Classwork 5.pdf - LMC

Materials Science and Engineering: 2 nd year, 2011: CRYSTALLOGRAPHY
Classwork 5: Interstitial Sites
1. Assuming that atoms are hard spheres in contact and letting r' be the radius of the
interstitial site and r the radius of the solvent atom, using diagrams:
i. show that for the two fcc interstitial sites (r'/r) equals 0.4142 and 0.2247.
ii. show that for the two bcc interstitial sites octahedral and tetrahedral
(r'/r) equals 0.1540 and 0.2910
answer
i)
r
4r 2a
2r
2r' a
2a
r
4
2a
2r' a
2 2a
4
r' a 2
2 4 a
r'
r
1 2
2 4
2
4
0.4142

The tetrahedral interstice from the figure are situated at 1/4 of the diagonal. The distance
between the center of a "corner" atom and the center of the tetrahedral interstice is: r+r'
So to express the whole diagonal as a function of r and r' we have to multiply the previous
distance by 4. Therefore we have the following system:
4(r r') 3a
r 2
4 a
ii)
r'
r
(as before)
3 2
2
0.2247
r
4r 3a
2r
2r' a
3a
r
4
2a
2r' a
2 3a
4
r' a 2
2 4 a
r'
r
1 3
2 4
3
4
0.1547

In the body centered cubic system the tetrahedral sites are located at 1/2 0 1/4.
Therefore we can draw the following scheme:
Using Pythagoras theorem we have the following system of equations:
4r 3a
(r r') 2 1
4 a
2
1
2 a
r
2
3a
4
5
r' a
r
4
r'
r
5 3
3
0.2910

2. At the transition in pure iron at 910°C, the lattice constant changes from,
a = 2.90 Å (bcc) to a = 3.64 Å (fcc).
i. Show that for both crystal forms and both sites, the carbon atom (atomic radius = 0.77
Å) distorts the lattice (i.e. it is larger than the site in which it resides).
ii. Carbon atoms in a-iron (bcc) occupy the octahedral interstices; why?
(hint consider the forces acting on the atoms and the symmetry of the two sites).
Answer
i) Consider the fcc lattice where a = 3.64 Å.
We know that th carbon atom resides in the interstitial sites, therefore r' = 0.77 Å.
For a fcc lattice we have the following relationship:
4 √2 from classwork 4
3.64 ∗ 1.414
4 1.29
Which means that: ´ .
0.5969
.
We notice that 0.5969 > 0.4142 > 0.2247, explains why the carbon atoms are preferably in the
octahedral sites (0.4142) because they are much larger and distort less the lattice than if they
were in the tetrahedral sites.
Consider now the bcc lattice where a = 2.90 Å and r' = 0.77 Å.
4 √3 from classwork 4
2.90 ∗ 1.732
1.256
4
Which gives us: ´ .
0.6132
.
Once again, we have 0.6132 > 0.2910 (tetrahedral) > 0.1547 (octahedral) which proves that
the carbon atoms distort the lattice. Note that the tetrahedral sites are larger than octahedral
ones.
ii) Considering the symmetry part of the octahedral sites we can see that the carbon interstitial
atoms have only 2 atoms to distort whereas in the tetrahedral sites they have 4 atoms making
it more difficult.

3. Write down the coordinates of all the tetrahedral and octahedral sites for the fcc and bcc
structures.
Fcc - octahedral sites:
(½ 0 0) (0 0 ½) (0 ½ 0)
(½ ½ ½).
Fcc - tetrahedral sites:
(¼ ¼ ¼)
(¾ ¼ ¼) (¼ ¾ ¼) (¼ ¼ ¾)
(¾ ¾ ¼) (¾ ¼ ¾) (¼ ¾ ¾)
(¾ ¾ ¾)
Bcc - octahedral sites:
(0 0 ½) (½ 0 0) (0 ½ 0)
(½ 0 ½) (0 ½ ½) (½ ½ 0)
Bcc - tetrahedral sites:
(½ 0 ¼) (¼ 0 ½) (0 ½ ¼) (0 ¼ ½) (½ ¼ 0) (¼ ½ 0)
(0 ½ ¾) (0 ¾ ½) (½ 0 ¾) (¾ 0 ½) (½ ¾ 0) (¾ ½ 0)