Practice question
Question
A nucleus has a radius of \( 3.6 \times 10^{-15} \, \text{m} \). What is its approximate mass number?
(Given \( R_0 = 1.2 \times 10^{-15} \, \text{m} \))
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.
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.