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

Hint: To answer this question you should recall the formula for calculation of density of a lattice from the solid-state. Here the number of atoms in one face-centered lattice will be four. Now arrange these values in the formula to answer this question.

Let’s find the correct answer to this question -

The formula for the calculation of density can be written as,
$\rho (density) = \dfrac { Z \times M }{ { N }_{ 0} \times a^{ 3 } }$
Where,
Z = number of atoms in a unit cell
M = molecular mass of an atom (in grams)
${ N }_{ 0 }$ = Avogadro number
$a$ = unit cell edge length (in cm)

We have the following values,
Z = 4 $\times$ 2 = 8 (because one lattice point has two atoms in fcc unit cell)
M = 12 g (atomic mass of carbon)
${ N }_{0 }$ = 6.022 $\times 10^{ 23 }$
$a$ = 3.569 $\times 10^{ -8 }$ cm

Now, we will put all these values in the equation of density,
$\rho = \dfrac { 8 \times 12g }{ 6.022\times { 10 }^{ 23 } \times 3.569\times { 10 }^{ -8 }cm)^{ 3 } }$
$\rho = \dfrac { 96g }{ 27.4 { cm }^{ 3 } }$
$\rho = 3.509 g/{ cm }^{ 3 }$

Therefore, we can conclude that the correct answer to this question is option A.

Note: We should know that the diamond lattice (formed by the carbon atoms in a diamond crystal) consists of two interpenetrating face-centered cubic Bravais lattices, displaced along the body diagonal of the cubic cell by one quarter the length of the diagonal.