Question

Iron crystallizes in several modifications. At about $911^{\circ}C$, the bcc $'\alpha'$ form undergoes a trasition to fcc $'\gamma'$ form. If the distance between the two nearest neighbours is the same in the two forms at the transition temperature, the ratio of the density of iron in fee form $\left(\rho_{2}\right)$ to that of iron of bcc form $\left(\rho_{1}\right)$ at the transition temperature

• A. $\frac{\rho_{1}}{\rho_{2}}=0.918$
• B. $\frac{\rho_{1}}{\rho_{2}}=0.718$
• C. $\frac{\rho_{1}}{\rho_{2}}=0.518$
• D. $\frac{\rho_{1}}{\rho_{2}}=0.318$

Answer 1

Answer: (A)

In $\alpha$ - form distance between nearest
neighbours atom is $\frac{\sqrt{3}a_{1}}{2}.$
In $\gamma$ form distance between nearest neighbours atom is $\frac{a_{2}}{\sqrt{2}} .$
$\therefore \frac{\sqrt{3}a_{1}}{2}=\frac{a_{2}}{\sqrt{2}}$ (given)
or $\frac{a_{2}}{a_{1}}=\sqrt{\frac{3}{2}}$
$\frac{\rho_{1}}{\rho}=\frac{Z_{1}}{Z_{2}} \left(\frac{a_{2}}{a_{1}}\right)^{3} =\frac{1}{2} \left(\sqrt{\frac{3}{2}}\right)^{3} =0.918$