The Nuclear Force

A Force of a Totally Different Kind

Atoms are held together by the familiar Coulomb force, with binding energies of a few eV. But the nucleus binds at ~8 MeV per nucleon — a million times stronger. And it must do something Coulomb cannot: hold positively charged protons together at femtometre range, where their mutual repulsion is ferocious, while also gripping neutral neutrons.

The conclusion: a strong attractive force of a totally different kind operates in the nucleus. Its properties were mapped by experiments between roughly 1930 and 1950:

Property 1: It is the strongest force

The nuclear force dominates the Coulomb repulsion between protons inside the nucleus — possible only because it is much stronger than the Coulomb force. (Gravity, for comparison, is weaker than even Coulomb by ~$10^{36}$ — utterly negligible here.) Strength ordering: nuclear ≫ Coulomb ≫ gravitational.

Property 2: It is short-ranged, hence saturating

The force between two nucleons falls rapidly to zero beyond a few femtometres. Consequence: in a medium or large nucleus, each nucleon bonds only with nearby neighbours — the force saturates, which is exactly why the binding energy per nucleon is constant across the plateau (Section 3's argument).

The Potential-Energy Curve and Charge Independence

The pair potential (NCERT Fig. 13.2)

A rough plot of the potential energy of two nucleons versus their separation shows:

• A minimum at $r_0 \approx 0.8$ fm.
• For r > $r_0$: potential rises with distance → the force is attractive (pulls them back together).
• For r < $r_0$: potential climbs steeply → the force is strongly repulsive (a hard core — nucleons resist being squeezed closer than ~0.8 fm).

This repulsive core is why nuclear matter has a fixed density: nucleons pack like marbles, not like collapsing dust.

Property 3: Charge independence

The nuclear force between neutron-neutron, proton-neutron and proton-proton is approximately the same — it does not depend on electric charge. The force sees nucleons, not charges.

Property 4 (the honest one): No simple formula

Unlike Coulomb's law or Newton's gravitation, there is no simple mathematical form of the nuclear force. Its properties are catalogued from experiment, not derived from an inverse-square-style law.

Key Point: Four claims to recite — strongest of all forces; short-range (few fm) with saturation; minimum of the pair potential at ~0.8 fm with attraction outside and strong repulsion inside; charge-independent. Plus the disclaimer: no simple closed formula exists.

[NEET Important] 'The nuclear force between p-p is weaker than n-n because of charge' — a tempting FALSE statement. The nuclear force is the same; the p-p pair additionally feels Coulomb repulsion, but that is a separate force, not a modification of the nuclear one.

Solved Examples

Example 1: Ranking the forces [Board Rapid]

Rank the gravitational, Coulomb and nuclear forces between two protons inside a nucleus.

Solution:

1. Nuclear — strongest (must overpower proton-proton Coulomb repulsion at fm range).
2. Coulomb — next (substantial at fm separations, but beaten).
3. Gravitational — weakest by ~36 orders of magnitude; irrelevant in nuclei.
4. Takeaway: nuclear ≫ Coulomb ≫ gravitational — the ordering NCERT states outright.

Example 2: Reading the pair potential [Board Graph]

From the potential-energy curve of a nucleon pair, state the force's character at r = 2 fm, r = 0.8 fm and r = 0.5 fm.

Solution:

1. r = 2 fm (> $r_0$): potential slopes upward with r → force attractive.
2. r = 0.8 fm (= $r_0$): the minimum — zero net force, equilibrium separation.
3. r = 0.5 fm (< $r_0$): steeply rising potential inward → strongly repulsive.
4. Takeaway: attraction outside, hard-core repulsion inside, equilibrium at ~0.8 fm — the whole curve in one sentence.

Example 3: Why doesn't the nucleus collapse? [NEET Conceptual]

The nuclear force is hugely attractive. Why don't nucleons collapse into a point?

Solution:

1. Below $r_0 \approx 0.8$ fm the force turns strongly repulsive — the hard core.
2. Nucleons therefore settle at ~fm spacings, like packed marbles.
3. This fixed packing is exactly why nuclear density is constant (Section 2) — every nucleus is an incompressible droplet.
4. Takeaway: attraction binds; the repulsive core sizes. Both halves of the curve are essential.

Example 4: Charge independence in action [Board Conceptual]

Compare the nuclear force in the pairs n-n, n-p and p-p, and the TOTAL force in each case.

