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Question
Obtain the binding energy (in MeV) of a nitrogen nucleus `(""_7^14"N")`, given `"m"(""_7^14"N")` = 14.00307 u.
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Solution
Atomic mass of nitrogen `(""_7^14"N")`, m = 14.00307 u
A nucleus of nitrogen `""_7^14"N"` contains 7 protons and 7 neutrons.
Hence, the mass defect of this nucleus, Δm = 7mH + 7mn − m
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn = 1.008665 u
∴ Δm = 7 × 1.007825 + 7 × 1.008665 − 14.00307
= 7.054775 + 7.06055 − 14.00307
= 0.11236 u
But 1 u = 931.5 MeV/c2
∴ Δm = 0.11236 × 931.5 MeV/c2
Hence, the binding energy of the nucleus is given as
Eb = Δmc2
Where,
c = Speed of light
∴ Eb = `0.11236 xx 931.5(("MeV")/"c"^2) xx "c"^2`
= 104.66334 MeV
Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.
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