Discussion
What is the binding energy of the nucleus for isobar \({}^{64}_{29}\mathrm{Cu}\)?
[Given : Mn = 1.008665 amu, Mp = 1.007825 amu, \({}^{64}_{29}\mathrm{Cu}\) = 63.9298 amu]
Answer: C
The binding energy is closest to 556.18 MeV.
The nucleus has 29 protons and 35 neutrons. The mass defect, using the given masses, is:
\[ \begin{aligned} \Delta m &= 29(1.007825) + 35(1.008665) - 63.9298 \\ &= 29.226925 + 35.303275 - 63.9298 \\ &= 0.6004\ \text{u} \end{aligned} \]
Multiplying by 931.5 MeV per u gives about 559 MeV, about 8.7 MeV per nucleon, typical of nuclei near iron. Of the options, 556.18 MeV is the only one within 1%; 450, 512 and 612 MeV are far from the mass defect.
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