NEET Solved Paper 2018 Question 44
Question: Iron exhibits bcc structure at room temperature. Above $ \text{900 }{}^\circ\text{ C} $ , it transforms to fcc structure. The ratio of density of iron at room temperature to that at $ \text{900 }{}^\circ\text{ C} $ (assuming molar mass and atomic radii of iron remains constant with temperature) is [NEET - 2018]
Options:
A) $ \frac{3\sqrt{3}}{4\sqrt{2}} $
B) $ \frac{4\sqrt{3}}{3\sqrt{2}} $
C) $ \frac{\sqrt{3}}{\sqrt{2}} $
D) $ \frac{1}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
For BCC lattice: $ Z=2,a=\frac{4r}{\sqrt{3}} $ For FCC lattice: $ \text{Z=4, a=2}\sqrt{2}r $
$ \therefore \frac{{d _{25{}^\circ C}}}{{d _{900{}^\circ C}}}=\frac{{{( \frac{ZM}{N _{A}a^{3}} )} _{BCC}}}{{{( \frac{ZM}{N _{A}a^{3}} )} _{FCC}}} $
$ =\frac{2}{4}{{( \frac{2\sqrt{2}r}{\frac{4r}{\sqrt{3}}} )}^{3}} $
$ =( \frac{3\sqrt{3}}{4\sqrt{2}} ) $
