Obtain the binding energy of the nuclei 5626Fe and 20983 Bi in units ofMeV from the following data:m (5626Fe ) = 55.934939 u m ( 20983 Bi ) = 208.980388 u
Given: The atomic mass of iron 26 56 Fe is 55.934939 u and the atomic mass of 83 209 Bi is 208.980388 u .
Case-1: For iron.
Since, the number of protons in the nucleus of 26 56 F e is 26 and the number of neutrons in the nucleus of 26 56 F e is 30 .
The mass defect of the nucleus is given as,
Δ m = 26 m p + 30 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 .
Since, 1 u = 931.5 MeV / c 2 .
By substituting the given values in the above formula, we get
Δ m = 26 × 1.007825 + 30 × 1.008665 − 55.934939 = 26.20345 + 30.25995 − 55.934939 = 0.528461 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 formula, we get
E = 0.528461 c 2 × 931.5 MeV / c 2 = 492.26 MeV
The average binding energy per nucleon is given as,
E 1 = E n
Where, the mass number is n and the average binding energy is E 1 .
By substituting the given values in the above formula, we get
E 1 = 492.26 56 = 8.79 MeV
Case-2: For Bi .
Since, the number of protons in the nucleus of 83 209 Bi is 83 and the number of neutrons in the nucleus of 83 209 Bi is 126 .
The mass defect of the nucleus is given as,
Δ m ′ = 83 m p + 126 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 formula, we get
Δ m ′ = 83 × 1.007825 + 126 × 1.008665 − 208.980388 = 83.649475 + 127.091790 − 208.980388 = 1.760877 u =1 .760877 × 931.5 MeV / c 2
The binding energy of the nucleus of iron is given as,
E = Δ m ′ c 2
By substituting the given values in the above formula, we get
E = 1 .760877 c 2 × 931.5 MeV / c 2 = 1640.26 MeV
The average binding energy per nucleon is given as,
E 1 = E n
Where, the mass number is n and the average binding energy is E 1 .
By substituting the given values in the above formula, we get
E 1 = 1640.26 209 = 7.848 MeV
Thus, the average binding energy per nucleon of Fe is 8.79 MeV and the average binding energy per nucleon of Bi is 7.848 MeV .
Given: The atomic mass of iron
Case-1: For iron.
Since, the number of protons in the nucleus of
The mass defect of the nucleus is given as,
Where, the mass of proton is
Since,
By substituting the given values in the above formula, 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 formula, we get
The average binding energy per nucleon is given as,
Where, the mass number is
By substituting the given values in the above formula, we get
Case-2: For
Since, the number of protons in the nucleus of
The mass defect of the nucleus is given as,
Where, the mass of proton is
By substituting the given values in the above formula, we get
The binding energy of the nucleus of iron is given as,
By substituting the given values in the above formula, we get
The average binding energy per nucleon is given as,
Where, the mass number is
By substituting the given values in the above formula, we get
Thus, the average binding energy per nucleon of
