Q.Heavy stable nuclei have more neutrons than protons. This is because of the fact that

The key idea is that in a stable nucleus, the repulsive electrostatic force between protons must be balanced by the strong nuclear force. Adding extra neutrons provides more attractive nuclear force without increasing electrostatic repulsion.

Reasoning:

1. Protons are positively charged and repel each other via the electrostatic force. This repulsion grows with the number of protons.
2. The strong nuclear force is attractive and acts between all nucleons (protons and neutrons), but it is short-ranged and saturates.
3. In heavy nuclei, the electrostatic repulsion becomes significant. Extra neutrons add more nuclear attraction without adding charge, thereby stabilising the nucleus.

Why this question matters — the concept of nuclear stability

A nucleus is a dense ball of protons and neutrons held together by the strong nuclear force. But protons are positively charged, so they also repel each other electrostatically. In a small nucleus (like helium or carbon), the strong force easily overpowers this repulsion. But as the nucleus gets larger, the Coulomb repulsion grows faster than the nuclear attraction — because the electrostatic force is long-range, while the strong force acts only between nearest neighbours.

To keep the nucleus stable, you need extra neutrons. Neutrons add strong-force binding without adding any extra Coulomb repulsion. So the neutron-to-proton ratio rises steadily as atomic number increases — from about 1:1 for light nuclei to about 1.5:1 for uranium.

Now let’s examine each option.

1. Option (A): neutrons are heavier than protons.

   True — a neutron is about 0.14% heavier than a proton. But this tiny mass difference has nothing to do with the number of neutrons in a stable nucleus. The mass difference matters in beta decay energetics, not in determining the neutron-proton ratio for stability. So this is a distractor.

2. Option (B): electrostatic force between protons is repulsive.

   This is the core reason. In a large nucleus, each proton repels every other proton via the Coulomb force. To counteract this, you need extra neutrons — they contribute strong attraction without adding to the repulsion. The larger the nucleus, the more neutrons you need per proton to keep it bound. This is exactly why the neutron-proton ratio increases with atomic number.

3. Option (C): neutrons decay into protons through beta decay.

   A free neutron does decay into a proton, an electron, and an antineutrino with a half-life of about 15 minutes. But inside a stable nucleus, a neutron is perfectly stable — it only decays if the resulting nucleus is more stable. This fact is a consequence of the stability balance, not the cause of the excess neutrons. In fact, in heavy nuclei, it’s often the proton that would prefer to turn into a neutron (via inverse beta decay or electron capture) to reduce Coulomb repulsion.

4. Option (D): nuclear forces between neutrons are weaker than that between protons.

   This is false. The strong nuclear force is charge-independent — it acts equally between any pair of nucleons (p-p, n-n, p-n) at the same separation. Small differences exist due to the Pauli exclusion principle and the fact that the p-n system has an extra bound state (the deuteron), but these are subtle and do not explain the systematic excess of neutrons in heavy nuclei.

Method: Weighing Coulomb Repulsion Against Nuclear Attraction to Judge Stability

Use this whenever a question asks why heavier nuclei need a particular proton-to-neutron composition to stay stable.

Steps

Step 1: Identify the two forces at play inside a nucleus

• The strong nuclear force: attractive, acts equally between any pair of nucleons (p-p, n-n, p-n), but only over a very short range (roughly the size of one or two nucleons).
• The electrostatic (Coulomb) force: repulsive, but acts only between protons, and is long-range — every proton feels the repulsion of every other proton in the nucleus, not just its nearest neighbours.

Step 2: Compare how each force scales with the number of nucleons

The strong force is short-range, so each nucleon only really "feels" its immediate neighbours — its total contribution grows roughly linearly with the number of nucleons $A$. The Coulomb repulsion, being long-range, grows faster — roughly like $Z^2$ (or $Z(Z-1)$) as the proton count $Z$ increases. So as a nucleus gets bigger, repulsion catches up with (and can overtake) attraction.

Step 3: See what extra neutrons buy you

A neutron contributes attractive strong-force bonds to its neighbours but, carrying no charge, adds zero Coulomb repulsion. So adding neutrons increases binding without increasing the destabilising repulsion — exactly what's needed to compensate for the fast-growing Coulomb term as $Z$ rises.

Step 4: Conclude the general trend and apply it

For light nuclei, where Coulomb repulsion is still small, $N \approx Z$ is enough for stability. For heavy nuclei, the repulsion is large enough that extra neutrons ($N > Z$) are required to keep the nucleus bound — this is the direct, general reason the stable neutron-to-proton ratio climbs with atomic number. Match this reasoning to whichever answer option correctly names "Coulomb repulsion between protons" as the underlying cause.