Physics · Nuclei

Nuclear Force

The force that holds protons and neutrons inside the tiny nuclear volume is the nuclear force, the strongest of the basic forces over its range. NCERT §13.5 and NIOS §26.2 summarise its character: very short range, charge-independent, saturating, attractive at the typical nucleon separation but strongly repulsive at very short distance. This subtopic is high-yield for NEET because its properties feed directly into the constancy of the binding energy per nucleon and the logic of fission and fusion.

Why a new force is needed

The force that determines the motion of atomic electrons is the familiar Coulomb force. It is attractive between the positive nucleus and the negative electrons, it obeys an inverse-square law, and it acts over the comparatively large atomic distances. When we turn from the atom to the nucleus, however, the Coulomb force becomes a problem rather than a solution. The nucleus packs protons and neutrons into a region only a few femtometres across, and the protons, all carrying positive charge, repel one another fiercely through the Coulomb force. Left to themselves the protons would fly apart.

Yet nuclei are stable. From the study of binding energy we know that, for average mass nuclei, the binding energy per nucleon is approximately 8 MeV. This is much larger than the binding energy in atoms, where electron binding energies are measured in electron volts. Therefore, to bind a nucleus together there must be a strong attractive force of a totally different kind. It must be strong enough to overcome the repulsion between the positively charged protons and to bind both protons and neutrons into the tiny nuclear volume.

NIOS frames the same conclusion as a process of elimination. The extremely small size of the nucleus, where protons and neutrons are closely packed, suggests that the binding force should be strong, short range and attractive. It cannot be electrostatic, because the electrostatic force between protons is repulsive; if only that force operated, the nucleons would fly away, which is contrary to experience. Gravitation is attractive between every pair of nucleons, but if the magnitude of the nucleon-nucleon force is taken to be unity, the gravitational force is only of the order of $10^{-39}$, far too weak to account for the binding. The conclusion is that the attractive force between nucleons is of a new type, the nuclear force, with no analogy in classical physics.

Figure 1

The three basic forces, ordered by strength over the nuclear range.

Nuclear force (strongest over its range) ~ 1 Coulomb force (much weaker) ~ 10⁻² Gravitational force (weakest) ~ 10⁻³⁹

Schematic ordering only; bar lengths are not to scale. Source ordering follows NCERT §13.5(i) and the NIOS estimate that gravity is about $10^{-39}$ of the nucleon-nucleon force.

Strength of the nuclear force

The first listed feature of the nuclear force in NCERT is its sheer magnitude. The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational force between masses. This is not an incidental property; it is a structural necessity. Inside the nucleus the nuclear binding force has to dominate over the Coulomb repulsive force between protons. That domination is possible only because the nuclear force is much stronger than the Coulomb force over the distances at which nucleons sit. The gravitational force, although it is always attractive, is much weaker than even the Coulomb force, and so plays no role in binding the nucleus.

The strength can be appreciated through the energy scale it produces. A nucleon held inside an average nucleus is bound with roughly 8 MeV, which is about a million times the few electron volts that bind an electron in an atom. A force that delivers binding energies of a few MeV per nucleon, against the Coulomb repulsion of the protons, must be intrinsically powerful. This is the sense in which the nuclear force is described as the strongest of the basic forces over its range.

NEET Trap

"Strongest" is only true over its short range

A common error is to read "the nuclear force is the strongest of the basic forces" as an unconditional statement. The strength claim holds over its range, that is, within a few femtometres. Beyond that range the nuclear force has fallen to zero, while the Coulomb force, following an inverse-square law, continues to act over atomic and macroscopic distances. So at large separations the Coulomb force is the relevant one, even though it is far weaker nucleon-for-nucleon at close range.

Rule: rank strengths only within the range where the nuclear force is non-zero. Outside a few fm, the nuclear force is simply absent.

Short range and saturation

The second defining feature is the very short range. The nuclear force between two nucleons falls rapidly to zero as their distance becomes more than a few femtometres. NIOS sharpens this picture by noting that the nucleons in a nucleus are very densely packed and that the force which holds them together must therefore exist between neighbouring nucleons; it is a short-range force operating over distances of the order of $10^{-15}\,\text{m}$. Once two nucleons are pulled apart by more than a few fm, the force between them is effectively switched off.

