91Ó°ÊÓ

Imagine a simple cubic array where the lattice points are occupied by spheres of radius \(r\) touching each other. Now, imagine that there is a space in the center, called the interior sphere (cubic hole), and we need to find its volume.

In a simple cubic array, the spheres in the array touch each other along the edge of the cubic unit cell. The diameter of a sphere is given by \(2r\). Since in this cubic array, the spheres are touching along the edge, the length of one edge of the cubic unit cell is equal to twice the sphere's radius: \(a = 2r\), where \(a\) represents the edge length of the unit cell.

We can now determine the relationship between the radius of the interior sphere (cubic hole) and the sphere radius in the array. In the cubic array, consider a tetrahedron formed by selecting a sphere at a corner and its three nearest neighbor spheres. The height of this tetrahedron intersects the center of the cubic hole, and it is perpendicular to one of the faces of the cube. Let's denote by \(R\) the radius of the cubic hole. The height of the tetrahedron can be found by using the Pythagorean theorem on one of the faces of the cube, which is an equilateral triangle with sides of length \(a = 2r\): \[h^2 = a^2 - (\frac{a\sqrt{3}}{2})^2\] Substitute \(a = 2r\): \[h^2 = (2r)^2 - (\frac{2r\sqrt{3}}{2})^2\] \[h = \sqrt{4r^2 - 3r^2}\] \[h = r\sqrt{1}\] \[h = r\] Now, we can see that the height of the tetrahedron from the vertex to the center is equal to the radius of the sphere in the array, \(r\). The height contains the radius of the sphere in the array and the radius of the interior sphere (cubic hole): \(h = r + R\). Since in our case, \(h = r\), the expression becomes: \[r = r + R\] Thus, \[R = 0\]

As we have found out that the radius of the interior sphere (cubic hole) is equal to 0, it means that there is no space for an interior sphere in the given simple cubic arrangement of spheres. Therefore, the volume of the interior sphere (cubic hole) is also 0.