Skip to content

Question

What is the binding energy of a nucleus with mass defect \( 0.12 \, \text{u} \)? (Given \( 1 \,
\text{u} = 931.5 \, \text{MeV/c}^2 \))

Options

Choose one · Correct answer highlighted

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. E_b = Δ M · c² . Δ M = 0.12 u . E_b = 0.12 × 931.5 = 111.78 MeV . 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 111.78 MeV, consistent with Bohr model and nuclear binding energy systematics.

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.