Question

Iron exhibits `bcc` structure at roomj temperature. Above `9000^(@)C`, it transformers to `fcc` structure. The ratio of density of iron at room temperature to that at `900^(@)C` (assuming molar mass and atomic radius of iron remains constant with temperature) is
A. `(3sqrt(3))/(4sqrt(2))`
B. `(4sqrt(3))/(3sqrt(2))`
C. `(sqrt(3))/(sqrt(2))`
D. `(1)/(2)`

Answer 1

Correct Answer - A
For BCC lattice : `Z=2,a=(4r)/(sqrt(3))`
For BCC lattice , `Z=4,a=2sqrt(2)r`
`therefore (d_(25^(@)C))/(d_(900^(@)C))=(((ZM)/(N_(A)a^(3)))_(B"CC"))/(((ZM)/(N_(A)a^(3)))_(F"CC"))`
`=(2)/(4)((2sqrt(2)r)/((4r)/(sqrt(3))))^(3)`
`=((3sqrt(3))/(4sqrt(2)))`