Nuclear Force

Nuclei · CBSE

Theory

Why a new force is needed

The force that governs atomic electrons is the familiar Coulomb force, and atomic binding energies are a few eV.
For middle-mass nuclei the binding energy per nucleon is about $8$ MeV — a million times larger. So a strong attractive force of a completely different kind must bind the nucleus.
It must be strong enough to overcome the Coulomb repulsion between the positively charged protons AND to hold both protons and neutrons within the tiny nuclear volume.
Everything known about this force comes from experiment (chiefly work done between 1930 and 1950); unlike Coulomb's law or Newton's law of gravitation, there is NO simple mathematical form for the nuclear force.

Properties of the nuclear force

1. It is the strongest force in nature

The nuclear force is much stronger than the Coulomb force between charges and vastly stronger than the gravitational force between masses (gravity is weaker even than the Coulomb force).
This has to be so: the nuclear attraction must dominate the Coulomb repulsion between protons packed a few fm apart, otherwise no nucleus with more than one proton could exist.
So the statement "the nuclear force is much weaker than the Coulomb force" is FALSE — a standard MCQ distractor.

2. It is short-ranged, and therefore saturates

The nuclear force between two nucleons falls rapidly to zero once their separation exceeds a few femtometres. It is strongly attractive over roughly $1$ fm to about $4$ fm and negligible beyond that.
Consequence — SATURATION of nuclear forces: a nucleon inside a medium or large nucleus is attracted only by the neighbours that lie within the range of the force, not by every nucleon in the nucleus.
Consequence of the consequence: adding more nucleons to a large nucleus does not increase the binding of the nucleons already inside, so the binding energy per nucleon stays nearly constant for $30 < A < 170$ . Saturation is the reason for the flat middle of the $E_{bn}$ curve.

3. It is charge independent

The nuclear force between a neutron–neutron pair, a proton–neutron pair and a proton–proton pair is approximately the same. The nuclear force does not depend on electric charge.
(The attraction is understood as arising from the exchange of $\pi$ -mesons — $\pi^{0}$ , $\pi^{+}$ , $\pi^{-}$ — between nucleons.)

4. It has a repulsive core

The nucleon–nucleon potential energy has a minimum at a separation $r_0 \approx 0.8$ fm.
For $r > r_0$ the force is ATTRACTIVE; for $r < r_0$ it is strongly REPULSIVE.
This repulsive core is what stops the nucleons collapsing into one another and fixes the size of the nucleus.
So "nuclear forces are always attractive" is FALSE — another standard MCQ answer.

The potential-energy curve, described in words

Axes: potential energy $U$ of a pair of nucleons in MeV (vertical, running from about $+100$ MeV down through zero to about $-100$ MeV) against their separation $r$ in fm (horizontal, 0 to about 4 fm).
Shape, reading from large $r$ inwards: $U$ is essentially zero beyond a few fm; as $r$ decreases the curve dips below zero into a deep attractive well, reaching a minimum of about $-100$ MeV at about $r \approx 1$ fm; it then turns and rises very steeply, crossing zero at $r_0 \approx 0.8$ fm and becoming large and positive for still smaller $r$ .
Mark $r_0 \approx 0.8$ fm on the axis at the point where the curve crosses $U = 0$ on its way up, and mark the minimum of the well.

Reading the force off the curve — the 3-mark justification

The force is related to the potential energy by $F = -\dfrac{dU}{dr}$ .
For $r > r_0$ : the potential energy increases as the separation increases (the curve climbs from the minimum back towards zero), so the slope is positive and the force is directed towards smaller $r$ . The nucleons are pulled together — the force is ATTRACTIVE in this region.
For $r < r_0$ : the potential energy rises very steeply as the nucleons are pushed closer, so the force strongly opposes further approach. The force is strongly REPULSIVE in this region.
The boundary between the two behaviours is $r_0 \approx 0.8$ fm, where the force is zero and the potential energy of the pair is at its minimum — this is the equilibrium separation of two nucleons.

The two conclusions CBSE asks to be drawn from the plot

(i) The nuclear force is much stronger than the Coulomb force between charges or the gravitational force between masses: the depth of the potential well, about $100$ MeV, is far larger than any Coulomb or gravitational potential energy at nuclear separations.
(ii) The nuclear force between two nucleons falls off rapidly to practically zero once their separation exceeds a few fermi — it is a SHORT-RANGE force.

Nuclear force contrasted with the Coulomb force

Point of comparison	Nuclear force	Coulomb force
Range	short, effective only to a few fm	long range, inverse square, never zero
Relative strength	strongest known force in nature	roughly 100 times weaker at nuclear separations
Dependence on charge	charge independent: same for n–n, p–n, p–p	acts only between charged particles
Nature	attractive for $r > 0.8$ fm, strongly repulsive for $r < 0.8$ fm	repulsive between like charges at all separations
Saturation	saturates, giving nearly constant $E_{bn}$	no saturation; every proton repels every other
Mathematical form	no simple formula	$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$

Why nuclear forces cannot be Coulomb forces (a 3-mark part): the Coulomb force between two protons is repulsive, yet a nucleus confines a large number of protons within a volume of radius a few fm. That is only possible if an entirely different force — extremely short-ranged, very strongly attractive and charge-independent — overwhelms the electrostatic repulsion at those separations.

