In a typical nuclear reaction e.g 21H + 21H → 32H + 10n + 3.27MeV, although the number of nucleons is conserved, yet energy is released. How?

In a nuclear reaction, the aggregate of the masses of the target nucleus (21H) and the bombarding particle may be greater or less than the aggregate of the masses of the product nucleus (32He) and the outgoing particle (10n). So, from the law of conservation of mass-energy, some energy (3.27 Mev) is evolved or involved in a nuclear reaction. This energy is called the Q-value of the nuclear reaction. Density of the nucleus = mass of nucleus / volume of nucleus Mass of the nucleus = A amu = A × 1.666 × 10⁻²⁷ kg Volume of the nucleus, V = (4/3)πR³ = (4/3)π(R₀A⅓)³ Where, R = R₀A¹/³ and V = (4/3)πR₀³A Thus, density = A × 1.66 × 10⁻²⁷ / ((4/3)πR₀³A) = 1.66 × 10⁻²⁷ / ((4/3)πR₀³) Which shows that the density is independent of mass number A.