Calculate the mass defect and binding energy per nucleon for an alpha particle (containing two protons and two neutrons) whose actual mass is 4.0028 amu (mass of proton = 1.00759 amu, mass of nuetron = 1.00898 amu).

Explanation

Mass defect , `Deltam=Zm_H+(A-Z)m_n-m_a`
[(2)(1.007825)+(4-2)(1.008665)-4.00260]u
`=(2xx1.007825+2xx1.008665-4.00260)u`
Mass defect, `Deltam`=0.03038 u
`therefore` Binding energy of the nucleus =`(Deltam)C^2`
`=(0.03038)uxxC^2`
`=0.03038xx931.5` MeV (`because 1uxxC^2` = 931.5 MeV)
=28.3 MeV
Binding Energy per nucleon = `28.3/4` MeV
Binding energy per nucleon = 7.075 MeV