This rapid fall to zero has a direct and important consequence called saturation. Because the force reaches only as far as the nearest neighbours, a given nucleon inside a sufficiently large nucleus is under the influence of only some of its neighbours, those that come within the range of the nuclear force. Any other nucleon farther away than the range has no influence on the binding of the nucleon under consideration. NIOS states the property compactly: each nucleon interacts only with neighbouring nucleons instead of with all nucleons from one end of the nucleus to the other.

Saturation explains one of the headline facts of nuclear structure, the constancy of the binding energy per nucleon for nuclei of middle mass number. If a nucleon can have at most $p$ neighbours within range, its binding energy is proportional to $p$, say $pk$ where $k$ is a constant with the dimensions of energy. Adding more nucleons to a large nucleus does not change the binding of a nucleon already buried inside, because the newcomers are out of range. Since most nucleons in a large nucleus reside inside it rather than on the surface, the binding energy per nucleon stays nearly constant at about $pk$.

Build the link

Saturation is the reason the binding-energy-per-nucleon curve is flat from $A \approx 30$ to $A \approx 170$. See how that curve is drawn and read in Nuclear Binding Energy.

NEET Trap

Saturation, not strength, explains the flat BE/A curve

When asked why the binding energy per nucleon is nearly constant for $30 < A < 170$, the correct reason is the short range of the nuclear force leading to saturation, because a nucleon only feels its few neighbours. It is not because "the force is strong" or because "the density is constant" on their own. A nucleon deep inside a large nucleus is already fully bound by its neighbours, so adding more distant nucleons leaves its binding unchanged.

Rule: constancy of BE per nucleon ⇐ short range ⇒ saturation ⇒ each nucleon binds only to nearest neighbours.

The potential-energy curve

The richest single piece of information about the nuclear force is the plot of the potential energy of a pair of nucleons against their separation. NCERT presents a rough version of this curve in Fig. 13.2. The potential energy is a minimum at a separation $r_0$ of about $0.8\,\text{fm}$. This minimum is the equilibrium point: it is the distance at which the two nucleons sit most comfortably, neither pulled together more tightly nor pushed apart.

The slope of the curve tells us the direction of the force. For separations greater than $r_0$, the potential energy rises as the nucleons move apart, so the force is attractive, pulling them back toward $r_0$. For separations less than $r_0$, the potential energy climbs very steeply, so the force is strongly repulsive, pushing the nucleons back out to $r_0$. This strongly repulsive core at small separation is what stops nucleons from collapsing into one another and keeps the average separation between nucleons roughly constant, which in turn keeps the nuclear volume proportional to the number of nucleons.

Figure 2

Potential energy of a pair of nucleons as a function of their separation $r$.

U(r) r (fm) 0 r₀ ≈ 0.8 fm minimum (equilibrium) strongly repulsive (r < r₀) attractive (r > r₀) force → 0 beyond a few fm

After NCERT Fig. 13.2. The force is attractive for separations larger than $r_0$ and strongly repulsive for separations smaller than $r_0$; beyond a few femtometres the force falls to zero.

It is worth stating precisely what "attractive for $r > r_0$ and repulsive for $r < r_0$" means physically. Two nucleons brought together from far away first feel almost nothing until they are within a few femtometres. As they approach, attraction grows and draws them toward the minimum at $r_0 \approx 0.8\,\text{fm}$. If something tries to squeeze them closer than $r_0$, the steeply rising repulsive core resists, and the nucleons are pushed back toward the equilibrium separation. This combination of an attractive outer region and a repulsive inner core is exactly what produces a constant average inter-nucleon spacing and the liquid-drop-like constant density of nuclear matter.

Charge independence

The third defining feature is charge independence. The nuclear force between neutron and neutron, between proton and neutron, and between proton and proton is approximately the same. The nuclear force does not depend on the electric charge of the interacting nucleons. A proton and a neutron attract one another through the nuclear force just as strongly as two neutrons or two protons do, once the Coulomb contribution between the two protons is set aside.

NIOS connects this property to the binding-energy evidence. Since the binding energy per nucleon is found to be roughly the same irrespective of the particular mix of neutrons and protons in the nucleus, we are justified in treating the force between any pair of nucleons as equivalent. The nuclear force must account equally for the attraction between a proton and a neutron, between two protons, and between two neutrons. That equality of behaviour across all three pairings is precisely what is meant by charge independence.

