Exercises
Obtain the binding energy (in MeV) of a nitrogen nucleus (14₇ N), given m(14₇ N) = 14.00307 u.
Given: m(14₇ N) = 14.00307 u, Z = 7 (protons), A = 14, N = 7 (neutrons).
Standard data: mH = 1.007825 u, mₙ = 1.008665 u, 1 u = 931.5 MeV/c².
Step 1 — Mass defect (using atomic masses; electron masses cancel automatically):
DeltaM = Z × mH + (A - Z) × mₙ - m(14₇ N)
DeltaM = 7 × 1.007825 + 7 × 1.008665 - 14.00307
DeltaM = 7.054775 + 7.060655 - 14.00307
DeltaM = 14.115430 - 14.00307 = 0.112360 u
Step 2 — Binding energy:
Eb = DeltaM × 931.5 MeV/u
Eb = 0.112360 × 931.5 = 104.66 MeV
Binding energy per nucleon = 104.66 / 14 = 7.48 MeV
Answer: Binding energy of 14₇ N nucleus = 104.66 MeV (≈ 7.48 MeV per nucleon)
