Q10
neet_2026_03_may_morning
Size of the Nucleus
An unknown nucleus has a nuclear density of 2.29 × 10¹⁷ ~kg/m³ and mass of 19.926 × 10⁻²⁷ ~kg. Its mass number A is approximately:
(Take R₀ = 1.2 × 10⁻¹⁵ ~m, 4π = 12.56)
Solution
Related Formula
R = R₀ A1/3
Density (ρ) = (M)/(V) = (M)/((4)/(3)π R³)Core Logic
Given data: ρ = 2.29 × 10¹⁷ ~kg/m³ M = 19.926 × 10⁻²⁷ ~kg R₀ = 1.2 × 10⁻¹⁵ ~m
Express volume in terms of mass and density:
V = (4)/(3) π R³ = (M)/(ρ)Substitute R = R₀ A1/3:
(4)/(3) π ( R₀ A1/3 )³ = (M)/(ρ) (4)/(3) π R₀³ A = (M)/(ρ)Step 1: Isolate Mass Number (A)
Rearranging for A:
A = (M)/(ρ) × (3)/(4π × R₀³)Substitute all values:
A = 19.926 × 10⁻²⁷2.29 × 10¹⁷ × 312.56 × (1.2 × 10⁻¹⁵)³ A = 19.926 × 10⁻²⁷ × 32.29 × 10¹⁷ × 12.56 × 1.728 × 10⁻⁴⁵ A = 59.778 × 10⁻²⁷49.7 × 10⁻²⁸ ≈ (59.778)/(4.97) ≈ 12Pattern Recognition
Nuclear density is essentially constant for all nuclei. The volume is directly proportional to mass number A. Algebraic isolation of A prevents calculation circularity.
Chapter Mix
Class 12 Physics: Nuclei
