A tetrahedral site in a close-packed lattice is formed by four spheres at the corners of a regular tetrahedron. This is equivalent to placing the spheres at alternate corners of a cube. In such a close-packed arrangement the spheres are in contact and if the spheres have a radius r, the diagonal of the face of the cube is 2r. The tetrahedral hole is inside the middle of the cube. Find the length of the body diagonal of this cube and then find the radius of the tetrahedral hole.

Now length of body diagonal of this cube = √3×√2×r =2.45r

Explanation:

a) length of body diagonal of this cube = √3×√2×r = 2.45 r

it can solved as following

length of body diagonal = √3a   (where a is edge length of cube)

length of face diagonal = √2a

But length of face diagonal = 2r

Therefore a = 2r /√2  = √2×r

Now length of body diagonal of this cube = √3×√2×r

metric time was an attempt to create a standard unit of time based off of 10, but it never took off an is not included in the metric system (si)