Nucleon pair Nuclear force present? Coulomb force present? Relative nuclear strength
neutron – neutron (n–n) Yes No (both neutral) Approximately equal
proton – neutron (p–n) Yes No (one neutral) Approximately equal
proton – proton (p–p) Yes Yes (mutual repulsion) Approximately equal
NEET Trap

Charge independence ≠ no Coulomb force between protons

Charge independence means the nuclear force is the same for n–n, p–n and p–p pairs. It does not mean the protons feel no Coulomb force. Two protons still repel through the Coulomb force in addition to their nuclear attraction; charge independence is a statement about the nuclear interaction alone. The nuclear force simply does not distinguish a proton from a neutron, even though the Coulomb force does.

Rule: nuclear force = charge-independent; Coulomb force = charge-dependent. Both can act between two protons at once.

Nuclear force vs Coulomb force

Placing the two forces side by side draws out every feature discussed above. The Coulomb force is the long-range, inverse-square, charge-dependent interaction that governs atomic electrons; the nuclear force is the short-range, charge-independent, saturating interaction that binds nucleons. Unlike Coulomb's law or Newton's law of gravitation, there is no simple mathematical form of the nuclear force; its character was pieced together from a variety of experiments carried out during 1930 to 1950 rather than expressed as one neat equation.

Property Nuclear force Coulomb force
Relative strength (close range) Strongest of the basic forces over its range Much weaker than the nuclear force
Range Very short; falls rapidly to zero beyond a few fm Long range; acts over atomic and macroscopic distances
Mathematical form No simple mathematical form Inverse-square law, F ∝ q₁q₂/r²
Charge dependence Charge-independent (n–n, p–n, p–p equal) Charge-dependent; acts only between charges
Nature with separation Attractive for r > r₀ (≈ 0.8 fm), strongly repulsive for r < r₀ Repulsive between like charges, attractive between unlike
Saturation Saturates; a nucleon binds only to nearest neighbours Does not saturate; every charge acts on every other charge

The contrast in the last two rows is the heart of the matter for NEET. Because the Coulomb force does not saturate, every proton in a heavy nucleus repels every other proton, and this cumulative repulsion grows quickly with the number of protons. The nuclear force, by saturating, cannot keep up indefinitely, since a nucleon binds only to its neighbours. This is the seed of the eventual instability of very heavy nuclei and the energetics that make fission favourable, themes developed in the binding-energy and nuclear-energy subtopics.

Figure 3

Reach of the two forces with separation: the nuclear force switches off beyond a few fm, the Coulomb force trails on.

|F| separation r nuclear force Coulomb force (∝ 1/r²) few fm nuclear force ≈ 0 beyond here

Schematic magnitudes, not to scale. The point is qualitative: the nuclear force has effectively zero reach past a few femtometres, whereas the Coulomb force tails off slowly and never quite vanishes.

Quick Recap

Nuclear force in one screen

  • Why needed: binding energy per nucleon is about 8 MeV, far above atomic binding, so a strong attractive force of a new kind must overcome proton-proton Coulomb repulsion and bind protons and neutrons.
  • Strength: much stronger than the Coulomb force; gravity is weaker than even the Coulomb force, of order $10^{-39}$ of the nucleon-nucleon force.
  • Range: very short; falls rapidly to zero beyond a few femtometres ($\sim 10^{-15}\,\text{m}$).
  • Saturation: short range means a nucleon binds only to nearest neighbours, which explains the constant binding energy per nucleon for $30 < A < 170$.
  • Potential-energy curve: minimum at $r_0 \approx 0.8\,\text{fm}$; attractive for $r > r_0$, strongly repulsive for $r < r_0$.
  • Charge independence: n–n, p–n and p–p nuclear forces are approximately equal; the force does not depend on electric charge.
  • Form: no simple mathematical expression, unlike Coulomb's law or Newton's gravitation.

NEET PYQ Snapshot — Nuclear Force

The NEET papers in our bank test nuclear-force ideas through the binding-energy and stability logic rather than as standalone "name the property" items, so these are framed as concept checks built on that evidence.

