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

$(a)\;9\qquad(b)\;10\qquad(c)\;11\qquad(d)\;12$

There are four body diagonals and atoms on the body diagonal are not shared by any other unit cell.
$\therefore$ Atoms at the corner = $8\times \large\frac{1}{8}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 1$
Atoms on the body diagonal = $2\times4$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; =8$
$\therefore$ Total number of atoms = 1+8 = 9