Crystal Structures in Practice
Linear Density and Planar Density
z
y
x
Example solutions for BCC
- Find Lattice Parameter
- Find Directions or Planes
- Calculate Linear or Planar Density
First, we should find the lattice parameter(a) in terms of atomic radius(R).
Then, we can find linear density or planar density.
z
Green ones touches each other.
y
x
Body-centered Cubic Crystal Structure (BCC)
z
𝑎2 + 𝑎2 + 𝑎2
4𝑅
→ 𝑎=
3
4𝑅 =
R
R
a
R
y
R
a
a
x
We found the lattice
parameter in terms of
atomic radius.
Linear Density
1. Draw the atoms on the direction, and use the
formula;
# 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑡𝑕𝑒 𝑙𝑖𝑛𝑒
𝐿𝐷 =
𝑡𝑕𝑒 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑙𝑖𝑛𝑒
z
1
1
2
3
4
Directions
[1 0 1]
[1 1 1]
[1 1 0]
[1 0 0]
2
y
4
x
3
Body-centered Cubic Crystal Structure (BCC)
Let’s apply the formula
below for [101] and [111].
# 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑡𝑕𝑒 𝑙𝑖𝑛𝑒
𝐿𝐷 =
𝑡𝑕𝑒 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑙𝑖𝑛𝑒
[1 0 1] direction for BCC and Linear Density
1
𝑎𝑡𝑜𝑚
2
z
𝐿=𝑎 2=
a
y
a
x
1
𝑎𝑡𝑜𝑚
2
4 2𝑅
3
# 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑡𝑕𝑒 𝑙𝑖𝑛𝑒
𝐿𝐷 =
𝑡𝑕𝑒 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑙𝑖𝑛𝑒
1 1
+2
3
2
𝐿𝐷 =
=
4 2𝑅 4 2𝑅
3
[1 1 1] direction for BCC and Linear Density
1
𝑎𝑡𝑜𝑚
2
z
1 𝑎𝑡𝑜𝑚
1
𝑎𝑡𝑜𝑚
2
𝐿 = 𝑎 3 = 4𝑅
a
y
a
x
# 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑡𝑕𝑒 𝑙𝑖𝑛𝑒
𝐿𝐷 =
𝑡𝑕𝑒 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑙𝑖𝑛𝑒
1
1
+
1
+
1
2
2
𝐿𝐷 =
=
4𝑅
2𝑅
Planar Density
1. Draw the atoms on the plane, and use the
formula;
# 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑡𝑕𝑒 𝑝𝑙𝑎𝑛𝑒
𝑃𝐷 =
𝑡𝑕𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑕𝑒 𝑝𝑙𝑎𝑛𝑒
z
Let’s apply the formula
below for (0 1 0).
# 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑡𝑕𝑒 𝑝𝑙𝑎𝑛𝑒
𝑃𝐷 =
𝑡𝑕𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑕𝑒 𝑝𝑙𝑎𝑛𝑒
y
x
Body-centered Cubic Crystal Structure (BCC)
(0 1 0) plane for BCC and Planar Density
z
1
𝑎𝑡𝑜𝑚
4
1
𝑎𝑡𝑜𝑚
4
1
𝑎𝑡𝑜𝑚
4
1
𝑎𝑡𝑜𝑚
4
𝑃𝐷 =
y
# 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑛 𝑡𝑕𝑒 𝑝𝑙𝑎𝑛𝑒
𝑡𝑕𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑕𝑒 𝑝𝑙𝑎𝑛𝑒
1 1 1 1
+ + +
4
4 4 4
𝑃𝐷 =
𝑎2
𝑃𝐷 =
x
Body-centered Cubic Crystal Structure (BCC)
1
4𝑅
3
2
3
=
16𝑅 2
Suggestion
• For the other directions and planes, also for all
crystal structures (FCC,SC), you can and you
should do this on your own.
• For any question, you can contact with me,
emreyilmaz@cankaya.edu.tr