Calculate the binding energy of Iron/Fe 56

As we have seen in a previous article, 56Fe is among the nuclides with one of the highest binding energy per nucleon (The highest is 62Ni which was not on the chart, see note below) but How was it determined?

First, from this table of nuclides, we found that the stable isotope of iron (Iron/Fe-56) has 26 protons and 30 neutrons.

Isotopes are atoms of the same element having same atomic number (same number of protons) but different mass number. The atomic number of all the isotopes of iron is 26. But their masses are different. Naturally occurring iron (Fe) consists of 4 isotopes: 5.845% of radioactive 54Fe, 91.754% of stable 56Fe, 2.119% of stable 57Fe and 0.282% of stable 58Fe. Of which, 56Fe is the more common because it is the more common endpoint of fusion chains inside extremely massive stars.

The mass of nuclides are measured with among other methods, mass spectroscopy. Mass spectroscopy is based on producing ions of the substance to analyze, accelerate the charged particles inside a magnetic field with a magnet and detect them with a detector, then the analyzer sort the ions by their mass-to-charge ratio. Of course, isotopes have different masses and because of F=ma, were a is constant inside the magnet, they will follow different trajectories.

The most common unit of mass at the nuclear level is not the kilogram, it is the atomic mass unit or dalton Da (u): defined as 1/12 of the mass of an unbound neutral atom of carbon/C-12 in its nuclear and electronic ground state.

$1u = \frac{1}{12} M(_{6}^{12}\textrm{C})$
$1u = 1.660538921\times10^{-27}kg$
$1u = 931.494061\frac{MeV}{c^{2}}$

Expressing atomic mass unit in terms of electronvolts or eV, requires the use of $E = mc^{2}$.

The mass of Iron/Fe-56 from the table of nuclides(1):
$M(_{26}^{56}\textrm{Fe}) = 55.9349375{\ u}$

NOTE: This value if for the whole atom, not just the nucleus, that is, it also includes the electron masses (and less significantly their binding energies). One can use the mass of hydrogen, which thus includes the mass of one electron, instead of the mass of a proton when calculating the binding energy of the nucleus in order to get the correct result.

The mass of a proton:
$m_p = 1.672621777\times10^{-27}{\ kg} = 1.007276467{\ u}$

The mass of a hydrogen atom:
$M(_{}^{1}\textrm{H}) = 1.007825{\ u}$

The mass of a neutron:
$m_n = 1.674927351\times10^{-27}{\ kg} = 1.008664916{\ u}$

The mass of an electron (it has a mass that is approximately 1/1836 that of the proton):
$m_e = 5.4857990943\times10^{-4}{\ u}$

$_{26}^{56}\textrm{Fe} + {Energy} \rightarrow 26\text{ p}^{+} + 30\text{ n}^{0}$

$26M(_{}^{1}\textrm{H}) + 30m_n = 56.46339748{\ u}$
$\Delta m = (56.46339748u - 55.9349375u) = 0.52845998{\ u}$

Converting this into energy:

$\Delta m c^{2} = 0.52845998\times931.494061\frac{MeV}{\ u} = 492.26\text{\ MeV}$

And thus, the binding energy per nucleon:
$\frac{492.26}{56} = 8.8\frac{MeV}{nucleon}$

NOTES:

* If you use the mass of the proton $m_p$ instead, you get $8.55\frac{MeV}{nucleon}$ which is not the correct value obviously

* The isotope 56Fe is the isotope with the lowest mass per nucleon (M(56Fe) in u/56 = 930.412 MeV/c2), though not the isotope with the highest nuclear binding energy per nucleon, which is nickel-62. However, because of the details of how nucleosynthesis works, 56Fe is a more common endpoint of fusion chains inside extremely massive stars and is therefore more common in the universe, relative to other metals which have a very high binding energy