Question

a. A${}^{238}\mathrm{U}$nucleus has a radius of $7.4\mathrm{fm}$. What is the density, in $\mathrm{kg}/{\mathrm{m}}^3$, of the nucleus?

b. A neutron star consists almost entirely of neutrons, created when electrons and protons are squeezed together under immense gravitational pressure, and it has the density of an atomic nucleus. What is the radius, in km, of a neutron star with the mass of the sun ?

Short answer

a. The density of the ${}^{238}U$nucleus is $2.33\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3.$

b.$\mathrm{r}=1.26\times {10}^4\mathrm{m}$

Step-by-step solution

Part (a) Step 1: Given Information 

We have given,

Radius of the nucleus = $7.4\mathrm{fm}$

We need to find out the density of the nucleus.

Simplify 

The${}^{238}U$nucleus has localid="1650684314290" $\mathrm{A}=238$nucleons, so its mass is

$\mathrm{m}=\mathrm{A}\mathrm{u}$

where $\mathrm{u}=1.66\times {10}^{-27}\mathrm{kg}$is the atomic unit of mass (approximate mass of each nucleon).

The radius is localid="1650684554678" $r=7.4\mathrm{fm}=7.4\times {10}^{-15}\mathrm{m}$.

Since, the volume of nucleus is

$\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3$

Then, The density is

localid="1650684706642" $$\begin{gathered} \rho =\frac{m}{v}=\frac{A\mathit{}u}{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}^3} \\ \mathrm{\rho }=\frac{3Au}{{\displaystyle 4r^3\mathrm{\pi }}} \end{gathered}$$

Putting the numerical values in this equation then,

localid="1650684745692" $\rho =2.33\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3$

Part (b) Step 1: Given information

We have given,

A neutron star of density similar the density of nucleus.

We have to find the radius of star if it has mass same as sun.

Simplify

We know that the mass of the sun is $1.9891\times {10}^{30}\mathrm{kg}$and the density of the nucleus in the $2.3\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3$.

Then the volume will be,

$$\begin{gathered} \mathrm{V}=\frac{\mathrm{m}}{\mathrm{\rho }} \\ \mathrm{V}=\frac{1.9891\times {10}^{30}\mathrm{kg}}{2.3\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3} \\ \mathrm{V}=8.65\times {10}^{12}{\mathrm{m}}^3 \end{gathered}$$

Then radius will be,

$$\begin{gathered} \mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3 \\ {\mathrm{r}}^3=\frac{3\mathrm{V}}{4\mathrm{\pi }} \\ {\mathrm{r}}^3=\frac{3(8.65\times {10}^{12})}{4(3.14)} \\ {\mathrm{r}}^3=2\times {10}^{12} \\ \mathrm{r}=1.26\times {10}^4\mathrm{m} \end{gathered}$$