Question

Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying this transformation is a change in the atomic radius of Fe—from RBCC = 0.12584 nm to RFCC = 0.12894 nm. The highest density planes in BCC structure is (110) and for FCC structure is (111). i. Compare the planar density of the two. (110) in BCC and (111) in FCC iron. EA = − 1.436 r ER = 5.8 × 10−6 r 9 ii. Do you think a (111) plane in FCC structure is more amenable to dislocation motion or (110) plane in BCC structure? What is an implication of that on the mechanical properties of materials.

Answer 1

Answer:

The description including its given problem is outlined in the following section on the clarification.

Explanation:

The given values are:

RBCC = 0.12584 nm

RFCC = 0.12894 nm

The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:

√3 ABCC = 4RBCC

⇒  ABCC = $\frac{4RBCC}{\sqrt{3} }$

⇒             = $\frac{4\times 0.12584}{\sqrt{3}}$

⇒             = $0.29062 \ nm$

Likewise AFCC as well as RFCC are interconnected by  

√2AFCC = 4RFCC

⇒  AFCC = $\frac{4RFCC}{\sqrt{2}}$

⇒             = $\frac{4\times 0.12894}{\sqrt{2} }$

⇒             = $0.36470 \ nm$

Now,

The Change in Percent Volume,

= $\frac{V \ final-V \ initial}{V \ initial}\times 100 \ percent$

= $\frac{(VFCC)unit \ cell-(VBCC)unit \ cell}{(VBCC)unit \ cell}\times 100 \ percent$

= $\frac{(aFCC)^3-(aBCC)^3}{(aBCC)^3}\times 100 \ percent$

= $\frac{(0.36470)^3-(0.29062)^3}{(0.29062)^3}\times 100 \ percent$

= $97.62 \ percent (approximately)$

Note: percent = %