Chapter 2, Problem 46E

A proton has a radius of approximately $\text{1}\times {\text{10}}^{\text{-13}}$ cm and a mass of $1.7\times {10}^{-24}$ g.

a. Determine the density of a proton. For a sphere, $V=(4/3)\pi r^3;{\text{ 1 cm}}^3=1\text{ mL; }\pi \text{=3}\text{.14}$.

b. By assuming that black holes have the same density as protons, determine the mass (in kilograms) of a black hole and the size of a sand particle. A typical sand particle has a radius of $1\times {10}^{-4}$ m.

Interpretation Introduction

Interpretation:

The density of a proton and the mass of a black hole is to be calculated.

Concept Introduction:

Density of a substance is a measure of compactness of that substance. It is defined as the mass of a substance present in per unit of its volume. The unit of density is derived when the unit of mass is divided by the unit of volume. Its most common units are ${\text{g/cm}}^{\text{3}}\text{ and g/mL}\text{.}$

$\text{Density = }\frac{\text{Mass }}{\text{Volume}}$

${\text{As 10}}^3\text{g = 1 kg}$, conversion factor for g into kg is as:

$\frac{\text{1 kg}}{{10}^3\text{g}}$

Answer to Problem 46E

Solution: a) $\text{4}\times {\text{10}}^{14}{\text{ g/cm}}^{\text{3}}.$ b) $\text{1}\text{.6}\times {\text{10}}^6\text{ kg}$.

Explanation of Solution

a) The density of a proton having radius of approximately $1\times {10}^{-13}\text{cm}$ and a mass $1.7\times {10}^{-24}\text{g}$.

Calculation of volume of a proton is as:

$\text{Volume of a proton = }\frac{4}{3}\pi {\text{r}}^3$

Substitute the given values in the above equation as:

$\begin{array}{l}\text{Volume of a proton = }\frac{4}{3}\times \text{3}\text{.14 }\times {(}^1\\ \text{Volume of a proton = 4}\times {10}^{-39}{\text{cm}}^3\end{array}$

Therefore, the volume of a proton is $\text{4}\times {10}^{-39}{\text{cm}}^3$.

Calculation of density of a proton is as follows:

$\text{Density of proton = }\frac{\text{Mass of proton}}{\text{Volume of proton}}$

Substitute the given values in the above equation as:

$\begin{array}{l}\text{Density of a proton = }\frac{\text{1}\text{.7}\times {\text{10}}^{-24}\text{ g}}{4\times {\text{10}}^{-39}{\text{ cm}}^{\text{3}}}\\ \text{Density of a proton = 4}\times {\text{10}}^{14}{\text{ g/cm}}^{\text{3}}.\end{array}$

Thus, the density of a proton is $\text{4}\times {\text{10}}^{14}{\text{ g/cm}}^{\text{3}}.$

b) The mass of a black hole and the size of a sand particle, radius of sand particle is $1\times {10}^{-4}\text{m}$.

The volume of a black hole is as follows:

$\begin{array}{l}\text{Radius of the black hole = }1\times {10}^{-4}\text{m}\\ \text{1m = 100 cm}\end{array}$

So,

$\begin{array}{l}\text{Radius of the black hole = }1\times {10}^{-4}\cancel{\text{m}}\times \frac{\text{100 cm}}{\text{1 }\cancel{\text{m}}}\\ \text{Radius of the black hole = }1\times {10}^{-2}\text{ cm}\end{array}$

$\text{Volume of the black hole = }\frac{4}{3}\pi {\text{r}}^3$

Substitute the given values in the above equation as:

$\begin{array}{l}\text{Volume of the black hole = }\frac{4}{3}\times \text{3}\text{.14 }\times {(}^1\\ \text{Volume of the black hole = 4}\times {10}^{-6}{\text{cm}}^3\end{array}$

Thus, the volume of the black hole is $\text{4}\times {10}^{-6}{\text{cm}}^3$.

The calculation of mass of the black hole is as follows:

$\begin{array}{l}\text{Density of the black hole = }\frac{\text{Mass of the black hole}}{\text{Volume of the black hole}}\\ \text{Mass of the black hole}=\text{Density of the black hole}\times \text{Volume of the black hole }\end{array}$

Substitute the values in the above equation as:

$\begin{array}{l}\text{ Mass of the black hole = 4}\times {\text{10}}^{-6}\cancel{{\text{cm}}^{\text{3}}}\times \frac{4\times {\text{10}}^{14}\text{ g}}{\cancel{1{\text{ cm}}^3}}\\ \text{Mass of the black hole = 1}\text{.6}\times {\text{10}}^9\text{ g}.\end{array}$

Conversion of grams into kilograms is as follows:

$\begin{array}{l}{\text{As 10}}^3\text{g = 1 kg}\\ \text{Mass of the black hole = 1}\text{.6}\times {\text{10}}^9\cancel{g}\times \frac{1\text{ kg}}{{\text{10}}^3\cancel{g}}\\ =\text{1}\text{.6}\times {\text{10}}^6\text{ kg}\end{array}$

Hence, the mass of the black hole is $\text{1}\text{.6}\times {\text{10}}^6\text{ kg}$.