Obtain the binding energy (in MeV) of a nitrogen nucleus ( )147 N ,given m ( )147 N =14.00307 u
Given: the atomic mass of nitrogen 7 14 N is 14.00307 u .
A nucleus of nitrogen atom contains 7 proton and 7 neutrons.
The mass defect of the nucleus is given as,
Δ m = 7 m p + 7 m n − m
Where, the mass of proton is m p , the mass of neutron is m n and the mass defect of the nucleus is Δ m .
By substituting the given values in the above equation, we get
Δ m = 7 × 1.007825 + 7 × 1.008665 − 14.00307 = 7.054775 + 7.06055 − 14.00307 = 0.11236 u
The binding energy of the nucleus of iron is given as,
E = Δ m c 2
Where, the speed of light is c and the binding energy is E .
By substituting the given values in the above equation, we get
E = 0.11236 c 2 × 931.5 MeV / c 2 ≈ 104.7 MeV
Thus, the binding energy of a nitrogen nucleus is 104.7 MeV .
Given: the atomic mass of nitrogen
A nucleus of nitrogen atom contains 7 proton and 7 neutrons.
The mass defect of the nucleus is given as,
Where, the mass of proton is
By substituting the given values in the above equation, we get
The binding energy of the nucleus of iron is given as,
Where, the speed of light is
By substituting the given values in the above equation, we get
Thus, the binding energy of a nitrogen nucleus is
