A metal crystallizes into two cubic system-face centred cubic (fcc) and body centred cubic (bcc) whose unit cell lengths are 3.5 and 3.0Å respectively. Calculate the ratio of densities of fcc and bcc.

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Solution

fcc unit cell length = $3.5 ˙ A$
bcc unit cell length = $3.0 ˙ A$
Density in fcc = $\frac{n_{1} \times s t o m i c w e i g h t}{V_{1}} \times$ Avogardo number
Density in bcc = $\frac{n_{2} \times s t o m i c w e i g h t}{V_{2}} \times$ Avogardo number
$\frac{D f c c}{D b c c} = \frac{n_{1}}{n_{2}} \times \frac{V_{2}}{V_{1}}$
$n_{1}$ for fcc = 4; Also $V_{1} = a_{3} = (3.5 \times 10^{- 8})^{3}$
$n_{2}$ for bcc = 2; Also $V_{2} = a_{3} = (3.0 \times 10^{- 8})^{3}$
= 1.259

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A metal crystallizes into two cubic system-face centred cubic (fcc) and body centred cubic (bcc) whose unit cell lengths are 3.5 and 3.0Å respectively. Calculate the ratio of densities of fcc and bcc.

fcc unit cell length = 3.5˙A bcc unit cell length = 3.0˙A Density in fcc = n1×stomic weightV1× Avogardo number Density in bcc = n2×stomic weightV2× Avogardo number DfccDbcc=n1n2×V2V1 n1 for fcc = 4; Also V1=a3=(3.5×10−8)3 n2 for bcc = 2; Also V2=a3=(3.0×10−8)3 = 1.259