = 1\ \text{fm}$ , this lies well inside the nuclear force's effective range of about $1$ to $3$ fm, and the nuclear force is far stronger than the Coulomb force at this scale, so the attractive nuclear force dominates the interaction.

Step 2. At $r = 10\ \text{fm}$ , this is well beyond the nuclear force's short range (effectively zero beyond about $3$ to $4$ fm), so only the Coulomb force survives; the interaction here is purely electrostatic repulsion.

Step 3. Since $U_{\text{Coulomb}} \propto 1/r$ from equation (1), the ratio of energies at the two separations is simply the inverse ratio of distances: $\dfrac{U(1\ \text{fm})}{U(10\ \text{fm})} = \dfrac{10\ \text{fm}}{1\ \text{fm}} = 10$ . Using the Step 1 result from the derivation above, scaling $0.72\ \text{MeV}$ at $2$ fm gives $U(1\ \text{fm}) \approx 1.44\ \text{MeV}$ and $U(10\ \text{fm}) \approx 0.144\ \text{MeV}$ .

Answer: At $1$ fm the attractive nuclear force dominates; at $10$ fm only the weaker Coulomb repulsion remains, and its potential energy is exactly ten times smaller than it would be at $1$ fm — $U(1\ \text{fm}) \approx 1.44\ \text{MeV}$ , $U(10\ \text{fm}) \approx 0.144\ \text{MeV}$ , ratio $10:1$ .

Competitive level - estimating the nuclear force's net contribution in an alpha particle

Question: The alpha particle ( $^4_2\text{He}$ , $2$ protons and $2$ neutrons) has a measured total binding energy of about $28.3$ MeV. Estimate, order-of-magnitude, the average net attractive contribution the nuclear force must supply per nucleon-nucleon pair inside it, given that only proton-proton pairs feel any Coulomb repulsion.

Step 1. Count the nucleon pairs. With $4$ nucleons, the total number of pairs is $\binom{4}{2} = 6$ : this splits into $1$ proton-proton pair, $1$ neutron-neutron pair, and $4$ proton-neutron pairs (each of the $2$ protons paired with each of the $2$ neutrons). Only the single proton-proton pair contributes a Coulomb repulsion term; charge-independence means all $6$ pairs contribute comparable nuclear attraction.

Step 2. Estimate that one Coulomb repulsion term using equation (1) at a close nucleon separation inside the tightly bound alpha particle, $r \approx 1\ \text{fm}$ : from the Board-level example above, $U_{\text{Coulomb}} \approx 1.44\ \text{MeV}$ .

Step 3. The total binding energy is what remains after the nuclear attraction has done its work and paid off this Coulomb repulsion, so the total nuclear attractive energy supplied must be $E_{\text{nuclear}} = 28.3 + 1.44 \approx 29.7\ \text{MeV}$ . Dividing this across the $6$ nucleon pairs gives an average of $29.7/6 \approx 4.95\ \text{MeV}$ of net attractive nuclear energy per pair.

Answer: roughly $5\ \text{MeV}$ of net attractive nuclear contribution per nucleon pair in the alpha particle — an order-of-magnitude estimate only (the true nucleon-nucleon potential well is deeper and this averages over the whole nucleus), but it already shows the nuclear force out-competing Coulomb repulsion by a healthy margin even in the smallest stable nucleus above hydrogen. Timing note: this is a two-minute estimate if you keep it structured — count pairs by $\binom{n}{2}$ , isolate the one type of pair that feels Coulomb repulsion, compute that one number from equation (1), then add it back to the given total binding energy before dividing by the pair count. The trap that costs marks under time pressure is dividing the raw binding energy by the pair count and forgetting to add back the Coulomb term first, which quietly understates the nuclear force's real job.

Common traps

Treating the nuclear force as "a stronger version of the electrostatic force." It is a completely separate interaction: charge-independent, with a totally different (short) range, and it does not appear anywhere in Coulomb's law.
Assuming the nuclear force reaches as far as gravity or electromagnetism. It is strictly short-range, negligible beyond about $3$ to $4$ fm — this is the property the whole "saturation" argument in Step 4 of the derivation depends on.
Believing the nuclear force acts only between protons. It acts on proton-proton, neutron-neutron, and proton-neutron pairs alike, which is exactly why neutrons are useful for stabilising heavy nuclei — they add nuclear attraction with zero Coulomb cost.
Forgetting that binding energy per nucleon does not keep rising with $A$ . Because of saturation, it rises, peaks near $A \approx 56$ , and then slowly falls — it never climbs forever, unlike the ever-growing Coulomb term.
In numeric estimates, mixing up which pairs feel Coulomb repulsion. Only proton-proton pairs do; neutron-neutron and proton-neutron pairs feel nuclear attraction alone.

