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What is the order of density of neutrons?
A) $\text{1}{{\text{0}}^{3}}\text{ kg/cc}$
B) $\text{1}{{\text{0}}^{6}}\text{ kg/cc}$
C) $\text{1}{{\text{0}}^{\text{12}}}\text{ kg/cc}$
D) $\text{1}{{\text{0}}^{\text{2}}}\text{ kg/cc}$

Last updated date: 25th Jun 2024
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Hint: The atomic nucleus is made of proton and neutron. Neutrons are the subatomic particles that are held together by nucleus forces in the nucleus. Neutron is considered as the perfectly spherical sum of atomic particles. We are interested in determining the neutron density. Density is defined as the ratio of mass to volume. It is written as,
$\text{ Density(d) = }\dfrac{\text{mass(m)}}{\text{Volume(V)}}\text{ }$
Individual or isolated neutron has a mass equal to $\text{ 9}\text{.39}\times \text{1}{{\text{0}}^{\text{8}}}\text{ eV/}{{\text{c}}^{\text{2}}}\text{ }$ or $\text{ 1}\text{.67 }\times \text{1}{{\text{0}}^{-27}}\text{ kg }$ and the mean radius of the neutron is $\text{ 1}\times \text{1}{{\text{0}}^{-13\text{ }}}cm\text{ }$ or $\text{ 1}\times \text{1}{{\text{0}}^{-15\text{ }}}m\text{ }$.

Complete Solution :
Neutrons are considered as perfectly sphere-shaped sub-particles. The volume of the sphere is given as follows,
$\text{ Volume of sphere = }\dfrac{4}{3}\text{ }\times \text{ }\pi {{\text{r}}^{\text{3}}}\text{ }$
Where r is the radius of the sphere.
- We are interested in determining the volume of the neutron. We know that the radius on neutron is $\text{ 1}\times \text{1}{{\text{0}}^{-13\text{ }}}cm\text{ }$. Let’s substitute the values of the radius in the volume relation to determine the volume of neutrons. It is calculated as follows,
  & \text{ Volume of neutron = }\dfrac{4}{3}\text{ }\times \text{ }\pi \times {{\left( 1.0\times {{10}^{-13}} \right)}^{\text{3}}}\text{ } \\
 & \Rightarrow \text{Volume of neutron = }\dfrac{4}{3}\text{ }\times \text{ }\pi \times {{10}^{-39}}\text{ } \\

- We want to determine the density of neutrons in the nucleus. Density is defined as the ratio of mass to volume. In other words, it is the amount of mass occupied by the matter in the given space. Mathematically density is written as follows,
$\text{ Density(d) = }\dfrac{\text{mass(m)}}{\text{Volume(V)}}\text{ }$
We know that mass of a neutron is $\text{ m = 1}\text{.6}\times \text{1}{{\text{0}}^{-27}}\text{ kg }$

Let's substitute the values of mass and volume in the density relation. the density of neutron can be calculated as:
$\text{ Density of neutron = }\dfrac{\text{mass}}{\text{Volume}}\text{ =}\dfrac{\text{1}\text{.6}\times \text{1}{{\text{0}}^{-27}}\text{ kg }}{\dfrac{4}{3}\text{ }\times \text{ }\pi \times {{10}^{-39}}\text{ }}\text{ }$

- On further simplification we have
$\text{ Density of neutron =}\dfrac{\text{1}\text{.6}\times \text{1}{{\text{0}}^{-27}}\text{ }}{\dfrac{4}{3}\text{ }\times \text{ }\pi \times {{10}^{-39}}\text{ }}\text{ = }\dfrac{\text{3}\times \text{1}\text{.6}\times \text{1}{{\text{0}}^{-27}}\text{ }}{\text{4 }\times \text{ }\pi \times {{10}^{-39}}\text{ }}\text{ = }\dfrac{4.8}{\text{4 }\times \text{ }\pi }\text{ = 1}{{\text{0}}^{\text{12}}}\text{ kg/cc }$
Therefore the order of density of neutrons is $\text{1}{{\text{0}}^{\text{12}}}\text{ kg/cc}$.
So, the correct answer is “Option C”.

Note: Note that, the neutron is a charge less subatomic has a mass, it occupies volume, and thus has a definite only contributes towards the mass of the nucleus but does not towards the charge. It remains affected by external fields. However, protons have a positive charge and mass. We are considered that neutron is a sphere to the simplified problems but the atom is a diffused mass.