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Question
Determine the binding energy per nucleon of the americium isotope \[\ce{_95^244Am}\], given the mass of \[\ce{_95^244Am}\] to be 244.06428 u.
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Solution
Given: Z = 95, N = 244 - 95 = 149, mp = 1.00728 u, mn = 1.00866 u, M = 244.06428 u, 1 u = 931.5 MeV/c2
To find: B.E. per nucleon `E_B/A = ?`
The binding energy per nucleon,
`"E"_"B"/"A" = (("Zm"_"p" + "Nm"_"n" - "M")"c"^2)/"A"`
`= ([95(1.00728) + 149(1.00866) - 244.06428]"uc"^2)/244`
`= ((95.6916 + 150.29034 - 244.06428)/244)(931.5)` MeV nucleon
∴ `E_B/A = ` 7.3209 MeV/nucleon
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