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Question
Obtain the binding energy of the nuclei `""_26^56"Fe"` and `""_83^209"Bi"` in units of MeV from the following data:
`"m"(""_26^56"Fe")` = 55.934939 u
`"m"(""_83^209"Bi")`= 208.980388 u
Solution
Atomic mass of `""_56^26"Fe"`, m1 = 55.934939 u
`""_56^26"Fe"` 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 `""_83^209"Bi"`, 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 binding energy per nucleon = `1640.26/209` = 7.848 MeV
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