SELECT * FROM question_mgmt as q WHERE id=1742 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=56 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) body-centred cubic
(iii) face-centred cubic (with the assumptions that atoms are touching each other).
(i) Simple cubic:
Suppose the edge length of the unit cell = a
&
Radius of the sphere = r
Then,since the sphere are touching each other along the edge,therefore a = 2r
Now there are 8 spheres at the corners of the cube & each sphere at the corner is shared by 8 unit cells & the contribution per unit cell is 1/8 so that
Number of spheres per unit cell is 8 x 1/8 = 1
Volume of sphere =4/3πr3 & volume of cube = a3 = (2r)3 = 8r3
Now packing efficiency = (volume of one sphere / total volume of cubic unit cell) x 100
Or
(4/3 πr3 / 8r3) x 100 = 52.4%
Therefore the volume occupied in simple cubic arrangement = 52.4%
(ii) Body centered cubic:
Let us suppose the edge leght = a & radius of each sphere = r then there are 8 spheres at the corners & 1 in the body of unit cell
Therefore number of spheres per unit cell = (8 x1/8) + 1 = 2
Now volume of unit cell = a3 = (4r / √3)3
and volume of a sphere = 4 / 3πr3
Total volume of two spheres = 2 x 4/3πr3
Packing efficiency = (volume of two spheres in unit cell/total volume of unit cell ) x 100
= (2 x 4/3πr3 / (4r/√3)3 ) x 100 = 68%
Therefore volume occupied in bcc arrangement = 68%
(iii) Face centered:
let us suppose the edge length of the unit cell = a
Radius of each sphere = r
Now there are 8 spheres at the corner & 6 at the faces
Therefore number of spheres in unit cell = (8 x 1/8 + 6 x1/2) = 4
From the arrangement of fcc, we get a = 2√2r
Now volume of a unit cell = a3 = (2√2r)3 = 16√2r3
Total volume of 4 spheres = 4 x 4/3 πr3 = 16/3 πr3
Packing efficiency = (volume of 4 spheres in the unit cell/total volume of unit cell) x 100
= (16/3 πr3 /16√2r3) x 100 = 74%
Therefore volume occupied in fcc = 74%
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