Question

What is the order of magnitude of the density of nuclear matter?
\[{\text{A}}{\text{. 1}}{{\text{0}}^4}\] kg/${{\text{m}}^3}$
\[{\text{B}}{\text{. }}{10^{17}}\] kg/${{\text{m}}^3}$
\[{\text{C}}{\text{. }}{10^{27}}\] kg/${{\text{m}}^3}$
\[{\text{D}}{\text{. }}{10^{34}}\] kg/${{\text{m}}^3}$

Answer 1

Hint: Here, we will proceed by defining the term ‘Density of nuclear matter’ and then we will assume an isotope of uranium i.e., 238U and we will find the nuclear radii and hence volume of its nucleus along with its mass inside the nucleus.

Formulas used: ${\text{r}} = {{\text{r}}_0}{\left( {\text{A}} \right)^{\dfrac{1}{3}}}$, ${\text{V}} = \dfrac{4}{3}\pi {{\text{r}}^3}$ and Nuclear density = $\dfrac{{{\text{Mass inside nucleus}}}}{{{\text{Volume inside nucleus}}}}$.

Complete Step-by-Step solution:
Nuclear matter is an idealised version of a system interacting nucleons (protons and neutrons) that exists in multiple phases that are not yet fully established. Within a nucleus, it's not matter, just a hypothetical material composed of a large number of protons and neutrons interacting only by nuclear forces and not by Coulomb forces. No surface effects and translational invariance (only differences in position matter, not absolute positions) implies infinite volume.
Nuclear Density is also called density of nuclear matter. Nuclear density represents the nucleus density of an atom. It is the ratio of mass to the volume which is there inside the nucleus. Since the atomic nucleus holds much of the mass of the atom, and the atomic nucleus is very small relative to the entire atom, the nuclear density is very high.
In a standard nucleus, the nuclear density can be determined roughly from the nucleus size and its mass. Nuclear radii is basically of the order ${10^{ - 14}}$ m.
Assuming spherical shape, nuclear radii can be calculated according to the following formula given under.
${\text{r}} = {{\text{r}}_0}{\left( {\text{A}} \right)^{\dfrac{1}{3}}}$ where A is the mass number and ${{\text{r}}_0} = 1.2{\text{ fm }} = 1.2 \times {10^{ - 15}}$ m
For example, natural uranium consists primarily of isotope 238U (99.28%), therefore the atomic mass of the uranium element is close to the atomic mass of 238U isotope (238.03u).
Nuclear radii of this isotope of uranium can be calculated using the formula ${\text{r}} = {{\text{r}}_0}{\left( {\text{A}} \right)^{\dfrac{1}{3}}}$,
${\text{r}} = \left( {1.2 \times {{10}^{ - 15}}} \right){\left( {{\text{238}}} \right)^{\dfrac{1}{3}}} = \left( {1.2 \times {{10}^{ - 15}}} \right)\left( {{\text{6}}{\text{.197}}} \right) = 7.44 \times {10^{ - 15}}$ m
Assuming it is spherical in shape, its volume will be given by ${\text{V}} = \dfrac{4}{3}\pi {{\text{r}}^3}$
Volume inside the nucleus, ${\text{V}} = \dfrac{4}{3}\left( {\dfrac{{22}}{7}} \right){\left( {7.44 \times {{10}^{ - 15}}} \right)^3} = 1.726 \times {10^{ - 42}}{\text{ }}{{\text{m}}^3}$
Mass inside the nucleus of the uranium (M) can be obtained by multiplying mass number (A) with Avogadro’s number ($1.66 \times {10^{ - 27}}$) as given under.
Mass inside nucleus, M = A$ \times 1.66 \times {10^{ - 27}} = 238 \times 1.66 \times {10^{ - 27}} = 3.95 \times {10^{ - 25}}$
Since, Nuclear density = $\dfrac{{{\text{Mass inside nucleus}}}}{{{\text{Volume inside nucleus}}}} = \dfrac{{\text{M}}}{{\text{V}}} = \dfrac{{3.95 \times {{10}^{ - 25}}}}{{1.726 \times {{10}^{ - 42}}}} = 2.288 \times {10^{17}}$ kg/${{\text{m}}^3}$
Clearly, the order of magnitude of density of nuclear matter (or nuclear density) is ${10^{17}}$ kg/${{\text{m}}^3}$.
Hence, option B is correct.

Note- The cumulative number of protons and neutrons in an atomic nucleus is the mass number, also called atomic mass number or nucleon number. Isotopes are variants of a specific chemical element that differ in the number of neutrons and therefore in the number of nucleons.