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Calcium crystallises in a face-centred cubic unit cell with a=0.556nm.Calculate the density if it contained 0.1% Schottky defects.

$\begin{array}{1 1}1.5440g/cm^3\\2.5440g/cm^3\\4.5440g/cm^3\\0.5440g/cm^3\end{array} $

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In case of Schottky defects =$4\times \big(1-\large\frac{0.1}{100}\big)$$=3.996$
Thus density can be determined using
$\rho=\large\frac{Mz}{N_0a^3}$
$a=0.556nm=0.556\times 10^{-9}m$
$\Rightarrow 0.556\times 10^{-7}cm$
$\rho$(with Schottky defects)=$\large\frac{40\times 3.996}{6.023\times 10^{23}\times (0.556\times 10^{-7})^3}$
$\Rightarrow 1.5440g/cm^3$
answered Aug 1, 2014 by sreemathi.v
 

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