Question 5
Face Centered Cubic Lattice, Part 2
For the face centered cubic crystal described above, i.e. a = 0.7\: nma=0.7nm, calculate the surface density of atoms (i.e. number of atoms per unit area) on the (100) plane in unit of cm^{-2}cm
−2
. Values within 5% error will be considered correct.
Answers
Answer:
Consider the top face of the unit cell in Fig. 1.4 (a) of Pierret, SDF. As shown below, there
are 5 atoms on the face, but the 4 on the corners are shared between 4 adjacent unit cells,
so the total number is N = 4 ×
1
4
+1= 2 per face of a cell. The density per unit area is
NS = 2
a
2 = 2
5.42 ×10−8
( cm)
2 = 6.81×1014 cm-2 NS = 6.81×1014 cm-2
Note that the corresponding answer for a (111) plane is NS = 7.86 ×1014 cm-2 , but the
geometry is a bit harder to visualize.
3b)
The volume of each sphere is Vsphere = 4
3
π R3
The radius is one-half the nearest neighbor distance, so
R = 1
2
3a
4
⎛
⎝
⎜
⎞
⎠
⎟ = 3a
8
Vsphere = 4
3
π R3 = 4
3
π
3a
8
⎛
⎝
⎜
⎞
⎠
⎟
3
= 4
3
π
3 3a3
64 = π 3a3
16
There are eight of these spheres in a unit cell, so the fraction of the unit cell volume filled
(the packing fraction, PF, ) is
PF = 8Vsphere
a3 = 3π
16 = 0.34