Chapter 8, Problem 20Q

(a)

To determine

The diameter of the disk as shown in the figure below using the scale bar provided below it.

Answer to Problem 20Q

Solution:

The diameter is, approximately, $\text{560 au}$.

Explanation of Solution

This can be calculated directly by calibrating our usual scale with the scale bar provided below the figure and measuring the diameter of the disk. This diameter comes to be around $\text{560 au}$.

Conclusion:

Hence, the diameter of the disk is $\text{560 au }$, approximately.

(b)

To determine

The volume of the disk if its thickness is $50\text{ au}$.

Answer to Problem 20Q

Solution:

Volume of the disk is $4.15\times {10}^{40}{\text{ m}}^3$.

Explanation of Solution

Given data:

The thickness of the disk is $50\text{ au}$ and diameter obtained above is $\text{560 au}$.

Formula used:

The expression for volume of the disk is written as,

$V=\pi r^{2}t$

Here, $V$, $r$ and $t$ are the volume, radius and thickness, respectively.

Use the conversion formula, $1{\text{ au}}^3=3.348\times {10}^{33}{\text{ m}}^3$.

Explanation:

The expression for the radius of the disk (r) is written as,

$r=\frac{d}{2}$

Here, d is the diameter of disk.

Substitute $\text{560 au}$ for d.

$\begin{array}{c}r=\frac{560\text{ au}}{2}\\ =280\text{ au}\end{array}$

Recall the expression for the volume of the disk.

$V=\pi r^{2}t$

Substitute $280\text{ au}$ for r and 50 au for t.

$\begin{array}{c}V=\pi \times {(}^{280}\times 50\text{ au}\\ =\left(1.23\times {10}^{40}{\text{ au}}^3\right)\left(\frac{3.348\times {10}^{33}{\text{ m}}^3}{1{\text{ au}}^3}\right)\\ =4.15\times {10}^{40}{\text{ m}}^3\end{array}$

Conclusion:

Thus, the volume of the disk is $4.15\times {10}^{40}{\text{ m}}^3$.

(c)

To determine

The number of hydrogen atoms and hydrogen atoms per cubic meter of the disk if the total mass of the disk is $2\times {10}^{28}\text{ kg}$ and the mass of a single hydrogen atom is $1.673\times {10}^{-27}\text{ kg}$.

Answer to Problem 20Q

Solution:

The total number of hydrogen atoms in the disk is $1.19\times {10}^{55}$

Explanation of Solution

Given data:

The thickness of the disk is $50\text{ au}$ and its total mass is $2\times {10}^{28}\text{ kg}$. It is composed only of hydrogen. Mass of a single hydrogen atom is $1.673\times {10}^{-27}\text{ kg}$.

Formula used:

The expression for the total number of hydrogen atoms $(N)$ is written as,

$N=\frac{M}{m}$

Here, $M\text{ and }m$ are the masses of the disk and the hydrogen atom, respectively.

Explanation:

If the total mass of the disk and the mass of each hydrogen atom are given, provided that the disk is composed entirely of hydrogen atoms, the number of hydrogen atoms is:

$N=\frac{M}{m}$

Substitute $2\times {10}^{28}$ for $M$ and $1.673\times {10}^{-27}$ for m.

$\begin{array}{c}N=\frac{2\times {10}^{28}\text{ kg}}{1.673\times {10}^{-27}\text{ kg}}\\ =1.19\times {10}^{55}\end{array}$

Conclusion:

Thus, the total number of hydrogen atoms in the entire disk is $1.19\times {10}^{55}$.

(d)

To determine

The number of hydrogen atoms per cubic meter of the disk. Also compare it with the density of air, which contains about $5.4\times {10}^{25}$ atoms per cubic meter.

Answer to Problem 20Q

Solution:

Number of hydrogen atoms per cubic meter is $1.28\times {10}^{14}$.

Explanation of Solution

Given data:

The thickness of the disk is $50\text{ au}$, total mass is $2\times {10}^{28}\text{ kg}$ and it is composed only of hydrogen. Mass of a single hydrogen atom is $1.673\times {10}^{-27}\text{ kg}$. Air contains about $5.4\times {10}^{25}$ atoms per cubic meter.

Formula used:

The number of atoms per cubic meter $(n)$ is written as,

$n=\frac{N}{V}$

Here, $V$ is volume of the disk and $N$ is total number of hydrogen atoms in the disk.

Explanation:

From part (c), total number of hydrogen atoms in the disk (N) is $1.19\times {10}^{55}$.

From part (b), volume of the disk is $4.15\times {10}^{40}{\text{ m}}^3$.

Recall the expression for number of atoms per cubic meter $(n)$.

$n=\frac{N}{V}$

Substitute $1.19\times {10}^{55}$ for Nand $4.15\times {10}^{40}{\text{ m}}^3$ for V.

$\begin{array}{c}n=\frac{1.19\times {10}^{55}\text{ atoms }}{4.15\times {10}^{40}{\text{ m}}^3}\\ =2.9\times {10}^{14}\frac{\text{atoms}}{{\text{m}}^3}\end{array}$

Thus, the disk has a density much less than that of the air we breathe, which is $5.4\times {10}^{25}$ atoms per cubic meter.

Conclusion:

Thus, the number of hydrogen atoms per cubic meter is $2.9\times {10}^{14}$.