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Question

A nucleus has a radius of \( 3.6 \times 10^{-15} \, \text{m} \). What is its approximate mass number?
(Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

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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. R = R₀ A¹/³ . 3.6 × 10⁻¹⁵ = 1.2 × 10⁻¹⁵ × A¹/³ . A¹/³ = (3.6/1.2) = 3 . A = 3³ = 27 . 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 u = 931.5 MeV, evaluation yields 27, consistent with Bohr model and nuclear binding energy systematics.

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