Chapter 30, Problem 20E

(a)

To determine

The density of oxygen nucleus.

Answer to Problem 20E

The density of oxygen nucleus is $2.32\times {10}^{14}\text{kg}\cdot {\text{m}}^{-3}$.

Explanation of Solution

Write the expression for the nuclear density of an atom.

    $n=\frac{A}{\frac{4}{3}\pi r^3}$        (I)

Here, $n$ is the nuclear density, $A$ is the mass number and $r$ is the radius of the nucleus.

Write the expression for the radius of a nucleus.

    $r=1.4A^{1/3}\text{fm}$        (II)

Write the expression for the nuclear density.

    $d=mNn$        (III)

Here, $d$ is the nuclear density, $m$ is the mass of each nucleon and $N$ is the total number of nucleon present within nucleus.

Conclusion:

Substitute $1.4A^{1/3}\text{fm}$ for $r$ in expression (I).

    $\begin{array}{c}n=\frac{A}{\frac{4}{3}\pi {\left(1.4A^{1/3}\text{fm}\right)}^3}\\ =\frac{3}{4\pi {\left(1.4\text{fm}\right)}^3}\\ =\frac{3}{4\pi {\left(1.4\text{fm}\right)}^3}{\left(\frac{1\text{fm}}{{10}^{-15}\text{m}}\right)}^3\\ =8.7\times {10}^{43}{\text{m}}^{-3}\end{array}$

Substitute $16$ for $N$ of oxygen, $1.67\times {10}^{-31}\text{kg}$ for $m$ and $8.7\times {10}^{43}{\text{m}}^{-3}$ for $n$ in expression (III).

    $\begin{array}{c}d=16\left(1.67\times {10}^{-31}\text{kg}\right)\left(8.7\times {10}^{43}{\text{m}}^{-3}\right)\\ =2.32\times {10}^{14}\text{kg}\cdot {\text{m}}^{-3}\end{array}$

Thus, the nuclear density of oxygen is $2.32\times {10}^{14}\text{kg}\cdot {\text{m}}^{-3}$.

(b)

To determine

The ratio of the oxygen nucleus density to the density of water.

Answer to Problem 20E

The ratio of oxygen nucleus density to the density of water is $2.32\times {10}^{11}$.

Explanation of Solution

The density of the water is ${10}^3\text{kg}\cdot {\text{m}}^{-3}$.

Write the expression for the ratio of two densities.

    $r=\frac{d}{{10}^3\text{kg}\cdot {\text{m}}^{-3}}$        (IV)

Here, $d$ is the density of the oxygen nucleus and $r$ is the ratio of the densities.

Conclusion:

Substitute $2.32\times {10}^{14}\text{kg}\cdot {\text{m}}^{-3}$ for $d$ in expression (IV).

    $\begin{array}{c}r=\frac{2.32\times {10}^{14}\text{kg}\cdot {\text{m}}^{-3}}{{10}^3\text{kg}\cdot {\text{m}}^{-3}}\\ =2.32\times {10}^{11}\end{array}$

Thus, the ratio of oxygen nucleus density to the density of water is $2.32\times {10}^{11}$.