Question

Packing fraction of diamond is
A) 0.74
B) 0.68
C) 0.52
D) 0.34

Answer 1

In the structure of diamond cubic crystal the unit cell present is fcc unit cell.The relation between r and a for diamond is $r=\dfrac{\sqrt{3}a}{8}$.

Complete answer:
So in the question we are supposed to answer the packing fraction of the diamond structure and we know that the diamond is having the cubic structure and hexagonal structure but cubic structure is the common one and we usually discuss about the cubic structure in which the unit cell dimension present in then is fcc lattice.
The coordination number of diamonds is four.
But there are 8 atoms in its cubic structure, four atoms contributed by the fcc lattice and four atoms are arranged alternatively in the tetrahedral voids.
$\text{Total number of atoms= fcc + atoms in tetrahedral voids}$
$\text{Total no}\text{. of atoms=8}\times \dfrac{1}{8}+6\times \dfrac{1}{2}+4$
$\text{Total no}\text{. of atoms=1+3+}4=8$
Hence the total number of atoms in the unit cell of the diamond is 8.
And now we have to calculate the packing fraction of the diamond.
Packing fraction gives the fraction of total space filled by the particles in a crystal lattice,
If we multiply the packing fraction with 100.then we will get packing efficiency, which is the percentage of the total space filled by the particles.
Packing fraction (P.F), is calculated by volume occupied by the number of spheres in the unit cell divided by volume of a unit cell.
$\text{P}\text{.F=}\dfrac{\text{Volume}\,\text{occupied}\,\text{by}\,\text{number}\,\text{atoms}\,\text{in}\,\text{the}\,\text{unit}\,\text{cell}}{\text{Volume}\,\text{of}\,\text{a}\,\text{unit}\,\text{cell(}{{\text{a}}^{\text{3}}}\text{)}}$
So we know there is 8 atoms and we assume that the atoms are spherical in shape and we give the volume of the one as,
$V=\dfrac{4}{3}\pi {{r}^{3}}$
$\text{P}\text{.F=}\dfrac{8\times \dfrac{4}{3}\pi {{r}^{3}}}{{{a}^{3}}}$
We know how a the edge length and r the atomic radius is related to by the equation,
$r=\dfrac{\sqrt{3}a}{8}$
Substitute the value of r in the P.F equation and we get,
$\text{P}\text{.F=}\dfrac{(8\times \dfrac{4}{3}\pi )\times \left( \dfrac{{{(\sqrt{3}a)}^{3}}}{8} \right)}{{{a}^{3}}}$
And by doing the calculations, we get a simpler equation in which the terms ‘r’ and ‘a’ are not present.
$P.F=\dfrac{\sqrt{3}\pi }{16}$
The value of $\sqrt{3}=1.7320$and the value of $\pi =3.14$
And hence substituting the values we get,
$P.F=\dfrac{1.7320\times 3.14}{16}=0.339905\approx 0.34$

So the correct answer for the question is option (D).

Note:
As we consider that diamond is having cubic structure and we know that the packing fraction for ccp and hcp is 0.74.
Hence there is a chance of opting the wrong answer, since for diamond the packing fraction is very much less than the normal ccp lattice, due to the atoms present in tetrahedral voids affects the packing of ccp and more space becomes vacant. Hence we should take care of this fact that the packing fraction of diamond is not similar to the packing fraction value for ccp.