Density of Black Holes and Atomic Nuclei - ChemPRIME

Density of Black Holes and Atomic Nuclei

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Although we know that atoms are mostly space, sometimes the significance of this statement is

underappreciated. Earlier we showed how unity factors can be used to express quantities in different units of the same parameter. For example, a density can be expressed in g/cm3 for ordinary objects or Pg/pm3 (petagrams (1015 g) per cubic picometer (10-12 m)). We might get more insight into the structure of an atom by exploring how conversion factors representing mathematical functions, like D = m/v, can be used to transform quantities into different parameters. For example, what is the volume of a black hole with a given mass, or what would the masses of common objects be if they were solid nuclear matter with the same density as a proton? Unity factors and conversion factors are conceptually different, and we'll see that the "dimensional analysis" we develop for unit conversion problems must be used with care in the case of functions.

When we are referring to the same object or sample of material, it is often useful to be able to convert one parameter into another. Conversion of one kind of quantity into another is usually done with what can be called a conversion factor, but the conversion factor is based on a mathematical function (D = m / V) or mathematical equation that relates parameters.

For example, suppose aluminum atoms were made of solid nuclear material with the same density as protons and neutrons. What would the mass of an atom be, and what would the mass of a 72.9 cm3 piece of aluminum which we know contains 6.02 x 1023 atoms (and weighs 27 g)? What would the volume of the Earth be if it were solid protons?

Although protons are not spherical because they're composed of three quarks, let's suppose that they can be approximated by a sphere of radius 1 × 10-13 cm, predicted by the general formula for estimating the radius of a nucleus: rnucleus = (1.2 × 10-13 m) A1/3, where A is the mass number of the atom.

Since the mass of a proton is about 1 amu, or 1.67 × 10-27 kg, its density is about the same as densities that may be found in some black holes:

\rho \text{ = }\frac{\text{m}}{\text{V}} = \frac {1.67 x 10^{-24} g}{4 x 10^{-39} cm^3} = 4 x 10^{14} \frac {g}{cm^3}

The radius of an atom can be determined by X-ray diffraction, and is about 143 pm (1.43 x 10-8 cm), so the volume is  V = \frac {4}{3} \pi r^3 = 1.2 x 10-23 cm3

We can estimate the mass of atoms made of solid nuclear material of this density by manipulating Eq. (1.1). If we multiply both sides by V, we obtain

\text{V}\times \rho =\frac{\text{m}}{\text{V}}\times \text{V}=\text{m}

m = V × ρ or mass = volume × density       (1.2)

So the atom would weigh

m = Vρ = = 1.2 x 10-23 cm3 × 4 x 1014 \frac {g}{cm^3} = 4.9 x 10-9 g.

If the 72.9 cm3 piece of aluminum were made of these atoms, it would have a mass of

m = Vρ = 72.9 cm3 × 4 x 1014 \frac {g}{cm^3} = 2.9 x 1016 g, which is 3.2 x 1010 tons.

It's clear that aluminum isn't as "solid" as it looks, if a 27 g piece of aluminum would weigh over 1010 tons if it were solid nuclear matter. The atoms are only about 10-12% solid matter in the nucleus, with very light electons in orbits with radii 105 times the radius of the nucleus.

The formula which defines density can also be used to convert the mass of a sample to the corresponding volume. If both sides of Eq. (1.2) are multiplied by 1/ρ, we have

  & \frac{\text{1}}{\rho }\times \text{m}=\text{V}\rho \times \frac{\text{1}}{\rho }=\text{V} \\ 
 & \text{         V}=\text{m}\times \frac{\text{1}}{\rho } \\ 
\end{align}       (1.3)

So if the Earth with its mass of 6 x 1024 kg were made of solid nuclear matter, its volume would be

\text{         V}=\text{m}\times \frac{\text{1}}{\rho } = \frac {6 \times 10^{24}}{4 x 10^{14} \frac {g}{cm^3}} = 1.5 x 1010 cm3, or a sphere about 15 meters (45 feet) in diameter.

Notice that we used the mathematical function D = m/V to convert parameters from mass to volume or vice versa in these examples. How does this differ from the use of unity factors to change units of one parameter?

An Important Caveat

A mistake sometimes made by beginning students is to confuse density with concentration, which also may have units of g/cm3. By dimensional analysis, this looks perfectly fine. To see the error, we must understand the meaning of the function

C = \frac{m}{V}.

In this case, V refers to the volume of a solution, which contains both a solute and solvent.

Given a concentration of an alloy is 10g gold in 100 cm3 of alloy, we see that it is wrong (although dimensionally correct as far as conversion factors go) to incorrectly calculate the volume of gold in 20 g of the alloy as follows:

20 g     x     \frac{100 cm^3}{10 g}     = 200 cm3

It is only possible to calculate the volume of gold if the density of the alloy is known, so that the volume of alloy represented by the 20 g could be calculated. This volume multiplied by the concentration gives the mass of gold, which then can be converted to a volume with the density function.

The bottom line is that using a simple unit cancellation method does not always lead to the expected results, unless the mathematical function on which the conversion factor is based is fully understood.

Example 1

Gold can be extracted from low grade gold ore by the "cyanide process". A gold ore with a concentration of 0.060 g / cm3 has a density of 8.25 g / cm3. What is the volume of gold (D = 19.32 g / cm3)in 100 cm3 of the ore?


The volume of 100 g of ore is

V = m / D = 100 g /8.25 g cm-3 = 12.12 cm3.