Concept

Which property of the nuclear force is directly responsible for the binding energy per nucleon being nearly constant for nuclei with $30 < A < 170$?

  • (1) Its great strength compared with the Coulomb force
  • (2) Its charge independence
  • (3) Its short range, leading to saturation
  • (4) Its repulsive core below $r_0$
Answer: (3)

A short-range force lets a nucleon interact with only its nearest neighbours (saturation). Adding nucleons does not change the binding of a nucleon already buried inside, so the binding energy per nucleon stays nearly constant across the middle-mass range.

Concept

On the potential-energy versus separation curve for two nucleons, the potential energy is a minimum at $r_0 \approx 0.8\,\text{fm}$. The nuclear force is therefore

  • (1) attractive for all separations
  • (2) repulsive for all separations
  • (3) attractive for $r > r_0$ and strongly repulsive for $r < r_0$
  • (4) repulsive for $r > r_0$ and attractive for $r < r_0$
Answer: (3)

The slope of $U(r)$ gives the force. Beyond the minimum the potential rises with separation, so the force is attractive; below the minimum it climbs steeply, so the force is strongly repulsive. The minimum at $r_0$ is the equilibrium separation.

Concept

Charge independence of the nuclear force implies that

  • (1) protons feel no Coulomb force inside the nucleus
  • (2) the nuclear force between n–n, p–n and p–p pairs is approximately equal
  • (3) only proton-neutron pairs experience the nuclear force
  • (4) the nuclear force follows an inverse-square law
Answer: (2)

Charge independence is the statement that the nuclear interaction is the same for all three nucleon pairings. Protons still experience Coulomb repulsion in addition; the nuclear force itself simply does not depend on charge, and it has no simple inverse-square form.

FAQs — Nuclear Force

Six points students most often confuse, answered straight from NCERT §13.5 and NIOS §26.2.

Why must the nuclear force be much stronger than the Coulomb force?
Inside the nucleus the positively charged protons repel one another through the Coulomb force, yet the nucleus stays bound with a binding energy of roughly 8 MeV per nucleon, far larger than the binding energy in atoms. To hold both protons and neutrons inside the tiny nuclear volume the attractive force must dominate over the Coulomb repulsion between protons. This is possible only because the nuclear force is much stronger than the Coulomb force; the gravitational force is much weaker than even the Coulomb force and cannot account for the binding.
What is the significance of the distance r0 about 0.8 fm in the nuclear potential-energy curve?
On the plot of the potential energy of a pair of nucleons against their separation, the potential energy is a minimum at a distance r0 of about 0.8 fm. For separations greater than r0 the force is attractive, and for separations less than r0 the force is strongly repulsive. The value r0 is therefore the equilibrium separation at which the two nucleons sit most comfortably.
What does charge independence of the nuclear force mean?
The nuclear force between neutron and neutron, between proton and neutron, and between proton and proton is approximately the same. In other words, the nuclear force does not depend on the electric charge of the nucleons. This is consistent with the binding energy per nucleon being roughly the same regardless of the mix of neutrons and protons in the nucleus.
What is the saturation property of the nuclear force?
Because the nuclear force is short ranged, a given nucleon interacts only with the few neighbouring nucleons that lie within the range of the force, not with every nucleon in the nucleus. This limited interaction is called saturation. It explains why the binding energy per nucleon stays nearly constant for nuclei of middle mass number, because adding more nucleons does not change the binding of a nucleon already buried inside the nucleus.
Is there a simple mathematical formula for the nuclear force like Coulomb's law?
No. Unlike Coulomb's law or Newton's law of gravitation, there is no simple mathematical form of the nuclear force. Its features, the great strength, the very short range, charge independence, saturation, and the change from attraction to strong repulsion below r0, were obtained from a variety of experiments carried out during 1930 to 1950 rather than from one tidy inverse-square expression.
How does the nuclear force differ from the Coulomb force in range and reach?
The Coulomb force follows an inverse-square law and acts over long distances, so it governs the motion of atomic electrons far from the nucleus. The nuclear force falls rapidly to zero once the separation between two nucleons is more than a few femtometres, so it acts only over very short distances. This rapid fall to zero is what leads to saturation of forces in a medium or large nucleus and to the constancy of the binding energy per nucleon.