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Question

What is the nuclear density of a nucleus with mass \( 8.35 \times 10^{-27} \, \text{kg} \) and radius
\( 2.7 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

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Explanation

**Nuclear density** nearly constant because R ∝ A^{1/3} so volume ∝ A, mass ∝ A, ratio constant, ~10¹⁷ kg/m³, 10¹⁴ times water density, shows nucleus compact, nuclear force short-range saturated, mass number 16 radius ~3×10⁻¹⁵ m, mass number from radius R=5.4×10⁻¹⁵ m => A=(R/R₀)³=(4.5)³=91. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.7 × 10⁻¹⁵)³ = 1.9683 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 1.9683 × 10⁻⁴⁴ ≈ 8.25 × 10⁻⁴⁴ m³ . Density = (8.35 × 10⁻²⁷/8.25 × 10⁻⁴⁴) ≈ 1.01 × 10¹⁷ kg/m³ . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n

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