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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?

Can you answer this question?

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