Question

Calculate the density of diamond from the fact that it has a face-centered cubic structure with two atoms per lattice point and unit cell edge length of $3.569\times {10}^{-8}$ cm.

• A. 3.509 g/cm${}^3$
• B. 7.012 g/cm${}^3$
• C. 10.12 g/cm${}^3$
• D. None of the above

Answer 1

The correct option is A 3.509 g/cm${}^3$

$\rho =\frac{Z\times M}{N_A\times a^3}$

The lattice points in the FCC lattice are at the corners of the cube and face center.

Since it is given in the question that at each lattice point there are $2$ atoms,

So, the $z$ will be-

$Z=2\times 8\times \frac{1}{8}+2\times 6\times \frac{1}{2}$

$Z=8$

$M=12g/mol$ for carbon, $N_A=6.023\times {10}^{23}/mol$

$a=3.569\times {10}^{-8},a^3=(3.569\times {10}^{-8})c{m}^3$

$a^3=4.546\times {10}^{-23}c{m}^3$

$\rho =\frac{8\times 12}{6.023\times {10}^{23}\times 4.546\times {10}^{-23}}=3.506g/c{m}^3$

$\rho =3.5g/c{m}^3$

Hence, the correct option is A.