Iron changes its crystal structure from body-centred to cubic close- packed structure when heated to916^(@)C. Calculate the ratio of the density of the bcc crystal to that of ccp crystal, assuming that themetallic radius of the atom does not change.

Iron changes its crystal structure from body-centred to cubic close- p

1.	Iron changes its crystal structure from body-centred to cubic close- packed structure when heated to916^(@)C. Calculate the ratio of the density of the bcc crystal to that of ccp crystal, assuming that themetallic radius of the atom does not change.
Answer» Solution :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,X ` (d(ccp))/(d(ccp)) = 68/74 = 0.919` Alternatively Density ` (p) = (Z xx M)/( a^(3) xx N_(0))` For bcc, ` Z = 2,r= ( sqrt3 a)/4 or a_("bcc") = (4R)/sqrt3` For fcc, ` Z = 4 ,r = a/(2sqrt2)or a_("fcc") = 2 sqrt2 r ` ` p_("bcc")= (2xxM)/((a_("bcc"))^(3) xx N_(0)) and p_("fcc") = ( 4 xx M)/((a_("fcc"))^(3)xx N_(0))` `p_("bcc")/(P_("fcc"))= 2/((a_("bcc"))^(3))xx ((a_("fcc"))^(3))/4 = 2/((4r//sqrt3)^(3))xx ((2sqrt2r)^(3))/4 = (2xx3sqrt3)/(64 r^(3)) xx (16sqrt2 r^(3))/4 = 3/8sqrt6 = 0.919`