The radius R of a nucleus of mass number A can be estimated by the formula
R = (1.3 × 10 –15 )A 1/3 m.
It follows that the mass density of a nucleus is of the order of :
(M prot. ≅ Mₙₑᵤₜ ≃ 1.67 × 10 –27 kg)

Solution

Physics sir · 3 steps

Mass goes as A and volume also goes as A, so A cancels and you get one density for every nucleus.

1. Mass and volume. A nucleus has A nucleons, each of mass m ≈ 1.67 × 10⁻²⁷ kg, so its mass is Am. With R = R₀ A^(1/3), the volume is 4/3π R³ = 4/3π R₀³ A, which is also proportional to A.

   ρ = Am/(4/3π R₀³ A) = m/(4/3π R₀³)

2. Put in numbers. R₀³ = (1.3 × 10⁻¹⁵)³ = 2.197 × 10⁻⁴⁵ m³, so 4/3π R₀³ = 9.20 × 10⁻⁴⁵ m³. Then ρ = (1.67 × 10⁻²⁷)/(9.20 × 10⁻⁴⁵) = 1.8 × 10¹⁷ kg m⁻³.
3. So. Nuclear density is of the order of 10¹⁷ kg m⁻³, which is option (c).

The idea to keep. Because R ∝ A^(1/3), nuclear volume grows as A, just like mass. So every nucleus has the same density, about 10¹⁷ kg m⁻³. Remember that number.

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