Question

In class we showed that the radius r of an interstitial that can just fit in a octahedral site (without causing lattice distortion) in an FCC crystal as a function of the atomic radius of the host atom R is r=0.441 R. Based on this relationship and taking into account the atomic radii for C and Fe given in the table of problem 4, do you expect some lattice distortion when Coccupies octahedral interstitial sites in FCC iron? Element Atomic Radius (nm) Valence Crystal Structure FCC Electronegativity 1.8 0.1246 FCC 0.071 0.046 0.060 0.1445 0.1431 0.1253 0.1249 0.1241 0.1387 0.1332 0.132 0.175 FCC HCP BCC BCC FCC HCP BCC FCC +5(+3)

Answer 1

FCC iron ans: radius of carbon = 0.071 nm radius of Iron = 0.1241mm Now for carbon to fit into octahedral site of radius of carbon = 0.441 x radius of iron R.H.S = 0.4418 radices of iron 20.441 X 0.1241 = 0.0547 nm radius of carbon which can fit into octahedral site=0.0547 nm carbon = 0.07lum which is greater than required to fit into octahedral interstitial sits. But radices of So it will cause distortion of