The mass of gold in this volume is

m = V x C = 12.12 cm3 x 0.060 g / cm3 = 0.727 g.

The volume of gold = m / D = 0.727 g / 19.32 g / cm3 = 0.0376 cm3.

Note that we cannot calculate the volume of gold by

 \frac {\frac{8.25 g}{cm^3} x 100 cm^3}{\frac {19.32 g}{cm^3}} = 42.70 cm^3

even though this is dimensionally correct.

Note that this result required when to use the function C = m/V, and when to use the function D=m/V as conversion factors. Pure dimensional analysis could not reliably give the answer, since both functions have the same dimensions.

Example 2 Find the volume occupied by a 4.73-g sample of benzene.

According to Table 1.4, the density of benzene is 0.880 g cm–3. Using Eq. (1.3),

\text{Volume = }V\text{ = }m\text{ }\times \text{ }\frac{\text{1}}{\rho }\text{ = 4}\text{.73 g }\times \text{ }\frac{\text{1 cm}^{\text{3}}}{\text{0}\text{.880 g}}\text{ = 5}\text{.38 cm}^{\text{3}}

(Note that taking the reciprocal of \tfrac{\text{0}\text{.880 g}}{\text{1 cm}^{3}} simply inverts the fraction ― 1 cm3 goes on top, and 0.880 g goes on the bottom.)

The two calculations just done show that density is a conversion factor which changes volume to mass, and the reciprocal of density is a conversion factor changing mass into volume. This can be done because of the mathematical formula, Eq. (1.1), which relates density, mass, and volume. Algebraic manipulation of this formula gave us expressions for mass and for volume [Eq. (1.2) and (l.3)], and we used them to solve our problems. If we understand the function D = m/V and heed the caveat above, we can devise appropriate converstion factors by unit cancellation, as the following example shows:

Example 3 A student weights 98.0 g of mercury. If the density of mercury is 13.6 g/cm3, what volume does the sample occupy?

We know that volume is related to mass through density.


V = m × conversion factor

Since the mass is in grams, we need to get rid of these units and replace them with volume units. This can be done if the reciprocal of the density is used as a conversion factor. This puts grams in the denominator so that these units cancel:

V=m\times \frac{\text{1}}{\rho }=\text{98}\text{.0 g}\times \frac{\text{1 cm}^{3}}{\text{13}\text{.6 g}}=\text{7}\text{.21 cm}^{3}

If we had multiplied by the density instead of its reciprocal, the units of the result would immediately show our error:

V=\text{98}\text{.0 g}\times \frac{\text{13,6 }g}{\text{1 cm}^{3}}=\text{1}\text{.333}{\text{g}^{2}}/{\text{cm}^{3}}\; (no cancellation!)

It is clear that square grams per cubic centimeter are not the units we want.

Using a conversion factor is very similar to using a unity factor—we know the factor is correct when units cancel appropriately. A conversion factor is not unity, however. Rather it is a physical quantity (or the reciprocal of a physical quantity) which is related to the two other quantities we are interconverting. The conversion factor works because of that relationship [Eqs. (1.1), (1.2), and (1.3) in the case of density, mass, and volume], not because it is equal to one. Once we have established that a relationship exists, it is no longer necessary to memorize a mathematical formula. The units tell us whether to use the conversion factor or its reciprocal. Without such a relationship, however, mere cancellation of units does not guarantee that we are doing the right thing.

A simple way to remember relationships among quantities and conversion factors is a “road map“of the type shown below:

\text{Mass }\overset{density}{\longleftrightarrow}\text{ volume   or   }m\overset{\rho }{\longleftrightarrow}V\text{ }

This indicates that the mass of a particular sample of matter is related to its volume (and the volume to its mass) through the conversion factor, density. The double arrow indicates that a conversion may be made in either direction, provided the units of the conversion factor cancel those of the quantity which was known initially. In general the road map can be written

\text{First quantity }\overset{\text{conversion factor}}{\longleftrightarrow}\text{ second quantity}

As we come to more complicated problems, where several steps are required to obtain a final result, such road maps will become more useful in charting a path to the solution.

Example 4 Black ironwood has a density of 67.24 lb/ft3. If you had a sample whose volume was 47.3 ml, how many grams would it weigh? (1 lb = 454 g; 1 ft = 30.5 cm).

The road map

V\xrightarrow{\rho }m\text{ }

tells us that the mass of the sample may be obtained from its volume using the conversion factor, density. Since milliliters and cubic centimeters are the same, we use the SI units for our calculation:

Mass = m = 47.3 cm3 × \frac{\text{67}\text{.24 lb}}{\text{1 ft}^{3}}

Since the volume units are different, we need a unity factor to get them to cancel:

m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\left( \frac{\text{1 ft}}{\text{30}\text{.5 cm}} \right)^{\text{3}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}

We now have the mass in pounds, but we want it in grams, so another unity factor is needed:

m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ }\times \text{ }\frac{\text{454 g}}{\text{ 1 lb}}\text{ = 50}\text{.9 g}

In subsequent chapters we will establish a number of relationships among physical quantities. Formulas will be given which define these relationships, but we do not advocate slavish memorization and manipulation of those formulas. Instead we recommend that you remember that a relationship exists, perhaps in terms of a road map, and then adjust the quantities involved so that the units cancel appropriately. Such an approach has the advantage that you can solve a wide variety of problems by using the same technique.

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