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.

Answer 1

Given: Z = 95, N = 244 - 95 = 149, m_p = 1.00728 u, m_n = 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