Question

For a simple cubic array, solve for the volume of an interior sphere (cubic hole) in terms of the radius of a sphere in the array.

Short answer

For a simple cubic array, the volume of an interior sphere (cubic hole) will be $\frac{4}{3}\pi {\left[\left(\sqrt{3}-1\right)x\right]}^3$ where x is the radius of a sphere in the array.

Step-by-step solution

Structure

Let, the radius of sphere in the array be x. Now we can assume for a simple cubic structure, the cube edge = 2x

 Step 2: Determine the volume

For a simple cubic array

${\left(Body diagonal\right)}^2={\left(Face diagonal\right)}^2+{\left(Cube edge length\right)}^2$

Face diagonal

$$\begin{gathered} =\sqrt{{\left(2x\right)}^2+{\left(2x\right)}^2} \\ =\sqrt{4x^2+4x^2} \\ =\sqrt{8x^2} \\ =x\sqrt{8} \end{gathered}$$

Therefore, we can write

$$\begin{gathered} {\left(Body diagonal\right)}^2={\left(x\sqrt{8}\right)}^2+{\left(2x\right)}^2 \\ {\left(Body diagonal\right)}^2=8x^2+4x^2 \\ Body diagonal=\sqrt{12x^2} \\ Body diagonal=2\sqrt{3}x \end{gathered}$$

The diameter of the hole

$$\begin{gathered} =Body Diagonal-2\times Atom radius at corner \\ =2\sqrt{3}x-2x \end{gathered}$$

Therefore, the radius of the hole will be

$$\begin{gathered} =\frac{2\sqrt{3}x-2x}{2} \\ =\left(\sqrt{3}-1\right) \end{gathered}$$

Volume of the hole will be

$\frac{4}{3}\pi {\left[\left(\sqrt{3}-1\right)x\right]}^3$