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An element having atomic mass 60 has face-centered cubic unit cells.The edge length of the unit cell is 400pm.Find the density of the element.

$\begin{array}{1 1}6.2gcm^{-3}\\5.2gcm^{-3}\\7.2gcm^{-3}\\8.2gcm^{-3}\end{array} $

1 Answer

Unit-cell edge length =400pm=$400\times 10^{-10}$cm
Volume of unit cell =$(400\times 10^{-10}cm)^3=64\times 10^{-24}cm^3$
Mass of the unit cell =Number of atoms in the unit cell $\times$ Mass of each atom
Number of atoms in the fcc unit cell =$8\times \large\frac{1}{8}$$+6\times \large\frac{1}{2}$
$\Rightarrow 4$
Mass of one atom =$\large\frac{\text{Atomic mass}}{\text{Avogadro number}}$
$\Rightarrow \large\frac{60}{6.023\times 10^{23}}$
$\therefore$ Mass of the unit cell =$\large\frac{4\times 60}{6.023\times 10^{23}}$
$\therefore$ Density of unit cell =$\large\frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}$
$\Rightarrow \large\frac{4\times 60}{6.023\times 10^{23}}\times \frac{1}{64\times 10^{-24}}$
$\Rightarrow 6.2gcm^{-3}$
answered Aug 1, 2014 by sreemathi.v

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