diameter of atoms and estimation of it spherical volume (iron and copper).. (see details)?

Question

Two metals, iron and copper, have densities 5.50 x 10^3 kg m^-3 and 8.93 x 10^3 kg m^-3 respectively.
1) estimate the diameter of an atom of copper given that the atomic mass of copper is 63.5
2) do as 1) for iron with Ar=56
3) Find the estimation of the ratio of their spherical volumes.
4) Explain whether or not your estimations in 1) and 2) is greater or less than the accepted values.
5) Do their crystalline structures matter for finding the ratio? Give reason/s...

Answer 1

(1) Atomic mass of Copper, M = 63.5

Density of copper is ρ = 8.93 x 10^3 kg m^-3

So to estimate the volume:

V = M/Nρ = (63.5)/(6.022 x 10^23)(8.93 x 10^3)

= 1.18 x 10^-26 m³

where N is the Avogadro's number.

Given that the volume of a sphere is (4/3)πr³, we get

r = cuberoot[(3/4π) (1.18 x 10^-26 m³)] = 1.41 x 10^-9 m
diameter of copper atom = 2r = 2.83 x 10^-9 m

(2) Atomic mass of Iron, M = 56

Density of iron is ρ = 5.50 x 10^3 kg m^-3

V = M/Nρ = (56)/(6.022 x 10^23)(5.50 x 10^3)

= 1.69 x 10^-26 m³

So r = 1.59 x 10^-9 m, hence d = 3.18 x 10^-9 m

(3) V(copper)/V(iron) = (1.18 x 10^-26 m³)/(1.69 x 10^-26 m³)

= 0.698

Answer 2

Use the densities and the weight per atom to calculate the volume of a cube that each atom occupies. Set the diameter equal to the side of the cube