Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is

Question

Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is (A) (1/2) (B) (√3/√2) (C) (3√3/4√2) (D) (4√3/3√2)

Tardigrade Admin · Accepted Answer

For BCC lattice: Z =2, a =(4 r /√3) For FCC lattice: Z =4, a =2 √2 r ∴ (d25° C /d900° C )=((( ZM / N A a 3)) BCC /(( ZM ) N A a 3) FCC )=(2/4)((2 √2 r /(4 r)√3))3=((3 √3/4 √2))

Q. Iron exhibits $b cc$ structure at room temperature. Above $90 0^{\circ} C$ , it transforms to fcc structure. The ratio of density of iron at room temperature to that at $90 0^{\circ} C$ (assuming molar mass and atomic radii of iron remains constant with temperature) is

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{3 3}{4 2}$

$\frac{4 3}{3 2}$

For $BCC$ lattice $: Z = 2, a = \frac{4 r}{3}$

For $FCC$ lattice $: Z = 4, a = 22 r$

$∴ \frac{d _{2 5^{\circ} C}}{d _{90 0^{\circ} C}} = \frac{( \frac{ZM}{N _{A} a ^{3}} ) _{BCC}}{( \frac{ZM}{N _{A} a ^{3}} ) _{FCC}} = \frac{2}{4} (\frac{2 2 r}{\frac{4 r}{3}})^{3} = (\frac{3 3}{4 2})$

Q. Iron exhibits bcc structure at room temperature. Above 900∘C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900∘C (assuming molar mass and atomic radii of iron remains constant with temperature) is

21​

2​3​​

42​33​​

32​43​​