SOLID STATE

space used
= V(particles) /V(crystal cell )

= (4/3)Ï€(a/2)3 / a3

= 52.36%

Coordination number: 8
space used
= V (particles) / V (crystal cell )
= [2×(4/3)πr3 ]/a3
= 2×4/3πr3
= 68.02%

Coordination number: 12

space used
= V (particles) /V (crystal cell )

= 2×(4/3)π(a/2)3 /(a3×1.633a×sin1200 )

= 74.05%

Chalcopyrite has a very similar structure to sphalerite. Cells with iron atoms (orange) alternate with cells containing copper (blue). This double cell means chalcopyrite is tetragonal, but the fact that the unit cell consists of two cubes means chalcopyrite mimics cubic forms very often. In particular it can occur as pseudo-tetrahedra (actually tetragonal disphenoids, but they look exactly like tetrahedra.)

In sphalerite, we have a face-centered cubic arrangement of sulfur (yellow) and zinc in the center of each tetrahedron (purple). At left is the structure illustrating the relationship to the tetrahedra. At right is the unit cell. Zinc atoms hidden behind sulfur atoms are shown in a lighter shade.

Is this really ZnS? We have four zinc atoms entirely within the unit cell. We have eight corner sulfur atoms, but they are shared among eight unit cells. There are six sulfur atoms in the centers of the faces, but they are each shared with a neighboring cell. So we have 8 x 1/8 + 6 x 1/2 = 4 sulfur atoms. There are an equal number of zincs and sulfur.

i couldnt get it you can also post at ag_2007@rediffmail.com

i had posted all
looking 4 more relevant

thanks for so clear visual scenes

in this type of unit cell the sphere at the center of the face is touching the end spheres
the radius of a sphere is r and the edge length is a
so the diagonal will be √2a

4r=√2a

=>a=2√2r

http://targetiit.com/iit_jee_forum/posts/18th_november_2008_583.html

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