Question

Iron changes its crystal structure from body-centred to cubic close-packed structure when heated to `916^@C`. Calculate the ratio of the density of the bcc crystal to that of ccp crystal, assuming that the metallic radius of the atom does not change.

Answer 1

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`