Practice question
Question
What is the binding energy of a nucleus with mass defect \( 0.14 \, \text{u} \)? (Given \( 1 \,
\text{u} = 931.5 \, \text{MeV/c}^2 \))
Explanation
**Binding energy calculation** from mass defect, for nucleus mass number 28 BE 224 MeV BE/A=8 MeV, A=18 BE 144 MeV BE/A=8 MeV, BE/A indicates stability, fusion of light nuclei and fission of heavy release energy because product has higher BE/A, difference released. E_b = Δ M · c² . Δ M = 0.14 u . E_b = 0.14 × 931.5 = 130.41 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 130.41 MeV, consistent with Bohr model and nuclear binding energy systematics.
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