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Question

What is the nuclear density of a nucleus with mass \( 6.68 \times 10^{-27} \, \text{kg} \) and radius
\( 2.4 \times 10^{-15} \, \text{m} \)? (Use \( \pi = 3.14 \))

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Explanation

**Nuclear size** order 10⁻¹⁵ m femtometer, atomic size 10⁻¹⁰ m, ratio 10⁵, nucleus contains protons and neutrons bound by strong force, density independent of A indicates incompressibility, R₀ determined from electron scattering experiments. Density = (mass/volume) , Volume = (4/3) π R³ . R³ = (2.4 × 10⁻¹⁵)³ = 1.382 × 10⁻⁴⁴ m³ . Volume = (4/3) × 3.14 × 1.382 × 10⁻⁴⁴ ≈ 5.79 × 10⁻⁴⁴ m³ . Density = (6.68 × 10⁻²⁷/5.79 × 10⁻⁴⁴) ≈ 1.15 × 10¹⁷ kg/m³ . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1

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