Chapter 21, Problem 13Q

(a)

To determine

The density of matter in a neutron and compare it with the average density of a neutron star. It is given that neutron has a mass of $1.7\times {10}^{-27}\text{ kg}$ and a radius of about ${10}^{-15}\text{ m}$.

Answer to Problem 13Q

Solution:

The density of neutron is $4.059\times {10}^{17}{\text{ kg/m}}^3$, which is lesser than the average density of neutron star.

Explanation of Solution

Given data:

The mass of a neutron is $1.7\times {10}^{-27}\text{ kg}$ and it has a radius of about ${10}^{-15}\text{ m}$.

Formula used:

Write the expression for density:

$\rho =\frac{M}{V}$

Here, $\rho$ is the density, $M$ is the mass, and $V$ is the volume.

Write the expression for volume of a sphere:

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

Here, $V^{\prime }$ is the volume of a sphere and $r$ is the radius of sphere.

Explanation:

Let’s assume that neutron is a sphere of radius $r$.

According to question, mass of neutron is $1.7\times {10}^{-27}\text{ kg}$ and radius is about ${10}^{-15}\text{ m}$.

Calculate the volume of a sphere.

Refer to the expression for volume of sphere,

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

Substitute ${10}^{-15}\text{ m}$ for $r$,

$\begin{array}{c}{V}^{\prime }=\frac{4}{3}\pi {\left({10}^{-15}\text{ m}\right)}^3\\ =4.188\times {10}^{-45}{\text{ m}}^3\end{array}$

Calculate density of neutron.

Refer to the expression of expression for density of a neutron star,

$\rho =\frac{M}{V}$

Substitute $4.188\times {10}^{-45}{\text{ m}}^3$ for $V$ and $1.7\times {10}^{-27}\text{ kg}$ for $M$:

$\begin{array}{c}\rho =\frac{\left(1.7\times {10}^{-27}\text{ kg}\right)}{\left(4.188\times {10}^{-45}{\text{ m}}^3\right)}\\ =4.059\times {10}^{17}{\text{ kg/m}}^3\end{array}$

Now, compare the density of matter in a neutron with average density of a neutron star.

As average density of neutron in neutron star is $7.3\times {10}^{17}{\text{ kg/m}}^3$ that is greater than calculated density of neutron.

Conclusion:

Hence, neutron has density of $4.059\times {10}^{17}{\text{ kg/m}}^3$, which is lesser than the density of a neutron star.

(b)

To determine

Neutrons within a neutron star are overlapping if density of neutron star is more than that of neutron, or are underlapping otherwise. Also, explain whether density at the center of a neutron star is higher than the average density of neutron star. It is given that neutron has mass of $1.7\times {10}^{-27}\text{ kg}$ and a radius of about ${10}^{-15}\text{ m}$.

Answer to Problem 13Q

Solution:

As the density of neutron is less than the average density of a neutron star, therefore, neutrons in the neutron star are in overlapping state and density is uniform throughout the neutron star.

Explanation of Solution

Given data:

The mass of neutron is $1.7\times {10}^{-27}\text{ kg}$ and radius is about ${10}^{-15}\text{ m}$.

Introduction:

A neutron star is composed of an incredibly dense sphere of neutrons formed by a supernova explosion. Its average density is about $7.3\times {10}^{17}{\text{ kg/m}}^3$.

Explanation:

Refer to part (a) of the question, the average density of neutron in a neutron star is $7.3\times {10}^{17}{\text{ kg/m}}^3$ which is greater than the calculated density of neutron, that is, $4.059\times {10}^{17}{\text{ kg/m}}^3$.

From the above observation, neutrons in the neutron star are overlapping as density of a neutron is less than the average density of a neutron star.

Also, in a neutron star, density is uniform throughout. In other words, density at the center is same as density at the surface of neutron star.

Conclusion:

Hence, neutrons in the neutron star are in overlapping state.