Question 24.7: Calculating the Binding Energy per Nucleon Problem Iron-56 i......
Calculating the Binding Energy per Nucleon
Problem Iron-56 is an extremely stable nuclide. Compute the binding energy per nucleon for ^{56}Fe and compare it with that for ^{12}C (mass of ^{56}Fe atom = 55.934939 amu; mass of ^1H atom = 1.007825 amu; mass of neutron = 1.008665 amu).
Plan Iron-56 has 26 protons and 30 neutrons. We calculate the mass difference, Δm, when the nucleus forms by subtracting the given mass of one ^{56}Fe atom from the sum of the masses of 26 ^1H atoms and 30 neutrons. To find the binding energy per nucleon, we multiply Δm by the equivalent in MeV (931.5 MeV/amu) and divide by the number of nucleons (56).
Solution Calculating the mass difference, Δm:
\text{Mass difference = }[(26 × \text{mass} ^1H\text{ atom}) + (30 × \text{mass neutron})] − \text{mass }^{56}Fe\text{ atom}
= [(26)(1.007825\text{ amu}) + (30)(1.008665\text{ amu})] − 55.934939\text{ amu}
= 0.52846\text{ amu}
Calculating the binding energy per nucleon:
An ^{56}Fe nucleus would require more energy per nucleon to break up into its nucleons than would a ^{12}C nucleus (7.680 MeV/nucleon), so ^{56}Fe is more stable than ^{12}C.
Check The answer is consistent with the great stability of ^{56}Fe. Given the number of decimal places in the values, rounding to check the math is useful only to find a major error. The number of nucleons (56) is exact, so we retain four significant figures.