(a) The density of the nuclear matter is tremendously larger than the physical density of the material. Explai

Question

Q. (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.

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Concept understanding — Nuclear Force Range

What is the Range of a Force?

Before we talk about the nuclear force, think about forces you already know. Gravity and electromagnetism have infinite range — a magnet on Earth still feels the pull of a magnet on the Moon, though it's unimaginably weak. The force just gets weaker with distance, but it never truly becomes zero.

Now imagine a force that simply does not exist beyond a certain distance. That's the nuclear force. It's like a rope that only works if you're within arm's length — step back, and the rope goes slack. This is what we mean by finite range.

The Nuclear Force: A Short-Range Glue

The strong nuclear force holds protons and neutrons together inside the nucleus. But here's the puzzle: protons are positively charged and repel each other violently. If the nuclear force had infinite range like electromagnetism, it would either pull everything together or push everything apart — but it doesn't. It only acts when particles are extremely close, about the size of a proton or neutron.

Note

The nuclear force range is roughly 1–2 femtometers (1 fm = $10^{-15}$ m). For comparison, a hydrogen atom is about 100,000 fm across. The nuclear force is a microscopic, neighbourhood-only force.

The Precise Statement

The strong nuclear force between two nucleons (protons or neutrons) is negligible when their separation exceeds about 2.5 fm. It becomes strongly attractive at distances around 1–2 fm, and then turns repulsive if they get closer than about 0.5 fm (this prevents the nucleus from collapsing).

V_{\text{nuclear}}(r) \approx 0 \quad \text{for } r > 2.5 \text{ fm}

This is why a nucleus is so dense — nucleons must be packed within this tiny range to feel the binding force. It's also why only certain combinations of protons and neutrons are stable: if the nucleus gets too large, protons on opposite sides are too far apart to feel the nuclear attraction, but they still feel the electric repulsion. That's why heavy elements need extra neutrons (which add attraction without repulsion) to stay stable.

Why Does It Have a Range?

The short answer: the nuclear force is mediated by particles called pions (pi mesons), which have mass. Unlike photons (massless, infinite range) or gravitons (massless, infinite range), massive exchange particles produce a force that dies out exponentially beyond a characteristic distance — the Compton wavelength of the pion. …

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Nuclear matter is incredibly dense because nucleons are packed at a fixed, universal density ( $\approx 2.3 \times 10^{17} \text{ kg/m}^3$ ), far exceeding ordinary matter. Nuclear forces are short-range, attractive at intermediate distances, and repulsive at very short distances — completely unlike the long-range Coulomb force.

Graph of the potential energy of a pair of nucleons as a function of their separation r: a steep repulsive rise for r below about r0, a deep attractive minimum of roughly -100 MeV near r0 (~0.8 fm, marked with a dashed line), and the potential energy tending to zero beyond about 3-4 fm.

(a) Why nuclear matter density is astronomically larger than ordinary material density

The key idea is that nucleons (protons and neutrons) are packed together at a nearly constant density inside any nucleus, regardless of its size. This is the nuclear saturation property.

Nuclear volume scales with mass number Experiments show that the radius of a nucleus is given by $R = R_0 A^{1/3}$ , where $R_0 \approx 1.2 \times 10^{-15} \text{ m}$ (1.2 fm). The volume is

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

So volume is directly proportional to $A$ , the number of nucleons.

Mass of the nucleus

The mass is roughly $m \approx A \times (1.67 \times 10^{-27} \text{ kg})$ (the mass of one nucleon).
Density calculation

\rho_{\text{nuclear}} = \frac{m}{V} = \frac{A \times 1.67 \times 10^{-27}}{\frac{4}{3}\pi (1.2 \times 10^{-15})^3 A} = \frac{1.67 \times 10^{-27}}{\frac{4}{3}\pi (1.2 \times 10^{-15})^3}.

The $A$ cancels out! Evaluating:

\rho_{\text{nuclear}} \approx 2.3 \times 10^{17} \text{ kg/m}^3.

Comparison with ordinary matter The density of water is $10^3 \text{ kg/m}^3$ , iron is about $7.8 \times 10^3 \text{ kg/m}^3$ . Nuclear density is roughly $10^{14}$ times larger. Why? Ordinary matter is mostly empty space — atoms are mostly vacuum between the nucleus and electrons. In nuclear matter, nucleons touch each other with no wasted space.

