Iron exhibits bcc structure at room temperatutre .Above `900^(@)C` .it transitions to fcc structure .The ratio of density of ions at room temperature to that at `900^(@)C`(assuming molar mass and atomic , radii of ion remain s constant with temperature ) is

Options

A. `sqrt(3)/sqrt(2)`
B. `(4sqrt(3))/(3sqrt(2))`
C. `(3sqrt(3))/(4sqrt(2))`
D. `1/2`

Answer

C

Explanation

For bcc lattice, `Z=2, a=(4r)/sqrt(3)`
For fcc lattice, `Z=4, a=2sqrt(2r)`
`(rho("room temp."))/(rho(90^(@)C))=(((ZM)/(N_(0)a^(3)))"bcc")/(((ZM)/(N_(0)a^(3)))"fcc")`
For the same metal iron, both M and r are constant
`(rho("room temp."))/(rho(190^(@)C))=2/4((2sqrt(2r))/(4r//sqrt(3)))^(3)=((3sqrt(3))/(4sqrt(2)))`