Standard results

Nuclear force: strongest known, short-ranged (a few fm), charge independent, saturating, attractive beyond $r_0 \approx 0.8$ fm and strongly repulsive within it, with no simple mathematical expression.
Potential well depth $\approx -100$ MeV at $r \approx 1$ fm; $U = 0$ crossing and force reversal at $r_0 \approx 0.8$ fm.
$F = -\dfrac{dU}{dr}$ is the tool for identifying the attractive and repulsive regions.
Saturation (from the short range) is the reason $E_{bn}$ is nearly constant for $30 < A < 170$ .
Statements to reject in MCQs: "always attractive"; "much weaker than the Coulomb force"; "depends on the charges of the nucleons"; "saturate as the separation increases" (they saturate with nucleon NUMBER, because of the short range).

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces.
2 marksCBSE Class 12 Physics — SA · AI 2015, Ap · Section B, 2 marksAnswer 1 ↓
2
Draw a plot showing the variation of potential energy of two nucleons as a function of the distance between them. Identify the regions in which the force between the nucleons is (i) attractive, and (ii) repulsive. Justify.
3 marksCBSE Class 12 Physics — SA · 2023 · Section C, 3 marksAnswer 2 ↓
3
(a) The density of the nuclear matter is tremendously larger than the physical density of the material. Explain. (b) The nuclear forces are not coulomb forces between nucleons. Explain. (c) Draw a plot of the potential energy between a pair of nucleons as a function of distance between them inside a nucleus.
3 marksCBSE Class 12 Physics — SA · 2020 · Section C, 3 marksAnswer 3 ↓
4
Write three characteristic properties of nuclear force.
3 marksCBSE Class 12 Physics — SA · AI 2015, R · Section C, 3 marksAnswer 4 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

5
Which of the following statements is not true for nuclear forces?
- (a)
  They are stronger than Coulomb forces.
- (b)
  They have about the same magnitude for different pairs of nucleons.
- (c)
  They are always attractive.
- (d)
  They saturate as the separation between two nucleons increases.
1 markCBSE Class 12 Physics — Multiple Choice · 2023 · Section A, 1 markAnswer 5 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
Potential energy vs separation: the curve dips to a deep minimum (about $-100$ MeV) near a separation of about $1$ fm, then rises steeply and becomes strongly positive (repulsive) for separations below about $0.8$ fm, and rises gently back toward zero for separations beyond a few fm. Conclusions: (i) The nuclear force is much stronger than the Coulomb force acting between charges, or the gravitational force acting between masses — the depth of the potential well ( ${\sim}100$ MeV) is far larger than typical Coulomb/gravitational potential energies at nuclear separations. (ii) The nuclear force between two nucleons falls off rapidly to (practically) zero once their separation is more than a few fermi — confirming that it is a short-range force.
2.↑ back to the question
Potential energy vs separation: the curve dips to a deep minimum (about $-100$ MeV) near $r\approx1$ fm, then rises very steeply for smaller $r$ (becoming strongly positive below about $0.8$ fm), and climbs gently back toward zero for larger $r$ . (i) For separations greater than $r_0\;(\approx0.8\text{ fm})$ , the potential energy INCREASES as the separation decreases (the curve slopes downward as separation increases toward the minimum) — i.e. the force $F=-\dfrac{dU}{dr}$ is directed so as to pull the nucleons together. So the force is ATTRACTIVE for $r>r_0$ . (ii) For separations less than $r_0$ , the potential energy rises very steeply as the nucleons are pushed closer, meaning the force strongly opposes further approach. So the force is STRONGLY REPULSIVE for $r<r_0$ — this repulsive core is what keeps the nucleons in a nucleus from collapsing into each other.
3.↑ back to the question
(a) Nuclear density is tremendously larger than the ordinary (bulk atomic) density of the material because almost the entire mass of an atom is concentrated in an extremely small nuclear volume (radius $\sim10^{-15}$ m, about $10^4$ – $10^5$ times smaller than the atomic radius), while most of the rest of the atom is empty space occupied only by the light, low-mass electrons. Packing essentially all the mass into such a tiny volume gives an enormously large density ( ${\sim}2.3\times10^{17}\text{ kg/m}^3$ ). (b) The Coulombic force between two protons is repulsive, yet within a nucleus a large number of protons (and neutrons) are held together in an extremely small volume. This is only possible because the force holding nucleons together is NOT the Coulomb force but an entirely different, extremely short-range, strongly ATTRACTIVE force (the strong nuclear force) that overwhelms the electrostatic repulsion between protons at nuclear separations, and is also charge-independent (acts equally between p-p, n-n and p-n pairs). (c) Potential energy of a pair of nucleons vs distance between them: the curve shows a deep attractive potential well (minimum around $-100$ MeV near $r\approx1$ fm), turning strongly repulsive (rising sharply above zero) for $r$ below about $0.8$ fm, and approaching zero for large $r$ .
4.↑ back to the question
(i) Nuclear forces are short-range forces — they are strongly attractive within a range of about 1 fermi to 4.2 fermi, and become negligible for separations beyond about 4.2 fermi. (ii) Nuclear forces are charge-independent — the same magnitude of nuclear force acts between a pair of protons, a pair of neutrons, and a proton-neutron pair. (The attractive nuclear force arises from the exchange of $\pi$ -mesons — $\pi^0,\pi^+,\pi^-$ — between nucleons.) (iii) Nuclear forces are the strongest known forces in nature (far stronger than the Coulomb or gravitational forces at nuclear separations), and they saturate — beyond a certain nucleon size, a nucleon interacts only with its nearest neighbours rather than with every nucleon in the nucleus.
5.Answer: (C)↑ back to the question
Nuclear forces are attractive only for separations greater than about $r_0\approx0.8$ fm; for separations smaller than $r_0$ they become STRONGLY REPULSIVE (this is what prevents the nucleus from collapsing). So statement (c), "they are always attractive", is NOT true. Option (c).