Discussion
What is the binding energy per nucleon for isotope \({}^{55}_{25}Mn\) ? (Given : \(M_{N} = 1.008665\) amu, \(M_{p} = 1.007825\) amu, \({}^{55}_{25}Mn = 54.938050\) amu)
Answer: A
The binding energy per nucleon is about 8.7 MeV.
\({}^{55}_{25}Mn\) has 25 protons and 30 neutrons. The mass defect is
\[ \begin{aligned} \Delta m &= 25(1.007825) + 30(1.008665) - 54.938050 \\ &= 25.195625 + 30.259950 - 54.938050 \\ &= 0.517525\ \text{amu} \end{aligned} \]
\[ \begin{aligned} B &= 0.517525 \times 931.5 = 482.1\ \text{MeV} \\ B/A &= 482.1/55 = 8.77\ \text{MeV} \end{aligned} \]
which is the 8.7 MeV option. The other values exceed the highest binding energy per nucleon of any nucleus, about 8.8 MeV near iron and nickel, so they cannot be right for manganese.
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