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

Answer

c

Explanation

In fcc atoms along face diagram touch each other i.e,`sqrt(2)a = 4r`
In fcc atoms along body diagram touch each other i.e,`sqrt(3)a = 4r`
BCC FCC
`4r = sqrt(3) a 4r = sqrt(2) a`
`4r = sqrt(3)a 4r = sqrt(2)a`
`a = (4r)/(sqrt(3)) a= (4r)/(sqrt(2))`
`(d_(BC C))/(d_(FC C))= ((Z_(BC C) xx M)/(N_(A)a^(3)))/((Z_(FC C) xx M)/(N_(Aa^(3)))) = ((Na((4r)/(sqrt(3)))^(3))/(4xxM))/(N_(A) xx((4r)/(sqrt(3)))^(3)) = (3)/(4) sqrt((3)/(2))`