Assuming that atoms are in a crystal structure and arranged as close-packed spheres, what is the ratio of the volume of the atoms to the volume available for the simple cubic structure? Assume a one-atom basis.

This answer is restricted. Please login to view the answer of this question.

Login Now

Solution: Each corner atom in a cubic unit cell is shared by a total number of eight unit cells so that the corner atom contributes only \( \frac{1}{8}\) of its effective part to a unit cell. Since there are in all 8 corners their total contribution is equal to \( \frac{8}{8}\) = 1.User Loaded Image | CSIT Guide

Therefore, a number of atoms per unit cell = 1.

From figure,

a = 2r

or, r = a/2

Therefore, volume occupied by the atom in the unit cell = \( \frac{4}{3} \pi r^{3}\) = \( \frac{4}{3} \pi \left( \frac{a}{2} \right)^3\)

Volume of unit cell = a3.

Thus, packing fraction = \( \frac{\frac{4}{3}\pi \left ( \frac{a}{2} \right)^3}{a^3} \) = \( \frac{\pi}{6} \) = 0.52 = 52%.

Therefore, 52% space of unit cell is occupied by the atoms. Here the atoms are loosely packed. Only Polonium at a certain temperature is known to exhibit such a structure. For example, KCL which has alternate icons of K and CL also behaves like a simple lattice as regards scattering of X-rays because the two icons are almost identical.

If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.
Leave your Answer:

Click here to submit your answer.

Discussion
0 Comments
  Loading . . .