Question

For lead 82Pb207 (atomic mass = 206.975880 u) obtain (a) the mass defect in atomic mass units, (b) the binding energy (in MeV), and (c) the binding energy per nucleon (in MeV).

Answer 1

Part A.

Mass defect is given by:

dm = Mass of Protons in lead+ mass of neutrons in lead+ Mass of electrons in lead- atomic Mass of lead

Given that

Atomic mass of lead= 206.975880 u

Mass of 1 proton = 1.007276467 u

Mass of 1 electron = 9.11*10^-31 kg = 0.000549 amu

Mass of 1 neutron = 1.008664916 u

Number of protons in lead= 82

Number of neutrons in lead= 207 - 82 = 125

Number of electrons in lead= 82

So,

dm = 82*mp + 125*mn + 82*me - M_Pb

dm = 82*1.007276467 + 125*1.008664916 +82*0.000549 - 206.975880

dm = mass defect = 1.748922794 amu

Part B.

Binding energy is given by:

E = dm*c^2

c^2 = 931.5 Mev/u

So,

E = (1.748922794 u)*(931.5 MeV/u) = 1.748922794*931.5

E = Binding energy = 1629.1215826 MeV = 1629.12 MeV

Part C.

Binding energy per nucleon will be:

BE per nucleon = BE/total number of nucleons

Total number of nucleons = 207

BE per nucleon = BE/total number of nucleons = 1629.12/207

Binding energy per nucleon = 7.87 MeV/nucleon

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