A metal crystallizes into two cubic phases, face-centred cubic and body-centred cubic, which have unit cell lengths `3.5` and `3.0 A`, respectively. Calculate the ration of densities of fcc and bcc.

Explanation

`rho = (n xx M_(m))/(N_(o)xx a^(3))`
For FCC
`n = 4, a = 3.5 Å`
`therefore rho_((fc c))=(4xx M_(m))/(N_(o)xx(3.5)^(3)) " "` ….(i)
For bcc lattice
`n = 2, a = 3.0 Å`
`rho_(bc c)=(2xx M_(m))/(N_(A)xx(3.0)^(3)) " "` ....(ii)
Comparing Eqs. (i) and (ii), we get
`therefore (rho_((fc c)))/(rho_((bc c)))=(4)/(2)xx(3^(3))/((3.5)^(3))=(4xx3xx3xx3)/(2xx3.5xx3.5)=1.259`