Physics conversion question?

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

Since Iron has a mass of 7.87 g per cubic centimeter,

Density of Iron = 7.87g/cm³

                      = 7.87 g/cm³ X (1000g/1Kg) X (100cm/1m)³

                      = 7870 Kg/m³

(a) Volume of an Iron atom = Mass of an Iron atom / Density of Iron

                                       = (9.27 X 10^-26 Kg) / (7870 K/m³)

                                       = 1.17 X 10^-29 m³

Since, each Iron atom can be considered as a cube,

(b) Distance between centers of adjacent cubes

     = Length of any one side of the cube

     = (Volume of Iron atom)^1/3

     = (1.17 X 10^-29m³)^1/3

     = 2.28 X 10^-10 m