When heated above `916^(@)C`, iron changes, its crystal structure from body centred cubic to cubic closed packed structure. Assuming that the metallic radius of an atom does not change, calculate the ratio of the density of the bcc crystal to that of ccp crystal.

Explanation

In the bcc packing, the space occupied is 68% of the total volume available while in ccp, the space occupied is 74%. This means that for the same volume , masses of bcc and ccp are in the ratio of 68:74 .As the volume is same, ratio of density is also same , viz, 68:74,
i.e., `"d(bcc)"/"d(ccp)"=68/74`=0.919
Alternatively, Density`(rho)=(ZxxM)/(a^3xxN_0)`
For bcc, Z=2 , `r=(sqrt3a)/4` or `a_"bcc"="4r"/sqrt3`
For bcc, Z=4 , r=`a/(2sqrt2)` or `a_"fcc"=2sqrt2r`
`therefore rho_"bcc"=(2xxM)/((a_"bcc")^3xxN_0)` and `rho_"fcc"=(4xxM)/((a_"fcc")^3xxN_0)`
`there rho_"bcc"/rho_"fcc"=2/(a_"bcc")^3xx(a_"fcc")^3/4=2/(4r//sqrt3)^3xx(2sqrt2r)^3/4=(2xx3sqrt3)/(64r^3)xx(16sqrt2r^3)/4=3/8sqrt6=0.919`