Chapter 15, Problem 1PQ

(a)

To determine

The density of an alpha particle’s nucleus.

Answer to Problem 1PQ

The density of an alpha particle’s nucleus is $\underset{\_}{4\times {10}^{17}{\text{kg/m}}^3}$.

Explanation of Solution

Given that the radius of the constituent particle of alpha particle is $1.00\text{fm}$.

Since nuclear density is uniform, the density of alpha particle’s nucleus is the same as the density of its constituent particles.

Write the expression for density.

$\rho =\frac{m}{V}$        (I)

Here, $\rho$ is the density, $m$ is the mas, and $V$ is the volume.

Assuming the particles to be of spherical shape, their the volume is given by the expression,

$V=\frac{4}{3}\pi r^3$        (II)

Here, $r$ is the radius.

Use equation (II) in (I).

$\begin{array}{c}\rho =\frac{m}{\frac{4}{3}\pi r^3}\\ =\frac{3m}{4\pi r^3}\end{array}$        (III)

Conclusion:

Substitute $1.67\times {10}^{-27}\text{kg}$ for $m$, and $1.00\text{fm}$ for $r$ in equation (III) to find $\rho$.

$\begin{array}{c}\rho =\frac{3\left(1.67\times {10}^{-27}\text{kg}\right)}{4\pi {\left(1.00\text{fm}\right)}^3}\\ =\frac{3\left(1.67\times {10}^{-27}\text{kg}\right)}{4\pi {\left(1.00\text{fm}\times \frac{1\text{m}}{1\times {10}^{15}\text{fm}}\right)}^3}\\ =4\times {10}^{17}{\text{kg/m}}^3\end{array}$

Therefore, the density of an alpha particle’s nucleus is $\underset{\_}{4\times {10}^{17}{\text{kg/m}}^3}$.

(b)

To determine

The comparison of the density of the nucleus of an alpha particle with the density of the osmium and what this comparison say about the space between nuclei in liquids and solids.

Answer to Problem 1PQ

The density of the nucleus is about ${10}^{13}$ times denser than the density of osmium and this says that an atom is mostly empty space. The space between the nuclei in liquids and solids are mostly empty space as well as gases.

Explanation of Solution

It is obtained that the density of alpha particle’s nucleus is $4\times {10}^{17}{\text{kg/m}}^3$. The density of osmium is $2.26\times {10}^4{\text{kg/m}}^3$.

Compare the values of the densities of nucleus and osmium.

$\begin{array}{c}\frac{{\rho }_{\alpha }}{{\rho }_{\text{Os}}}=\frac{4\times {10}^{17}{\text{kg/m}}^3}{2.26\times {10}^4{\text{kg/m}}^3}\\ =1.76\times {10}^{13}\\ \approx {10}^{13}\end{array}$

Conclusion:

Therefore, the density of the nucleus is about ${10}^{13}$ times denser than the density of osmium and this says that an atom is mostly empty space. The space between the nuclei in liquids and solids are mostly empty space as well as gases.