Watch out

A common mistake is to think nuclear density varies with element. It does not — all nuclei have nearly the same density because the strong force saturates. A sugar-cube-sized piece of nuclear matter would weigh about 200 million tonnes.

(b) Why nuclear forces are not Coulomb forces between nucleons

The Coulomb force acts between charged particles and is long-range ( $1/r^2$ ). Nuclear forces are fundamentally different.

Range

Coulomb force between two protons extends to infinity. Nuclear force drops to zero beyond about $2\text{–3}$ fm (femtometers). It is a short-range force.
Charge independence

The nuclear force between two protons, two neutrons, or a proton and a neutron is approximately the same (ignoring the small Coulomb repulsion between protons). Coulomb force only acts between charged particles — neutrons feel no Coulomb force at all.
Strength

At distances around 1 fm, the nuclear attractive force is about 100 times stronger than the Coulomb repulsion between two protons. This is why nuclei can stay bound despite the electrostatic repulsion.
Nature of the force

Coulomb force is described by quantum electrodynamics (exchange of photons). Nuclear force is a residual effect of the strong interaction (exchange of mesons like pions), which binds quarks inside nucleons.

Tip

Think of the nuclear force like a very strong, very short-range glue that only works when nucleons are practically touching. Coulomb force is like a weak, long-range rubber band that works even at a distance.

(c) Plot of potential energy between a pair of nucleons

The potential energy $V(r)$ between two nucleons as a function of separation $r$ has a characteristic shape: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 55/6/11 markMCQ
$30 < A < 170$
›Reveal solutionSolution
The binding energy per nucleon is indeed nearly constant in the range $30 < A < 170$ because nuclear forces are short-range and saturate, meaning each nucleon interacts only with its immediate neighbors. The Reason incorrectly states that nuclear forces are "not short-range," making it false. The correct option is (C).

The binding energy curve is one of the most important graphs in nuclear physics, and understanding why it has the shape it does reveals the fundamental nature of the strong nuclear force.

Why binding energy per nucleon plateaus

The binding energy per nucleon, $\frac{B}{A}$ , measures how tightly bound each nucleon is on average. For medium-mass nuclei (roughly $A = 30$ to $170$ ), this quantity hovers around $8$ to $8.8$ MeV per nucleon, staying remarkably constant. This plateau exists because of two competing effects:

The saturation property of nuclear forces. The strong nuclear force is extremely short-range—it acts only over distances of about $1$ to $2$ fm (roughly the size of a nucleon itself). Each nucleon in the interior of a nucleus interacts with only its nearest neighbors, typically $8$ to $12$ nucleons in a close-packed arrangement. Adding more nucleons to the nucleus doesn't increase the number of interactions per nucleon in the bulk; it just adds more nucleons experiencing the same local environment.

Think of it like a crowd of people holding hands: each person can only hold hands with their immediate neighbors. Doubling the crowd size doesn't double the number of hands each person holds—it stays constant.

Volume vs. surface effects. The total binding energy grows roughly as $A$ (volume term in the semi-empirical mass formula), while the number of nucleons is also $A$ , so $\frac{B}{A}$ remains approximately constant. Surface nucleons are slightly less bound (they have fewer neighbors), but for large $A$ , the fraction of surface nucleons becomes negligible.
$\frac{B}{A} \approx a_V - a_S A^{-1/3} - a_C \frac{Z^2}{A^{4/3}} - a_A \frac{(A-2Z)^2}{A^2} \pm \frac{\delta(A)}{A}$
For medium-mass nuclei, the volume term $a_V$ dominates, and the correction terms are small, yielding a nearly flat curve.

Evaluating the Assertion and Reason

Assertion (A): The binding energy per nucleon is practically constant for $30 < A < 170$ .

This is true. Experimental data shows that $\frac{B}{A}$ varies only slightly in this range, peaking near iron-56 at about $8.8$ MeV/nucleon and remaining within $\pm 5\%$ across the entire interval. This is the stable plateau region of the binding energy curve.

Reason (R): Nuclear forces are not short-range for $30 < A < 170$ .

