How does the nuclear force behave, when the spin of nucleons are parallel ?

Question

UnlockIAS · Accepted Answer

Correct — option (1), "The force is stronger". The nuclear force between two nucleons is spin-dependent: it is stronger when their intrinsic spins are parallel than when they are antiparallel. This is one of the defining properties of the force, listed alongside its short range, its charge independence and its saturation, and it is not a theoretical nicety — it is read directly off the simplest nuclear system there is. The evidence is the deuteron. A deuteron is one proton bound to one neutron, and it is the only bound two-nucleon system in nature. Its total spin is 1 in units of ħ, which means the proton and neutron spins are parallel — the triplet configuration — and its binding energy is about 2.22 MeV, small but real. Now take the same proton and neutron with their spins antiparallel, the singlet configuration. That system has no bound state at all. The two particles are identical in every respect except the relative orientation of their spins, and one arrangement binds while the other does not. Nothing but a spin dependence in the force itself can account for that, and the direction of the dependence is fixed by which arrangement binds: parallel spins, stronger attraction, a bound nucleus. The point is reinforced from the other side. There is no bound di-proton and no bound di-neutron. In each of those pairs the two nucleons are identical fermions, so the Pauli exclusion principle bars them from occupying the same lowest state with parallel spins; they are forced into the antiparallel arrangement, and in that arrangement the attraction is too weak to bind. Only the neutron-proton pair, which is free to take the parallel-spin configuration because the two particles are distinguishable, forms a bound two-body nucleus. The pattern of what exists and what does not in the lightest nuclei is itself the measurement. So the answer is option (1). The force is stronger when the spins are parallel — which is why the deuteron exists, why its spin is 1, and why the periodic table has a starting point at all.

Why the other options are wrong — (b) Option (2) reverses the dependence. If the attraction were weaker with spins parallel, the bound two-nucleon state would be the singlet, with proton and neutron spins antiparallel, and the deuteron would have total spin zero. It does not — the deuteron's spin is 1, its nucleons are in the parallel-spin triplet configuration, and it is the singlet neutron-proton system that fails to bind. The observed facts of the lightest bound nucleus point squarely against this option.  (c) Option (3) denies that the nuclear force depends on spin at all. That is a substantive claim and it is contradicted by the plainest evidence available: a proton and a neutron bind when their spins are parallel and do not bind when they are antiparallel. A spin-independent force could not distinguish the two cases, so both would bind or neither would. Spin dependence is a standard listed property of the nucleon-nucleon interaction precisely because this comparison forces it.  (d) Option (4) attaches a real property of the nuclear force to the wrong variable. The force does turn strongly repulsive — it has a hard repulsive core at very short separations, below roughly 0.7 femtometre, which is what stops nuclear matter from collapsing and gives nuclei their almost constant density. But that repulsion is a function of the DISTANCE between nucleons, not of the relative orientation of their spins. Parallel spins strengthen the attraction; they do not turn it into a repulsion, and if they did the deuteron could not exist.

Concept: The nuclear force is the attraction that binds protons and neutrons into a nucleus against the electrostatic repulsion of the protons. Its standard properties form a compact list that examiners return to. It is the strongest of the known interactions at nuclear distances, roughly a hundred times stronger than the electromagnetic force. It is extremely short-ranged, effective over one or two femtometres and negligible beyond about two and a half, which is why nuclear forces play no part in chemistry or in everyday physics. It is charge-independent: the neutron-neutron, proton-proton and neutron-proton attractions are essentially equal once the Coulomb repulsion between protons is set aside. It saturates — each nucleon interacts only with its nearest neighbours rather than with every other nucleon in the nucleus, which is why the binding energy per nucleon stays close to 8 MeV across most of the periodic table instead of growing with mass number. It has a repulsive hard core at very small separations, which keeps nuclear density nearly constant. It is spin-dependent, stronger for parallel spins, which is the property tested here. And it is non-central: the deuteron has a small but non-zero electric quadrupole moment, showing that its ground state is not perfectly spherical and that the force contains a tensor component depending on the orientation of the spins relative to the line joining the nucleons.

State PCS 2026 Q56 — Science & Technology PYQ with Answer

State PCS 2026 Q56 — Science & Technology PYQ with Answer