Solid State Question 86
Question: Iron crystallizes in several-modifications. At about $ 910{}^\circ C $ ‘bcc’ form (called $ \alpha - $ form) undergoes transitions, to $ \gamma - $ form with ‘fcc’ lattice. Assuming that the distance between the nearest neighbors is the same in the two forms at the transition temperature, calculate the ratio of the density of $ \gamma - $ ron to that of $ \alpha - $ iron at the transition temperature.
Options:
A) 0.0887
B) 1.0887
C) 1.546
D) 1.544
Show Answer
Answer:
Correct Answer: B
Solution:
[b] In bcc structure, body diagonal $ =4r(Fe)=\sqrt{3}a $
$ \therefore $ $ a=\frac{4}{\sqrt{3}}r(Fe) $ $ Z=2 $ atoms per unit cell
$ \therefore $ $ d(\alpha -form)=\frac{Zm}{N_0a^{3}}=\frac{56\times 2}{(6.02\times 10^{23}){{( \frac{4}{\sqrt{3}} )}^{3}}} $ In fee structure, face diagonal $ =4r(Fe)=\sqrt{2}a^{1} $
$ \therefore $ $ a^{1}=2\sqrt{2}R(Fe) $ $ Z=4 $ atoms per unit cell
$ \therefore $ $ d(\gamma -form)=\frac{mZ}{N_0{{(a^{b})}^{3}}} $ $ =\frac{56\times 4}{(6.02\times 10^{23}){{(2\sqrt{2}r)}^{3}}} $
$ \Rightarrow $ Density ratio of $ \gamma - $ form to $ \alpha - $ form $ =\frac{2\times {{( \frac{4}{\sqrt{3}}r )}^{3}}}{{{(2\sqrt{2}r)}^{3}}} $
BETA
