For lead (atomic mass = 205.974440 u) obtain (a) the mass defect in atomic mass

Question

For lead (atomic mass = 205.974440 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/nucleon). For lead (atomic mass = 205.974440 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/nucleon). For lead (atomic mass = 205.974440 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/nucleon).

Explanation / Answer

Solution: We have given For lead (atomic mass = 205.974440 u)

Pb has 82 protons and 124 neutrons

(a) the mass defect in atomic mass units
So the mass defect = (82*1.007825 + 124*1.008665 - 205.97440) = 1.74171u

b) the binding energy (in MeV)

=> EB = 1.74171*c^2 = 1.74171*931.5MeV/u = 1622MeV

c)the binding energy per nucleon (in MeV/nucleon).

=>EB/206 = 7.876MeV/u

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.