Question

The fraction of total volume occupied by the atoms in a simple cube is:
A.$\dfrac{\pi }{4}$
B.$\dfrac{{\sqrt 2 \pi }}{8}$
C.$\dfrac{{\sqrt 2 \pi }}{6}$
D.$\dfrac{\pi }{6}$

Answer 1

Hint: We know that the location of atoms in the simple cubic lattice is only on the corners of the cube. So, the particle touches each other along the edge. The number of atoms in a simple cubic unit cell is one.

Complete step by step answer:
Let's first understand the number of atoms in a simple cubic unit cell. We know that each cubic unit cell possesses eight atoms on its corners. So, the number of atoms in one unit cell is equal to $8 \times \dfrac{1}{8} = 1$ atom.

Now, we consider a as the edge length of the cube and r as the radius of the atom.

So, the volume occupied by the atom=$1 \times \dfrac{4}{3}\pi {r^3}$

Also, we know that in simple cube particles touch each other at corners. So, we can say that,

 $ \Rightarrow a = 2r \Rightarrow r = \dfrac{a}{2}$

Now, we have put the value of r in the volume of the atom.

So, the volume of the atom=$1 \times \dfrac{4}{3}\pi {\left( {\dfrac{a}{2}} \right)^3}$

The volume of the cube will be = ${a^3}$

So, the fraction of the volume occupied by the atom in the cube can be calculated by dividing the volume of the atom by the cubic volume.

Fraction of volume occupied by the atom=$\dfrac{{{\rm{Volume}}\,{\rm{of}}\,{\rm{atom}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{cube}}}}$

$ \Rightarrow $ Fraction of volume occupied by the atom$ = \dfrac{{\dfrac{4}{3}\pi {{\left( {\dfrac{a}{2}} \right)}^3}}}{{{a^3}}} = \dfrac{{4\pi {a^3}}}{{3 \times 8{a^3}}} = \dfrac{\pi }{6}$

Therefore, the atom of the simple cubic unit cell occupies $\dfrac{\pi }{6}$ of the cube.

Hence, the correct choice is option D.

Note: Always remember that there is always the existence of some free space in whatever way constituent particles such as atoms or molecules are packed. The free space is termed as voids. Packing efficiency gives the percentage of total space occupied by the particles.