Number of atoms per unit cell if atoms are present at the corner of the unit cell and two atoms at each body diagonal

Ans: 9

how the answer came

I want step by step detailed explanation

Answer is 9 Nine.

First imagine a cube.

8 cubes can share a corner.

1 atom in one corner.

So 8 unit cells surround this corner (ie- tis one cell lies in eight unit cells)

So This unit cell has 1/8 cells at corner.

There are eight corners and same in each case.

So number of cells in this unit cell by corner atom is (1/8)×8= 1

Imagine body diagonals of a cube.

If you are perfect in imagination there are 4body diagonals each having 2 atoms(given in question)

Now atoms present don't share with any other unit cells since these are inside unit cell.

Total atoms by body diagonals are 4×2= 8

Total atoms by unit cell is 8( body diagonals) +1(corners)

=9

Or in simple chemistry,

Each corner of unit cell is occupied by 1/8th of an atom.

Total atoms in unit cell due to 8 corners = 8 × 1/8 = 1 atom

1 unit cell contain 4 body diagonals. Therefore total atoms due to body diagonals = 2 × 4 = 8 atoms

Total atoms in this unit cell = 1 + 8 = 9 atom