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A metal of density $7.5\times 10^3kgm^{-3}$ has an fcc crystal structure with lattice parameter $a=400$pm.Calculate the number of unit cells present in 0.015kg of the metal

$\begin{array}{1 1}2.125\times 10^{22}\\3.125\times 10^{22}\\1.125\times 10^{22}\\4.125\times 10^{22}\end{array} $

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Let $V_1$ be the volume of the unit cell and $V_2$ that of the metal sample.
Given,$V_1=(400\times 10^{-12}m)^3=64\times 10^{-30}m^3$
$\quad\;=\large\frac{0.015}{7.5\times 10^{-3}}$
$\quad\;=2\times 10^{-6}m^3$
Number of unit cells =$\large\frac{2\times 10^{-6}}{64\times 10^{-30}}$
$\Rightarrow 3.125\times 10^{22}$
answered Aug 1, 2014 by sreemathi.v

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