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Question

What is the approximate mass number of a nucleus with radius \( 3.3 \times 10^{-15} \, \text{m} \)?
(Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))

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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. R = R₀ A¹/³ . 3.3 × 10⁻¹⁵ = 1.2 × 10⁻¹⁵ × A¹/³ . A¹/³ = (3.3/1.2) = 2.75 . A = (2.75)³ ≈ 20.8 ≈ 21 . 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 21, consistent with

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