The mass of the nucleus of an 
Michael S. answered • 08/05/26
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
Heads up: the neutron mass dropped out of your post - it reads "The mass of a neutron is" and then jumps to the next sentence. I will use the standard value, 1.675 x 10^-24 g. Swap in whatever number your table gives if it differs slightly.
Binding energy is the energy equivalent of the mass defect: a nucleus weighs less than its separated pieces do, and that missing mass is what holds it together.
Step 1 - mass of the separated nucleons
Fe-56 has 26 protons and 56 - 26 = 30 neutrons.
protons: 26 x 1.673 x 10^-24 g = 4.3498 x 10^-23 g
neutrons: 30 x 1.675 x 10^-24 g = 5.0250 x 10^-23 g
total: 9.3748 x 10^-23 g
Step 2 - mass defect
9.3748 x 10^-23 g - 9.286 x 10^-23 g = 8.88 x 10^-25 g
Convert to kilograms, because c is in m/s and E = mc^2 only gives joules in SI units:
8.88 x 10^-25 g x (1 kg / 1000 g) = 8.88 x 10^-28 kg
Forgetting this conversion is the most common error here, and it puts you off by exactly a factor of 1000.
Step 3 - Einstein
E = mc^2 = (8.88 x 10^-28 kg)(2.998 x 10^8 m/s)^2
E = 8.88 x 10^-28 x 8.988 x 10^16 = 7.98 x 10^-11 J
Worth checking
Divide by 56 nucleons: 1.43 x 10^-12 J per nucleon, which is about 8.9 MeV per nucleon. The accepted figure for iron-56 is 8.79 MeV per nucleon, and the small gap comes from the rounded masses you were given. Iron-56 sits at the peak of the binding-energy-per-nucleon curve, which is why it is the endpoint of fusion in stars - past iron, fusing costs energy instead of releasing it.
Two notes on setup. Your problem gives the mass of the nucleus, so you correctly compare against protons and neutrons only. If a problem instead gives the mass of the neutral atom, use hydrogen-1 atom masses in place of proton masses so the electrons cancel. And keep every digit until the final subtraction - the mass defect is a small difference between two close numbers, so rounding early destroys your answer.
