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An element having atomic mass 60 has face centred cubic unit cell. The edge length of the unit cell is 400 pm. Find out the density of the element?

1 Answer

Unit cell edge length = 400 pm = $400×10^{-10}$ cm
Volume of unit cell = $(400×10^{-10})^3 = 64×10^{-24} cm^3$
Mass of the unit cell = No. of atoms in the unit cell $\times$ mass of each atom
No. of atoms in fcc unit cell = $8\times \large\frac{1}{8}$$+6\times \large\frac{1}{2}$$=4$
$\therefore$ Mass of unit cell =$\large\frac{4\times 60}{6.023\times 10^{23}}$
Density of unit cell =$\large\frac{\text{mass of unit cell}}{\text{volume of unit cell}}=\large\frac{4\times 60}{6.023\times 10^{23}\times 64\times 10^{-24}}$$=6.2g/cm^3$
Hence (A) is the correct answer.
answered Jun 2, 2014 by sreemathi.v
 
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