The density of nickel (FCC unit cell) is 8.94 g/cc. at 20°C. What is the diameter of the atom ? (atomic weight = 59)

I got #"249 pm"#. This is exactly twice the metallic radius of #"124 pm"#, which makes sense.

A face-centered cubic (FCC) unit cell looks like this:

where #r# is the radius of the atom.

The first thing we want to do is find the number of atoms in one FCC unit cell. That would later allow us to relate the volume of ALL the nickel atoms in the unit cell.

• Each of 8 corners has #1/8# of an atom.
• Each of 6 faces has #1/2# of an atom.

That accounts for all the atoms in the unit cell, which gives #8 cdot 1/8 + 6 cdot 1/2 = 4# total atoms in one unit cell.

Looking at the face, the total diagonal length is #r + 2r + r = 4r = 2D#. This is a 45-45-90 triangle that has diagonal #asqrt2#. That gives a side length of

#a = 2/sqrt2 D#.

This makes the volume:

#V = a^3 = 2^(3"/" 2)D^3#

Knowing the density (the mass of nickel atoms per unit volume), the volume of the unit cell is:

#V = "cc"/(8.94 cancel"g") xx (58.6934 cancel"g Ni")/cancel"1 mol" xx cancel"1 mol"/(6.0221413 xx 10^23)#

#= 1.090 xx 10^(-23) "cc/atom"#

#= 4.360 xx 10^(-23) "cc/unit cell"# (containing 4 atoms)

As a result, the diameter of the atom would be:

#color(blue)(D) = (V/2^(3//2))^(1//3) = 1/sqrt2(4.360 xx 10^(-23) "cc")^(1//3)#

#= color(blue)ul(2.49 xx 10^(-8) "cm")#

Or in terms of #"pm"#,

#color(blue)(D) = 2.49 xx 10^(-8) cancel"cm" xx cancel"1 m"/(100 cancel"cm") xx (10^12 "pm")/(cancel"1 m") = color(blue)ul("249 pm")#