Question

The fraction of the total volume occupied by the atoms present in the simple cube is:
A) $\dfrac{\pi }{4}$
B) $\dfrac{\pi }{6}$
C) $\dfrac{\pi }{{3\sqrt 2 }}$
D) $\dfrac{\pi }{{4\sqrt 2 }}$

Answer 1

Hint: A simple cube has a structure in which there is only one lattice point present at each corner of the cubic shaped unit cell. One can use the volume formula for the cube and put in the relevant values needed for the cube to get the volume.

Complete step by step answer:
1) First of all let's understand the simple cube structure where all the sides of a cube are equal. Now let's consider the edge length of the cube ${\text{a}}$ and the radius of the atom as ${\text{r}}$.
2) As in the structure of a simple cube the number of atoms is present in it as $ = \dfrac{1}{8} \times 8 = 1$.
3) Hence the volume occupied by an atom in a simple cube is $ = 1 \times \dfrac{4}{3}\pi {r^3}$.
4) As there are atoms present in the corners of the simple cube and they will be touching each other we can say that ${\text{a = 2r}}$. We can modify this equation as ${\text{r = }}\dfrac{{\text{a}}}{2}$.
As we know the formula for the volume of the simple cube is ${a^3}$ we can say the formula for the fraction of the simple cube occupied is a follows,
${\text{The fraction of the simple cube = }}\dfrac{{\dfrac{4}{3}\pi {r^3}}}{{{a^3}}}$
Now let's put the value of ${\text{r}}$ which we have got earlier and do calculation,
${\text{The fraction of the simple cube = }}\dfrac{{\dfrac{4}{3}\pi {{\left( {\dfrac{a}{2}} \right)}^3}}}{{{a^3}}}$
Now doing the cube of bracket value we get,
${\text{The fraction of the simple cube = }}\dfrac{{\dfrac{4}{3}\pi \times \dfrac{{{a^3}}}{8}}}{{{a^3}}}$
As there are the same values of ${a^3}$ in numerator and denominator we can cancel them out and get,
${\text{The fraction of the simple cube = }}\dfrac{4}{3} \times \pi \times \dfrac{1}{8}$
Now by doing the calculation we get the value as,
${\text{The fraction of the simple cube = }}\dfrac{\pi }{6}$
Therefore, the fraction of the total volume occupied by the atoms present in the simple cube is $\dfrac{\pi }{6}$ which shows option (B) as the correct choice.

Note:
The value of the fraction of the total volume occupied by the atoms present in the simple cube is also called the packing fraction of the simple cube. The value of $\dfrac{\pi }{6}$ when calculated by putting the pi value comes ${\text{0}} \cdot {\text{52}}$.