# How to Calculate Nuclear Density

Nuclear Density = Mass Density of the Nuclear Matter.

Nuclear Density is Constant.

Value of Nuclear Density is $2.3 \times 10^{17} kg / m^3$

This density also occurs in case of Neutron Stars. A neutron star is a compact star and very dense (rather densest) star known in the Universe. They are entirely composed on Neutrons.

Neutrons are sub-atomic particles which have mass comparable to proton but unlike proton they do not have any electric charge.

Mass of neutron = $1.00866 u ~(1.674 927 \times 10^{-27} kg)$

## Why is Nuclear Density Constant?

Nuclear radius (R) is proportional to one-third power of atomic mass number (A)

$R = R_0 A^{1/3}, ~R_o ~=~ 1.2 \times 10^{-15} m \\ \\ \therefore ~ R \propto A^{1/3} \\ \\ \therefore ~ R^3 ~\propto ~A \\ \\ \therefore ~ Volume ~ ({\frac {4} {3}} \pi R^3) ~\propto ~A \\ \\ \therefore ~ Volume ~\propto ~ A . m_n \\ \\ where~( m_n ~=~ mass~of~neutron, ~ 1.00866 u ) \\ \\ \therefore ~Volume \propto Mass ~of ~Nucleus \\ \\ \therefore ~Density ~\rho ~=~ ~{\frac {Mass}{Volume}} ~=~ constant \\ \\ \therefore ~Density ~\rho ~=~ 2.3 \times 10^{17} kg / m^3$

## How to calculate density of nuclear matter?

$Nuclear~Density ~\rho ~=~ ~{\frac {Mass}{Volume}} \\ \\ \\ \rho ~=~ ~\frac {A . m_n}{ {\frac {4} {3}} \pi R_0^3 A} \\ \\ \\ A ~get~cancelled~out.~This~mean~that \\ \\ \\ density~is~independent~of ~atomic~mass~number. \\ \\ \\ \rho ~=~ ~\frac {1.67 \times 10^{-27}}{ {\frac {4} {3}} \pi (1.2 \times 10^{-15})^3} \\ \\ \\ \rho ~=~ 2.3 \times 10^{17} kg / m^3$

Compare this density with the density of water: 1000 kg/m3

Consider a hypothetical case: If the Earth was suddenly to acquire this density, what would be the size of the Earth, if its mass is kept same?

$Mass~of~the~Earth ~=~ 5.972 \times 10^{24} kg \\ \\ \\ Now~if~it~acquires~density~\rho ~=~ 2.3 \times 10^{17} kg / m^3, ~then, \\ \\ \\ Volume ~=~ \frac {Mass}{Density} \\ \\ \\ \therefore V~=~ \frac {5.972 \times 10^{24}}{2.3 \times 10^{17}} \\ \\ \\ \therefore V~=~ 2.5965 \times 10^{7} \\ \\ \\ But~V~=~ {\frac {4} {3}} \pi R^3 \\ \\ \therefore R~=~ \root 3 \of {\frac {3 V}{4 \pi}} \\ \\ \\ \therefore R~=~ \root 3 \of {\frac {3 \times 2.5965 \times 10^{7}}{4 \pi}} \\ \\ \\ \therefore R~=~ 183.7 m$

The Earth will squeeze to a sphere of just 184 m, were it to acquire nuclear density!

## Structure of a Nucleus

An atomic nucleus (plural nuclei) is a compact collection of nucleons. A nucleon may be a proton or a neutron. Nucleus is very small in comparison to the electron orbits and atomic sizes. The radius of electron orbit is approximately 10,000 more than that of a nucleus. Nucleus is very small. If nucleus is 1 cm in diameter, then, an electron’s orbit is 100 m in diameter. So practically an atom is just an “empty space”

However, nucleus hold 99.9% of the mass of the atom. They hold together positively charged protons and neutral neutrons. One may ask how nucleus holds protons together without they repelling each other? The answer is the Nuclear-Force.

## What is the Nuclear Force?

Nuclear force is another kind of force, which is different from Gravitational and Coulomb forces.

• Nuclear force is much stronger than Coulomb force, which in-turn is much stronger than gravitational force.
• The range of nuclear force is of the order of femtometer (fm), which is much smaller than range of coulomb force, which in-turn is, again, much smaller than the range of gravitation force.
• Nuclear force is independent of the charge.
• Potential energy due to this nuclear force is at a minimum (potential well) at range r0 = 0.8 fm, above which it increases and saturates and becomes attractive. Below this r0, it rapidly increases and becomes repulsive.