You should be able to…

State and explain all four qualitative properties of the nuclear force: strongest known force at nuclear range, short-range, charge-independent, attractive then sharply repulsive at very short distance.
Explain, using Coulomb's law, why a separate attractive force must exist to hold protons together inside a nucleus.
Use equation (1) to estimate the Coulomb potential energy between two protons at a given nuclear separation, in joules or MeV, and compare it sensibly against typical binding energies.
Explain nuclear force saturation and connect it to why binding energy per nucleon rises, peaks, and then slowly falls with mass number $A$ , rather than growing without limit.
Explain why the stable neutron-to-proton ratio increases for heavier elements, in terms of the different growth rates of total Coulomb repulsion ( $\sim Z^2$ ) and total nuclear attraction ( $\sim A$ ).
Recognise which claims about the nuclear force are examinable at CBSE Class 12 level (the four qualitative properties, Coulomb's-law estimates) and which are out of syllabus (any explicit nuclear-force potential formula, meson-exchange models).

Ask the tutor: want to see how this same short-range-versus-Coulomb balance sets the exact shape of the binding-energy-per-nucleon curve, or why it makes fission of heavy nuclei and fusion of light nuclei both release energy? Ask the AI tutor to walk through it with the actual curve.

If the nuclear force is so much stronger than the Coulomb force, why do we even need extra neutrons in heavy nuclei?

Because "much stronger" is only true pair by pair, at short range, not overall. Each nucleon's nuclear pull is strong but reaches only its nearest handful of neighbours, so total nuclear attraction in a nucleus grows roughly in step with the total number of nucleons, $A$ . Coulomb repulsion, though weaker pair by pair, acts between every single proton and every other proton no matter how far apart they sit inside the nucleus, so the total repulsive effect grows faster, roughly as $Z^2$ . In a light nucleus like carbon this difference barely matters. In a heavy nucleus like lead, the sheer number of proton pairs makes the cumulative Coulomb repulsion large enough that extra neutrons — which add nuclear attraction with no Coulomb penalty at all — become necessary to keep the whole thing bound.

Is the nuclear force the same thing as the strong force I might have heard about in particle physics?

They are closely related but not identical, and CBSE only needs the nuclear (nucleon-level) version. The fundamental strong force acts between quarks inside protons and neutrons, glued together by particles called gluons. The nuclear force you study in this chapter is a residual effect of that deeper force leaking out between whole nucleons, somewhat like how electrically neutral atoms can still weakly attract each other through residual electric effects (van der Waals forces). For your syllabus, you only need the nucleon-level description: short-range, charge-independent, strongest at nuclear distances, attractive then repulsive at very short range. The quark-level machinery is well beyond Class 12 and is not examined.

Why doesn't the nuclear force just pull every nucleon in the nucleus into a single point?

Two separate reasons stop that. First, the nuclear force itself turns sharply repulsive if two nucleons are pushed closer than about $0.7$ fm, acting like a hard core that resists further squeezing — this is the fourth qualitative property listed above. Second, quantum mechanics (the Pauli exclusion principle, beyond this syllabus in detail) also prevents identical fermions like protons or neutrons from all crowding into the same lowest-energy state. Together these keep the nucleus at a roughly constant density rather than collapsing, which is also why the nuclear radius formula $R = R_0A^{1/3}$ (from the "size of the nucleus" topic in this same chapter) gives a nuclear volume that grows in direct proportion to the number of nucleons, rather than shrinking as more nucleons are packed in.

Do I need to memorise the Yukawa potential or meson-exchange picture of the nuclear force for my board exam?

No. That model, which explains the nuclear force in terms of particles called mesons being exchanged between nucleons, belongs to university-level nuclear and particle physics and is explicitly outside the CBSE Class 12 syllabus for 2025-26. Your board exam expects the four qualitative properties covered in this article (strongest known force, short-range, charge-independent, attractive-then-repulsive at very short range), the ability to explain nuclear stability and the rising neutron-to-proton ratio in heavy nuclei using those properties, and comfort applying ordinary Coulomb's law for any numeric estimate of proton-proton repulsion that a question might ask for. Nothing about the mechanism that produces the nuclear force itself is examinable.