This is false. Nuclear forces are always short-range, regardless of mass number. The range of the strong force is approximately $1$ – $2$ fm and does not change with $A$ . In fact, it is precisely because nuclear forces are short-range that the binding energy per nucleon saturates and remains constant. …
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CBSE 2024Set 55/5/11 markMCQ
$Q. The potential energy between two nucleons inside a nucleus is minimum at a distance of about ______. (A) 0.8 fm (B) 1.6 fm (C) 2.0 fm (D) 2.8 fm$
›Reveal solutionSolution
The nuclear force is attractive at intermediate distances and strongly repulsive at very short distances. The potential energy minimum occurs at the equilibrium separation where the net force is zero, which is approximately 0.8 fm — option (A).

Why this distance matters

The nuclear force between two nucleons (protons or neutrons) is not like gravity or electromagnetism. It has a very short range — it only acts over distances comparable to the size of a nucleus itself. The key feature is that the force is strongly repulsive when nucleons are too close (to stop them from collapsing into each other) and attractive at slightly larger separations, binding the nucleus together.

The potential energy curve for the nucleon-nucleon interaction therefore has a deep minimum at some equilibrium separation. That minimum is the distance where the attractive and repulsive contributions balance — the "sweet spot" where a pair of nucleons would sit if undisturbed.

Step-by-step reasoning

Recall the shape of the nuclear potential.

The nuclear force is modelled by potentials like the Yukawa potential or the Reid soft-core potential. All such models show a repulsive core below about $0.5\ \text{fm}$ and an attractive well that peaks (minimum potential) near $0.8\ \text{fm}$ to $1.0\ \text{fm}$ .

Eliminate the larger distances.

At $1.6\ \text{fm}$ , the nuclear force is still attractive but much weaker — the potential is already climbing back toward zero. At $2.0\ \text{fm}$ and $2.8\ \text{fm}$ , the interaction is negligible (the force range is about $2\ \text{fm}$ ). The minimum cannot be there because the potential is nearly zero, not a deep well.

Identify the correct value. …
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CBSE 2023Set 55/4/11 markMCQ
$Q. Which of the following statements is not true for nuclear forces ? (A) They are stronger than Coulomb forces. (B) They have about the same magnitude for different pairs of nucleons. (C) They are always attractive. (D) They saturate as the separation between two nucleons increases.$
›Reveal solutionSolution
Nuclear forces are short-range, charge-independent, and saturate — but they are not always attractive; they become repulsive at very short distances (hard-core repulsion), making statement (C) false.

The key to this question lies in understanding the range and nature of the strong nuclear force. Unlike the Coulomb force, which is long-range and can be either attractive or repulsive, the nuclear force has a very specific behaviour: it is strongly attractive at typical nucleon separations (around 1–2 fm), but becomes strongly repulsive when nucleons are pushed too close together (below about 0.5 fm). This repulsive core is what prevents nuclei from collapsing.

Let’s examine each statement carefully.

Statement (A): "They are stronger than Coulomb forces."

This is true. At distances of about 1 fm, the nuclear force between two protons is roughly 100 times stronger than the electrostatic repulsion. That’s why nuclei with many protons can stay bound despite the Coulomb repulsion.

Statement (B): "They have about the same magnitude for different pairs of nucleons."

This is also true. The strong nuclear force is charge-independent — it acts almost identically between two protons, two neutrons, or a proton and a neutron (ignoring the small electromagnetic correction). This is why the deuteron (p–n) and the diproton (p–p) would have similar binding if not for Coulomb repulsion.

Statement (C): "They are always attractive."

This is false. While nuclear forces are attractive at typical internucleon distances, they become repulsive at very short separations (the “hard core”). This repulsion is essential to explain nuclear stability and the fact that nucleons do not coalesce into a point. So the force is not always attractive — it has a repulsive core.

Statement (D): "They saturate as the separation between two nucleons increases." …
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Q.(a) The density of the nuclear matter is tremendously larger than the physical density of the material. Explain.

(a) Why nuclear matter density is astronomically larger than ordinary material density

(b) Why nuclear forces are not Coulomb forces between nucleons

(c) Plot of potential energy between a pair of nucleons

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