Iron has an FCC structure at temperatures just above 912°C, and the lattice parameter is 0.3589 nm. Assume that BCC iron’s atomic radius is equal to 0.1241 nm as given in Appendix C, and for carbon it is 0.0077 nm. The atomic radius of iron should not increase with temperature and with changes in crystal structure; only the size of the lattice changes.

Question

(a) Calculate the radius ratio of carbon to iron to see if it is appropriate for octahedral coordination.
(b) Calculate diameter of the space available for the octahedral site at the center of the FCC unit cell, and compare this to the diameter of a carbon atom.
(c) Comment on the diameter available relative to the diameter of the carbon atom and upon the possibility of solubility of carbon in FCC iron at octahedral interstitial sites. Also comment on the expected relative solubility of carbon in octahedral sites in FCC and BCC iron (see Problem 3.3).

Answer 1

From Figure 2.27 the radius ratio range for octahedral coordination is 0.414 to 0.732, thus carbon in iron is in the octahedral coordination range.



(b) By inspection of Figure 3.2 the minimum space available for an interstitial at the body centered position is between atoms at the face centered positions. These atoms are separated by the lattice parameter a. In this distance of 0.3589 nm there are two atomic radii equal to 0.2482 nm and the remaining space is available for the interstitial.

Diameter available = 0.3589 nm – 0.2482 nm = 0.1107 nm

(c) This diameter is smaller than the diameter of the carbon atom of 0.154 nm, thus there will be limited solubility and there will be lattice strain. There is much more space available in the octahedral sites in FCC lattice than in the BCC lattice. Thus there should be more solubility of carbon in the FCC phase at temperatures above 912°C than in BCC iron at room temperature.