Problem: What is the nuclear binding energy of a lithium-7 nucleus in units of kJ/mol and eV/nucleus? (Mass of a lithium-7 atom = 7.016003 amu.)

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FREE Expert Solution

First, calculate mass defect: 

 Li–7 (atomic mass = 7.016003 amu)

Atomic number→ 3 → 3 protons

Mass number → 7 

Neutrons → 7-3 = 4


Mass values needed: 

Proton = 1.007277 amu

electron = 0.000548597 amu

neutron = 1.008665 amu

Lithium = 7.016003 amu


Calculate mass defect: 

m = [Z(mp +me)+(A-Z)mn]-matomm = [3(1.007277 amu +0.000548597 amu)+(7-3)(1.008665 amu)]           -7.016003 amum = [3(1.007826)+(4)(1.008665 amu)]-7.016003 amu

Δm = 0.042105 amu


Substitute the equation below: 

1 amu c2 = 931.49 MeV

E = mc2E =(0.042105 amu)c2 ×931.49 MeV1 amu c2 ×106 eV1  MeV

E = 3.92 x 107  eV/nucleus


Convert to kJ/mol: 


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Problem Details

What is the nuclear binding energy of a lithium-7 nucleus in units of kJ/mol and eV/nucleus? (Mass of a lithium-7 atom = 7.016003 amu.)