How many atoms are there in a cube based unit cell having one atom on each corner and two atoms on each body diagonal?

A cube has 8 corners, hence 8 atoms are positioned at each corner of the cube. However, these 8 atoms are distributed equally across 8 different units. Therefore, 
Atomic contribution at the corner = $8\times \frac{1}{8}=1$
In a cube, there are 4 body diagonals, and there are 2 atoms on each of them. These atoms are not shared by any other units because they are contained within the unit cell, Therefore,
Atomic contribution at the body diagonal = $2 \times 4 = 8$

So, the number of atoms in the cube-based unit cell is = $8 + 1 = 9$

Therefore, the correct option is D 9