We all know that the physical properties and processes are described in terms of quantities such as mass, time, energy, momentum, Area, etc. And all these quantities are measured in units. So, we can say that a physical quantity describes what you are going to measure, whereas a unit describes how much is that quantity.
The international scientific community has agreed to adopt SI units as the standard for scientific communication. But the events occurring at the nuclear level, involve amounts of mass and energy that are very small when described in SI or other conventional units. Therefore they often are described in terms of specially defined units that are more convenient for the nuclear scale.
Unit of Energy
The binding energy of atomic electrons by which they are connected with their nucleus is quite small and it is expressed in electron Volts (\rm eV). But what is an electron volt? And how do you define it? An electronvolt is basically the amount of kinetic energy gained by a single electron when it accelerates in a vacuum through an electric potential difference of one volt.
Its relationship with the SI unit of energy is;
1\rm \ eV = 1.6022 \times 10^{-19} \ J
So, we can say that \rm eV is the unit of energy that can be specially used in atomic physics.
Just like electrons of an atom, nucleons also have binding energy because nucleons are closely packed inside a nucleus due to strong nuclear force.
Now, if you calculate the binding energy of a nucleon inside a nucleus then it comes out to be almost 1 million times the binding energy of an atomic electron, and hence it is expressed in Mega electron Volt. That is \rm MeV. So, \rm MeV is the unit of energy that can be specially used in nuclear Physics.
Unit of mass
If we talk about mass then the basic unit of mass used in nuclear Physics is the unified atomic mass unit. It is abbreviated as \rm u or sometimes \rm amu.
One \rm amu is the one-twelfth of the mass of the unbound C-12 atom at rest and in its ground state. The conversion from SI units to atomic mass units is.
\rm 1 \ u = 1.66054 \times 10^{-27} kg
From Einstein’s mass-energy equivalence, we have E = mc^2 . That is,
m = \dfrac{E}{c^2}
So, if the energy is in \rm MeV, then the unit of mass will be \rm MeV/c^2.
If you convert \rm 1 \ MeV/c^2 into \rm kg then it will be calculated as;
\begin{aligned}\rm 1\ MeV/c^2 &=\dfrac{(1\times 10^6)\times(1.6022 \times 10^{-19}\rm \ J) }{(3 \times 10^8\rm \ m/s)^2}\\&=1.782662 \times 10^{-30}\rm \ kg\\ \end{aligned}
And from here we can calculate as;
{1}\rm \ kg = 5.60959 \times 10^{29}\ MeV/c^2
So,
\begin{aligned}{1}\rm \ u &= 1.66054 \times 10^{-27}\rm \ kg\\&=1.66054 \times 10^{-27}\times (5.60959 × 10^{29}\rm \ MeV/c^2)\\&\approx 931.5\rm \ MeV/c^2\\ \end{aligned}
Mass of an electron
The mass of an electron is;
\begin{aligned}M_e &= 9.10938 \times 10^{-31}\rm \ kg\\&=9.10938 \times 10^{-31}\times (5.60959 × 10^{29}\rm \ MeV/c^2)\\&\approx 0.511 \rm \ MeV/c^2\\ \end{aligned}
Mass of a proton
The mass of a proton is;
\begin{aligned}M_p &=1.672622 \times 10^{-27}\rm \ kg\\&=1.672622 \times 10^{-27}\times (5.60959 × 10^{29}\rm \ MeV/c^2)\\&\approx 938.27 \rm \ MeV/c^2\\ \end{aligned}
Mass of a neutron
And the mass of neutron is;
\begin{aligned}M_n &=1.67493 \times 10^{-27}\rm \ kg\\&=11.67493 \times 10^{-27}\times (5.60959 × 10^{29}\rm \ MeV/c^2)\\&\approx 939.56 \rm \ MeV/c^2\\ \end{aligned}
Unit of length
Now can you compare the size of an atom with the thickness of your hair? Actually, the size of an atom is of the order of 10^{-12}\rm \ m which is around one million times smaller than the thickness of your hair. But the size of a nucleus is quite smaller than the size of an atom. Actually, it’s around ten thousand times smaller. So, nuclear sizes are on the order of femtometer, which in the nuclear context are usually called fermi in honor of Enrico Fermi.
So, we have
\rm 1 \ fermi=1\ fm=10^{-15}\ m
Unit of area
In nuclear physics, the cross-section is a measure of the probability that a specific reaction will take place when a beam of incident particles collides with the target nuclei. The unit of cross-section used in nuclear physics is the barn.
\rm 1 \ barn=10^{-28}\ m^2
Although cross-section is expressed in terms of the area, it is not the same as the geometric cross-sectional area of the target nucleus. It depends on the energy of the beam, target nuclei, the kind of reaction, and many more.
Unit of radioactivity
The number of nuclei that decay per unit time from a radioactive substance is called radioactivity. The unit of radioactivity is disintegration per second but the internationally accepted unit of radioactivity is Becquerel abbreviated as Bq in honor of Henri Becquerel. The other unit of radioactivity that is used worldwide is curie.
\rm 1\ curie=37\times10^{9}\ bq
Unit of spin
In Nuclear Physics, spin is an intrinsic form of angular momentum associated with a subatomic particle or nucleus and measured in \rm h or \rm \hbar.
Unit of the Quadrupole Moment
The electric quadrupole moment is a property of a nucleus in its excited states. It’s a parameter that tells about the structure of the nucleus. Its value depends on the arrangement of nucleons inside a nucleus. That is how the nuclear charge is distributed in a nucleus. The SI unit of electric quadrupole moment is \rm Cm^2. But in nuclear physics, a special unit is used for it and that unit is eb. Where e is an electronic charge which is equal to 1.6\times 10^{-19}\rm\ C and b is barn.
Unit of Planck constant
We know the value of the Planck constant in the SI unit. It is;
h=6.626\times10^{-34}\rm \ Js
Also, we have,
\begin{aligned}\rm 1\ eV &=1.6022 \times 10^{-19}\rm \ J\\ \Rightarrow \rm 1\ J&=6.24\times 10^{18}\rm \ eV\\&=6.24\times 10^{12}\rm \ MeV\\ \end{aligned}
So, if you convert the unit Joule into MeV then the Planck constant will be
\begin{aligned}h &=6.626\times10^{-34}\rm \ J.s\\&=6.626\times10^{-34}\times(6.24\times10^{12}\rm \ MeV).s\\&=4.1356\times10^{-21}\rm \ MeV.s\\ \end{aligned}
Unit of reduced Planck Constant
And the value of reduced Planck constant will be
\begin{aligned}\hbar &=\dfrac{h}{2\pi}\\&=1.05457\times10^{-34}\rm \ J.s\\&=6.5820\times10^{-22}\rm \ MeV.s\\ \end{aligned}
During the solving of nuclear physics problems, you see that many times you need to plug the value of \hbar c. So let’s calculate its value in \rm MeV.fm.
\begin{aligned}\hbar c &=6.5820\times 10^{-22}\rm \ MeV.s \times (3\times 10^8\ m/s)\\&=1.9746\times10^{-13}\rm \ MeV.m\\&=1.9746\times10^{-13}\rm \ MeV.(1\times 10^{15}\ fm)\\&=197.46\rm \ MeV.fm\\ \end{aligned}