Q.(a) Distinguish between isotopes and isobars, giving one example for each.

A nucleus's actual measured mass is always slightly less than the sum of the masses of its separate protons, neutrons and electrons; this shortfall is the mass defect, $\Delta m = [Zm_H + (A-Z)m_n] - m$. By Einstein's relation $E=mc^2$, that 'missing' mass corresponds to energy that was released when the nucleus formed -- the nuclear binding energy, $B.E. = \Delta m(u) \times 931.4$ MeV per u (using nuclear masses in unified mass units u, where 1u = 1.66x10^-27 kg). Dividing by the number of nucleons gives binding energy per nucleon, $\bar{B} = B.E./A$, the real measure of a nuclide's stability. Plotted against mass number, $\bar{B}$ peaks sharply at light nuclides that are multiples of helium-4, climbs through medium-mass nuclides, and reaches its overall maximum (~8.79 MeV/nucleon) at iron-56, the single most tightly bound nuclide known, before falling off again for heavy nuclides. Because both fusing light nuclei and splitting heavy nuclei move the products toward that high-$\bar{B}$ peak, both processes release energy. …

Nuclear Density: Why All Nuclei Are Almost Equally Dense

Imagine you have a bag of marbles. If you pack them tightly, the density of the bag depends only on the marbles themselves — not on how many you put in. The nucleus behaves the same way. That's the core idea.

The Intuition

An atom's nucleus is made of protons and neutrons (collectively called nucleons). These nucleons are held together by the strong nuclear force, which is extremely short-ranged. Think of it like magnets: each nucleon only "feels" its immediate neighbours. So adding more nucleons doesn't compress the inner ones — it just adds a new layer on the outside.

This means the nucleus grows in volume proportionally to the number of nucleons. Double the number of nucleons, double the volume. And since mass also doubles, the density stays constant.

The Precise Statement

The nuclear radius $R$ is experimentally found to follow:

where:

• $A$ = mass number (total protons + neutrons)
• $R_0 \approx 1.2 \times 10^{-15} \text{ m}$ (a constant, about 1.2 femtometres)

This is the nuclear radius formula. It's not a guess — it comes from scattering experiments where high-energy electrons or alpha particles bounce off nuclei.

Deriving the Density

The nucleus is roughly spherical, so its volume is:

Notice: volume is directly proportional to $A$. The mass of the nucleus is approximately $m \approx A \times (1.67 \times 10^{-27} \text{ kg})$ (mass of one nucleon). So density:

The $A$ cancels out completely. The density is a constant — independent of the nucleus size.

The Numerical Value

Plug in the numbers:

• $m_{\text{nucleon}} \approx 1.67 \times 10^{-27} \text{ kg}$
• $R_0 \approx 1.2 \times 10^{-15} \text{ m}$

That's about 230 million tonnes per cubic centimetre. To put it in perspective: a sugar-cube-sized piece of nuclear matter would weigh as much as 230 million cars.

Why This Matters

This constancy of density tells us something profound: the strong nuclear force saturates. Each nucleon only interacts with its nearest neighbours, not with the whole nucleus. If the force were long-range (like gravity), density would increase with size. It doesn't — so the force is short-range. …