Question

(a) For BCC iron, calculate the diameter of the minimum space available in an octahedral site at the center of the (010) plane, and compare this to the diameter of a carbon atom. Assume that the iron atoms in the BCC structure are hard spheres that touch along the [111] direction.
(b) Compare the space available with the size of a carbon atom from Appendix C, and comment about the potential solubility of carbon in BCC iron in octahedral interstitial sites.

Answer 1

Answer:

a) diameter available = 0.0384 nm

b)The space  is smaller than the carbon atom which has a  radius of  0.077 nm and this simply means that the carbon atom will not conquer these sites

Explanation:

For BCC iron

From Appendix B given,select the lattice parameter ( a ) as = 0.2866 nm

The BCC iron has 4 atomic radii and therefore the body diagonal length = $a(3)^\frac{1}{2}$

expressing the atomic radius of the BCC iron

4r = $a(3)^\frac{1}{2}$

insert the value of (a) from appendix B which is = 0.2866 nm

4r = $0.2866 nm (3)^\frac{1}{2}$

therefore  r =  0.4964 nm / 4 = 0.1241 nm

Refer again to appendix C given select the atomic radius of the BCC iron as = 0.1241 nm   assuming the atomic radius of the iron are the same

then the radius ratio = 0.62

Refer to the Figure 3.2 given, the amount of space required for an interstitial at the BCC position is between the atoms at the FCC position and also in this space there are two atoms that are equal to a radius of 0.2482 nm

The diameter of the minimum space available

$d_{a} = a - r_{a}$

$r_{a} = atomic radii$ = 0.2482 nm

a = 0.2666 nm

therefore

d$_{a}$ = 0.2866 nm - 0.2482 nm = 0.0384 nm

comparing this to the diameter of a carbon atom

The space  is smaller than the carbon atom which has a  radius of  0.077 nm and this simply means that the carbon atom will not conquer these sites