Question

Iron has a mass of 7.87 g per cubic centimeter of volume, and the mass of an iron atom is 9.27 × 10-26 kg. If you simplify and treat each atom as a cube, (a) what is the average volume (in cubic meters) required for each iron atom and (b) what is the distance (in meters) between the centers of adjacent atoms?

Answer 1

I like the first part of the question, it's a lovely way of saying density. So density of Iron is 7.87\[\frac{ g }{ cm^3 }\].

Mass of one Fe atom is \[9.27 \times 10^{-26} kg\].

So the volume that one Fe atom would take up would require rearranging the density formula: \[\rho = \frac{ m }{ V } \rightarrow V = \frac{ m }{ }\]

However, density is given in \[\frac{ g }{ cm^3 }\], and need to be in \[\frac{ kg }{ m^3 } \] to give the answer in the correct units.

So 7.87\[\frac{ g }{ cm^3 } \] divided by 1000, would convert the grams into kilograms as 1000g = 1kg,

so \[\frac{ 7.87 }{ 1000 } = 7.87 \times 10^{-3} \frac{ kg }{ cm^3 }\]

Also, \[1cm^3 = 1\times 10^{-6}m^3 \] therefore multiplying this answer by 10^{-6} will yield:

\[7.87 \times 10^{-9} \frac{ kg }{ m^3 }\]


So now we simply apply the formula:

\[V = \frac{ m }{ \rho } \rightarrow \frac{ 9.27 \times 10^{-26} kg }{ 7.87 \times 10^{-9} kg/m^3} \rightarrow 1.177 \times 10^{-17} m^3\]

The units are \[m^3 \] as the kilogram units are cancelled in the division.

Now we have the volume of one iron atom, of \[1.177 \times 10^{-17}\]

We are told that each atom should be imagined to be a cube is, so finding the distance between is pretty straight forward.





Therefore to get the distance between, simply cube root the volume of each atom. This makes sense if you think about it dimensionally.

\[\sqrt[3]{1.177 \times 10^{-17}m^3} = 2.7528 \times 10^{-6}m \approx 2.8 \times 10^{-6}m\]

\[10^{-6}m \rightarrow \mu m\]

So:

a) \[1.18 \times 10^{-17}m^3 \]

&

b) \[2.8 \mu m\]