Obtain the binding energy of the nuclei `""_56^26Fe` and `""_83^209Bi`in units of MeV from the following data:
`m(""_26^56Fe)` = 55.934939 u `m(""_83^209Bi)`= 208.980388 u
Solution
Atomic mass of `""_56^26Fe`, m1 = 55.934939 u
`""_56^26Fe` nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus, Δm = 26 × mH + 30 × mn − m1
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∴Δm = 26 × 1.007825 + 30 × 1.008665 − 55.934939
= 26.20345 + 30.25995 − 55.934939
= 0.528461 u
But 1 u = 931.5 MeV/c2
∴Δm = 0.528461 × 931.5 MeV/c2
The binding energy of this nucleus is given as:
Eb1 = Δmc2
Where,
c = Speed of light
∴Eb1 = 0.528461 × 931.5 `((MeV)/c^2) xx c^2`
= 492.26 MeV
Average binding energy per nucleon = `492.26/56 = 8.79 MeV`
Atomic mass of `""_83^209Bi`, m2 = 208.980388 u
`""_83^209 Bi` nucleus has 83 protons and (209 − 83) 126 neutrons.
Hence, the mass defect of this nucleus is given as:
Δm' = 83 × mH + 126 × mn − m2
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∴Δm' = 83 × 1.007825 + 126 × 1.008665 − 208.980388
= 83.649475 + 127.091790 − 208.980388
= 1.760877 u
But 1 u = 931.5 MeV/c2
∴Δm' = 1.760877 × 931.5 MeV/c2
Hence, the binding energy of this nucleus is given as:
Eb2 = Δm'c2
= 1.760877 × 931.5 `((MeV)/c^2) xx c^2`
= 1640.26 MeV
Average bindingenergy per nucleon = `1640.26/209 = 7.848 MeV`
