What is the binding energy (in MeV per nucleon) for the ⁶₃Li nucleus?

Explanation:

⁶₃Li will have 3 protons and 3 neutrons .

mass of proton in amu = 1.00727 amu

mass of neutron in amu = 1.00866 amu

mass of lithium nucleus in amu = 6.01512 amu

mass defect = 3 (  1.00727  + 1.00866 ) - 6.01512 amu

= .03267 amu

Binding energy = mass defect in amu x 931 Mev

= 30.41 MeV

binding energy per nucleon

no of nucleon = 3 + 3 = 6

binding energy per nucleon = 30.41 / 6 Mev

= 5.068 MeV .

0.11 m/s and 0.28 m/s should be correct