For a nucleus of mass number A and radius R, the mass density of nucleus can be represented as

Solution

Physics sir · 3 steps

Mass grows like A and volume also grows like A, so their ratio leaves no A behind.

1. Mass of nucleus. A nucleus has A nucleons, each of mass about m (roughly 1.67 × 10⁻²⁷ kg). So its mass is M ≈ mA.
2. Radius and volume. The radius follows R = R₀ A^(1/3) with R₀ ≈ 1.2 fm. So the volume is 4/3π R³ = 4/3π R₀³ A, which is directly proportional to A.

   ρ = mA/(4/3π R₀³ A) = 3m/(4π R₀³)

3. So. A cancels and only constants remain, giving about 2.3 × 10¹⁷ kg m⁻³ for every nucleus. The density is independent of A, option (b).

The idea to keep. Since R ∝ A^(1/3), nuclear volume goes as A, just like mass. So all nuclei have nearly the same density, the way drops of one liquid do.

Answer from the NTA final answer key. Solution by PrepWiser.