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In a cubic lattice each edge of the unit cell is 400pm.Atomic weight of the element is 60 and its density is $6.25g/c.c$.Avogadro number =$6.023\times 10^{23}$.What will be the crystal lattice.

$\begin{array}{1 1}\text{Face-centred cubic}\\\text{Simple cubic}\\\text{Body centred cubic}\\\text{Face centred tetragonal}\end{array} $

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Let the number of atoms in a unit cell =$x$
mass of x-atoms i.e one unit cell $=\large\frac{60\times x}{6\times 10^{23}}$
Volume of the unit cell =(edge length)$^3$
$\Rightarrow (400\times 10^{-12}\times 100)^3$
$\Rightarrow (400\times 10^{-10})^3$
$\Rightarrow (4\times 10^{-8}cm)^3=64\times 10^{-24}cm^3$
Density =6.25=$\large\frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}$
$\therefore 6.25=\large\frac{60\times x}{6.023\times 10^{23}\times 64\times 10^{-24}}$
$x=\large\frac{6.25\times 6\times 64\times 10^{-1}}{60}$$=4$
Since the unit cell contains 4 atoms,so it is face centred cubic unit cell.
answered Aug 1, 2014 by sreemathi.v
 

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