### Theory:

Atomic mass unit:
Atomic mass unit (u) is defined as $$1/12$$$$^{th}$$ of the mass of a neutral carbon atom $$_{6}C^{12}$$ (carbon – 12 atom). This unit is used to measure the mass of an atom.

Thus, the mass of a carbon atom is $$12\ u$$.

Estimation of mass of helium nucleus:
Consider a helium atom $$_{2}He^{4}$$, which has $$2$$ electrons, $$2$$ protons and $$2$$ neutrons. Helium atom

Mass of a proton $$=\ 1.0078\ u$$

Mass of a neutron $$=\ 1.0087\ u$$

$\begin{array}{l}\mathit{Total}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{nucleus}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\left(2\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}1.0078\right)\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\left(2\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}1.0087\right)\\ \\ =4.033\phantom{\rule{0.147em}{0ex}}u\end{array}$

But experimentally, the actual mass of a helium nucleus is $$4.0026\ u$$. So, what happened to the remaining mass of $$0.0304\ u$$?

To know that, we need to look at the concept of the mass defect.
Mass defect:
The mass of the daughter nucleus formed during a nuclear reaction (fission and fusion) is lesser than the sum of the masses of the two-parent nuclei. This difference in mass is called the mass defect.
In the above example, the mass defect is found to be $$0.0304\ u$$.

Unit of energy:
In nuclear physics, the electron volt ($$eV$$) is the unit used to measure the energy of tiny particles. It is the energy of an electron when it is accelerated by an electric potential of one volt.
$$1\ eV$$ $$=$$ $$1.602 \times{10^{-19}}$$ $$joule$$

$$1\ million\ electron\ volt$$ $$=$$ $$1\ MeV$$ $$=$$ $$10^6\ eV$$ ($$mega\ electron\ volt$$)

Generally, the energy of about $$200\ MeV$$ is released in a nuclear fission process.

Einstein's mass-energy equivalence:
According to the mass-energy equivalence, the mass is converted into energy and vice versa. Albert Einstein proposed the concept of mass-energy equivalence in $$1905$$.

The relation between mass and energy is

$$E = mc^2$$

Where $$E$$ is the energy, $$m$$ is the mass, and $$c$$ is the velocity of light in a vacuum, which is equal to $$3 \times {10^8}\ ms^{-1}$$.

According to Einstein, mass and energy are not independent but are mutually convertible. A body that changes its energy '$$E$$' undergoes a change in its mass, '$$m$$'. For example, if the energy of a particle is increased, its mass also increases.
Binding energy:
Binding energy holds the nucleons (protons and neutrons) of a nucleus together, even from the loss in the total mass of the nucleons (mass defect). It is measured in $$MeV$$.
$\mathit{Binding}\phantom{\rule{0.147em}{0ex}}\mathit{energy}=\mathrm{\Delta }\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}×931.5\phantom{\rule{0.147em}{0ex}}\mathit{MeV}$

Where $$Δm$$ is the mass defect of the nucleus in atomic mass units ‘$$u$$’.

$\begin{array}{l}\mathit{Binding}\phantom{\rule{0.147em}{0ex}}\mathit{energy}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{helium}\phantom{\rule{0.147em}{0ex}}\mathit{nucleus}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}0.0304\phantom{\rule{0.147em}{0ex}}u\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}931.5\phantom{\rule{0.147em}{0ex}}\mathit{MeV}\\ \\ =\phantom{\rule{0.147em}{0ex}}28.32\phantom{\rule{0.147em}{0ex}}\mathit{MeV}\end{array}$

The higher the binding energy per nucleon, the greater is the stability of the atom. Also,

$\mathit{Binding}\phantom{\rule{0.147em}{0ex}}\mathit{energy}\phantom{\rule{0.147em}{0ex}}\mathit{per}\phantom{\rule{0.147em}{0ex}}\mathit{nucleon}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{\mathit{Binding}\phantom{\rule{0.147em}{0ex}}\mathit{energy}}{A}$

Where $$A$$ is the mass number of an element.
Reference:
https://live.staticflickr.com/5005/5351051490_483f3695b0.jpg