Solution:

1. Nuclear force: approximately equal for all three pairs — charge-independent.
2. Total force: the p-p pair additionally suffers Coulomb repulsion; n-p and n-n do not.
3. So p-p is the least bound pairing overall — but by Coulomb's doing, not the nuclear force's.
4. Takeaway: separate the two forces in your head; exams probe exactly this distinction.

Example 5: Why more neutrons in heavy nuclei? [JEE/NEET Link]

Light stable nuclei have N/Z ≈ 1, heavy ones N/Z ≈ 3/2. Explain using force properties.

Solution:

1. Nuclear attraction is short-range: each nucleon binds only neighbours.
2. Coulomb repulsion is long-range: every proton pair in the nucleus repels, however far apart.
3. As Z grows, total repulsion grows fast (~$Z^2$ pairs) while attraction grows only ~A; extra neutrons add glue without adding repulsion.
4. Takeaway: hence N/Z drifts from 1 : 1 (light) to about 3 : 2 (heavy) — NCERT Points to Ponder 8, and the seed of beta-decay stability logic.

Example 6: Range from the binding plateau [Conceptual]

What feature of the binding-energy curve proves the nuclear force is short-ranged?

Solution:

1. If the force were long-range (every nucleon binding every other), total binding would grow like the number of pairs ~$A^2$, so $E_{bn} \propto A$ — rising steadily.
2. Observed: $E_{bn}$ is flat across 30 < A < 170.
3. Flatness ⇒ each nucleon binds only a fixed number of neighbours ⇒ the force reaches only a few fm: short range and saturation.
4. Takeaway: the plateau is the fingerprint of short range — curve and force explain each other.

Example 7: Coulomb vs nuclear at 2 fm [JEE Numerical]

Estimate the Coulomb repulsion between two protons 2 fm apart, and compare with the ~MeV scale of nuclear binding.

Solution:

1. Coulomb energy: $U = \dfrac{e^2}{4\pi\varepsilon_0 r} = \dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2 \times 10^{-15}}$ J.
2. Compute: $U = 1.15 \times 10^{-13}$ J ≈ 0.72 MeV.
3. Compare: nucleon binding ~8 MeV — the nuclear attraction outweighs this repulsion roughly tenfold at these separations.
4. Takeaway: the MeV scale of Coulomb energy at fm range shows why only a super-strong force can build nuclei — and why the p-p Coulomb tax matters for heavy-nucleus stability.

Example 8: No inverse-square law [Board Conceptual]

A student writes $F_{nuclear} = \dfrac{k}{r^2}$ for the nuclear force. What's wrong?

Solution:

1. NCERT is explicit: there is no simple mathematical form of the nuclear force, unlike Coulomb's or Newton's laws.
2. An inverse-square force is long-range — it would forbid saturation and make $E_{bn}$ grow with A, contradicting the plateau.
3. The real force: attractive beyond ~0.8 fm, repulsive within, dying off entirely beyond a few fm — no single power law captures that.
4. Takeaway: the absence of a formula is itself a listed property; don't invent one.

Example 9: Which properties explain which observations? [Matching]

Match: (a) constant nuclear density, (b) flat $E_{bn}$ plateau, (c) mirror-nuclei similarity (e.g. $^3$H vs $^3$He binding) — with force properties.

Solution:

1. (a) ← repulsive core at r < 0.8 fm: nucleons keep fixed spacing → incompressible droplet.
2. (b) ← short range/saturation: neighbour-only bonding caps each nucleon's binding.
3. (c) ← charge independence: swapping protons ↔ neutrons barely changes nuclear binding (differences trace to Coulomb effects).
4. Takeaway: each experimental fact is one force property wearing data's clothing.

Example 10: The equilibrium separation [NEET Numerical]

Two nucleons sit at their potential minimum. State the separation, and what happens if they are nudged (i) closer, (ii) apart.

Solution:

1. Separation: $r_0 \approx 0.8$ fm — the minimum of the pair potential.
2. (i) Nudged closer (r < $r_0$): strong repulsion pushes them back out.
3. (ii) Nudged apart (r > $r_0$): attraction pulls them back in.
4. Takeaway: the minimum is a stable equilibrium — nucleon pairs oscillate about ~0.8 fm like masses on a spring, the microscopic basis of the liquid-drop picture.