Problem 2
Nickel has the FCC crystal structure. Find the density and atomic mass of nickel and
determine the radius of atoms in a single crystal and also the lattice parameter of the FCC
cell. How many atoms are there in a single crystal shaped as a cube with side = 1 mm?
From http://en.wikipedia.org/wiki/Nickel:
Density = 8.908 g/cm3
Atomic mass = 58.6934 g/mol
Avogadro’s number = 6.022 x 1023 mol-1
Mass of one Ni atom = standard atomic weight / Avogadro’s number
𝑔
58.6934
𝑔
𝑚𝑜𝑙
−23
𝐴𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 =
=
9.746
𝑥
10
6.022 𝑥 1023 𝑚𝑜𝑙 −1
𝑎𝑡𝑜𝑚
Ni has a FCC crystal structure: (http://mrsec.wisc.edu/Edetc/SlideShow/slides/unit_cells/face_centered_cubic.html)
or
(3D)
(looking straight on at one side)
1
1
Therefore, there is 2 an atom on each of the six sides of the cube ( = 3 atoms), and 8 an atom on
each of the eight corners (= 1 atom), for a total of 4 atoms per unit cell. The volume of the unit
cell is:
(9.746 𝑥 10−23
𝑔
𝑔
) ∗ (4 𝑎𝑡𝑜𝑚𝑠) = 3.899 𝑥 10−22
𝑎𝑡𝑜𝑚
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
𝑔
3
(3.899 𝑥 10−22
)
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 = 4.377 𝑥 10−23 𝑐𝑚
𝑔
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
8.908
𝑐𝑚3
Since the volume of a cube is side3, the length of a side is:
3
√4.377 𝑥 10−23
𝑐𝑚3
= 3.524 𝑥 10−8 𝑐𝑚
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
Drawing a diagonal line through the cube, we see that there are 4 radii of Ni atom along the
hypotenuse of a triangle, as shown below:
Using the Pythagorean theorem, the hypotenuse ( = 4 radii) is:
𝑠𝑖𝑑𝑒 2 + 𝑠𝑖𝑑𝑒 2 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 2
2𝑠𝑖𝑑𝑒 2 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 2
2(3.524 𝑥 10−8 𝑐𝑚)2 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 2
2.484 𝑥 10−15 𝑐𝑚 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 2
4.984 𝑥 10−8 𝑐𝑚 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 4 𝑟𝑎𝑑𝑖𝑖
Thus, 1 𝒓𝒂𝒅𝒊𝒖𝒔 𝒐𝒇 𝑵𝒊 𝒂𝒕𝒐𝒎 =
𝟒.𝟗𝟖𝟒 𝒙 𝟏𝟎−𝟖 𝒄𝒎
𝟒
= 𝟏. 𝟐𝟒𝟔 𝒙 𝟏𝟎−𝟖 𝒄𝒎
The lattice parameters are the unit cell side lengths and angles between them. For a FCC crystal
structure, all sides are the same length, and all angles between the sides are the same.
(http://www.doitpoms.ac.uk/tlplib/crystallography3/parameters.php)
𝒂 = 𝒃 = 𝒄 = 𝟑. 𝟓𝟐𝟒 𝒙 𝟏𝟎−𝟖 𝒄𝒎
𝜶 = 𝜷 = 𝜸 = 𝟗𝟎°
𝑐𝑚3
If there are 4 atoms in a cube with volume = 4.377 𝑥 10−23 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙, then for a cube with side of
𝑐𝑚3
1 mm, or volume = 1 𝑚𝑚3 = 0.001 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙, there will be:
4 𝑎𝑡𝑜𝑚𝑠
4.377 𝑥 10−23
𝑐𝑚3
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
=
𝑋 𝑎𝑡𝑜𝑚𝑠
𝑐𝑚3
0.001
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
𝑋 = 𝒂𝒕𝒐𝒎𝒔 𝒊𝒏 𝒂 𝒄𝒖𝒃𝒆 𝒘𝒊𝒕𝒉 𝒂 𝒔𝒊𝒅𝒆 𝒐𝒇 𝟏𝒎𝒎 = 𝟗. 𝟏𝟑𝟗 𝒙 𝟏𝟎𝟏𝟗 𝒂𝒕𝒐𝒎𝒔