# The Mass Defect

*Mass defect is the difference between the mass of a composite particle and the sum of the masses of its parts*. By plugging this mass into $E=mc^2,$ Einstein's famous equation to find the binding energy.

*Nuclear binding energy is the minimum amount of energy required to disassemble the nucleus of an atom*. It can also be defined as the energy holding the nucleons(protons and neutrons) of an atom together in the nucleus.*The nuclear binding energy is also known as the strong nuclear force.*
*This disproves the law of conservation of mass. However, this is very negligible and the law of conservation of mass is a great approximation.*

If binding energy gets freed when two light nuclei fuse together or when a heavy nucleus splits into two lighter nuclei the energy will be released.

**Nuclear Binding Energy Curve**:

*In the periodic table, elements from hydrogen up to sodium show an increase in binding energy(BE). This region of increasing binding energy is followed by a stable BE from magnesium to xenon. In this region, the nucleus has become large enough so that nuclear forces do not extend their reach sufficiently across the nucleus. Attractive nuclear forces are nearly balanced by the repulsion between the protons. In elements heavier than xenon there is a fall in BE. This happens because as the atomic number grows, more protons are present in the nucleus, thus creating a repulsive force negating the BE.
At the curves peak there is nickel 62, having the greatest BE, followed by iron 58 and iron 56.*

**End products in Supernovae**:
*Iron and nickel are very common metals in planetary cores as they are the end products of supernovae. The conclusion is that at the high pressures and temperatures in stars, energy is released by converting all matter to Fe 56.*

**Mass Energy Equivalence Formula**:
$\bigtriangleup$E=$\bigtriangleup$m$c^2$.

Where$\bigtriangleup$E:energy release
and$\bigtriangleup$m:mass defect

**Decrease in Binding Energy due to Increase in Distance**:
The strong nuclear force is very powerful at 1 femtometer, repulsive below 0.7 femtometers and above 2.5 femtometers it is considered insignificant.
(Note:1femtometer=1$\times$$10^{-15}$)

Case Study:The mass of a helium atom can be calculated using the knowledge we have about the mass of protons and neutrons. We also know that the in a Helium 4 atom has 2 protons and 2 neutrons (electrons will not be considered as they have negligible mass). The mass of a proton will be taken as 1.00727647 amu, while the mass of a neutron will be taken as 1.00866490. The total mass of the protons will be 1.00727647 $\times$ 2 = 2.01455294 amu. The total mass of the neutrons will be taken as 1.00866490 $\times$2 = 2.0173298. Therefore, the total mass of the helium 4 atom should be 2.01455294 + 2.0173298 = 4.03188274.However, the observed mass of the helium 4 atom is actually 4.00150608

The difference between the observed and the predicted mass is 0.03037666. This is the mass defect. If you plug this mass into Einstein's famous equation, $E=mc^2$, you will get 4.53346$\times$$10^{-12}$ Joules.

Therefore the nuclear binding force is 4.53346 $\times$ $10^{-12}$ joules.