Iron has a density of 7.87 g/cm^3, and the mass

Iron has a density of 7.87g/cm^{3}, and the mass of an iron atom is 9.27\times 10^{-26} kg. If the atoms are spherical and tightly packed, (a) what is the volume of an iron atom and (b) what is the distance between the centers of adjacent atoms?

Solution:

Density is \rho=\frac{m}{V} where \rho is density, m is mass and V is the volume. Let us first convert the given density of iron into the units the mass of our question is given in, which is kg.

7.87g/cm^{3} is equal to 7878kg/m^{3} .

a) The volume of our atom, if we ignore the spaces between the spheres is V=\frac{M}{\rho}

V=\frac{9.27\times 10^{-26} kg}{7878kg/m^{3}}

V=1.18 \times 10^{-29} m^{3}

b) To find the distance between the centers of adjacent atoms, we will use the volume of a sphere formula, which is \frac{4\pi R^{3}}{3} where R is the radius, in our case, the radius of an atom.

Isolating for R gives us,

R=\left(\frac{3V}{4 \pi}\right)^{\frac{1}{3}}

R=\left(\frac{3(1.18 \times 10^{-29} m^{3})}{4 \pi}\right)^{\frac{1}{3}}

R= 1.41 \times 10^{-10}m

This result gives us the radius of one atom, however, we need the distance between two atoms (as seen in the figure), which means we must multiply our answer by 2. This gives us, R=2.82 \times 10^{-10}m

This question can be found in Fundamentals of Physics, 10th edition, chapter 1, question 27.