Answer to Question #173485 in Mechanical Engineering for MASOOD HANIF KASU

Question

Metallic iron changes from BCC to FCC form at 9100C. At this temperature,

the atomic radii of the iron atom in the two structures are 0.1258 nm and

0.1292 nm respectively. Calculate the volume change in percentage during

this structural change. Also, calculate the percentage change in density.

Answer 1

The relation between radius(r) and lattice constant (a) is given as,

For BCC structure: $a_{BCC}=\frac{4×r}{\sqrt3}=0.2958 nm$

For FCC structure: $a_{FCC}=\frac{4×r}{\sqrt2}=0.3652nm$

The percentage change in volume is given by,

$\frac{V_{FCC}-V_{BCC}}{V_{BCC}}×100=\frac{a_{FCC}^2-a_{BCC}^2 }{a_{BCC}^2}×100$ =52.42%

Percentage change in density is given by,

$\frac{\frac{Mass}{V_{FCC}} -\frac{Mass}{V_{BCC}}}{\frac{Mass}{V_{BCC}}} =\frac{\frac{1}{V_{FCC}} -\frac{1}{V_{BCC}}}{\frac{1}{V_{BCC}}}$ = -34.395% ( Reduction in the density)