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JEE Advanced
Exam
JEE Advanced
Subject
Chemistry
Chapter
Solid State

JEE Advanced · Chemistry · Solid State

Iron crystallizes in several modifications. At about 910^ C , the...

Iron crystallizes in several modifications. At about 910∘C910^{\circ} \mathrm{C}, the body centred cubic α\alpha-form undergoes a transition to the face centred cubic γ\gamma-form. Assuming that the distance between nearest neighbours is same in the two forms at the transition temperature. Calculate the ratio of density of γ\gamma-iron to that of α\alpha iron at the transition temperature.

Correct answer

1.0887

Step-by-step solution

In bcc structure ( α\alpha form) of iron 4r=3a4 r=\sqrt{3} a n=2\mathrm{n}=2 atoms per unti cell from (one atom from 8 corners, one atom from the body centre) So, ρ(γ\rho(\gamma-form )=n×Mma3×N0=2×Mm(43r)3×N0)=\frac{n \times M_m}{a^3 \times N_0}=\frac{2 \times M_m}{\left(\frac{4}{\sqrt{3}} r\right)^3 \times N_0} ....(i) In fcc structue ( r form) of iron 4r=2a4 r=\sqrt{2} a ∴a=22r\therefore a=2 \sqrt{2} r n=4\mathrm{n}=4 atoms per unit cell ∴ρ(γ\therefore \rho(\gamma form )=4×Mm(22r)3×N0...(ii))=\frac{4 \times M_m}{(2 \sqrt{2} r)^3 \times N_0}...(ii) Dividing Eq. (ii) by Eq. (i) ρ(y form )ρ(α form )=2(43r)3×(22r)34=1.0887\frac{\rho(y \text { form })}{\rho(\alpha \text { form })}=\frac{2}{\left(\frac{4}{\sqrt{3}} r\right)^3} \times \frac{(2 \sqrt{2} r)^3}{4}=1.0887
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