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A metal crystallises into two cubic phases,face centred cubic (FCC) and body centred cubic (BCC),whose unit cell length are 3.5 and $3.0A^{\large\circ}$,respectively.Calculate the ratio of densities of FCC and BCC.

$\begin{array}{1 1}1.36\\1.26\\1.46\\1.16\end{array} $

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Answer : 1.26
The expression of density is
$\rho =\large\frac{N}{a^3}\big(\frac{M}{N_A}\big)$
For face-centred cubic unit cell,N=4.For body-centred cubic unit cell,N=2
Hence,$\large\frac{\rho_{FCC}}{\rho_{BCC}}=\frac{N_{FCC}a^3_{BCC}}{a^3_{FCC}N_{BCC}}=\frac{4}{(3.5A^{\large\circ})^3}\frac{(3.0A^{\large\circ})^3}{2}$$=1.26$
answered Jul 17, 2014 by sreemathi.v
 

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