The unit cell in a body centered cubic lattice is given. Each sphere has a radius r and the cube has a side a. What fraction of the total cube volume is empty?


closed as off-topic by jerepierre, Klaus-Dieter Warzecha, MaxW, ron, orthocresol Mar 31 '16 at 17:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.


Assuming the $a$ is large enough to accommodate a BCC lattice with spheres of radius $r$, we may define

$V_s = \frac{4}{3}\pi r^3$ is the volume of the sphere

$V_c = a^3$ is the volume of the cube.

in a BCC lattice, there are 8 corner points, and one central. Because the lattice points on the corners are defined with respect to the center of each of the spheres, we know that each of the corner spheres will contribute $\frac{1}{8}$ of their total volume to the occupied volume of the unit cell. In addition, the sphere at the center will contribute its entire volume to the occupied volume. We may then write the total volume occupied in the unit cell as

$V_{occ} = 8\left(\frac{1}{8}\right)V_s + V_s =2V_s$

We know that the total volume of the unit cell is that of the cube, meaning that the unoccupied (empty) volume of the unit cell may be written

$V_{unocc} = V_c - V_{occ} = a^3 - \frac{8}{3}\pi r^3$

Wherefore we may write the empty fraction as

$\frac{V_{unocc}}{V_c} = \frac{a^3 - \frac{8}{3}\pi r^3}{a^3} = 1 - \frac{8\pi r^3}{3a^3}$


Not the answer you're looking for? Browse other questions tagged or ask your own question.