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).
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Let ‘a’ be the edge length of the unit cell and r be the radius of sphere.
As sphere are touching each other
Therefore a = 2r
No. of spheres per unit cell = 1/8 × 8 = 1
Volume of the sphere = 4/3 πr3
Volume of the cube = a3= (2r)3= 8r3
∴Fraction of the space occupied =1/3πr3/ 8r3= 0.524
∴% occupied = 52.4 %
Body-centred cubic unit cell

In body centred cubic unit cell
In Triangle EFD
Let DF= b
and we know that
ED=EF= a (edge length)
Now,
b2= a2+ a2= 2a2
In [\Delta AFD]
Let, AF = c
We know that
FD = b
& AD = a (edge length)
Now,
c2= a2+ b2=a2+ 2a2= 2a2
or c =√3 a
we know that c is body diagonal.As the sphere at the centre touches the sphere at the corner. Therefore body diagonalc = 4r
i.e.√3 a = 4r
or r = (√3/4)a
or a = 4r /√3
∴Volume of the unit cell = a3= (4r / √3)3= 64r3/ 3√3
No. of spheres in bcc = 2
∴volume of 2 spheres = 2×4/3πr3

Face-Centred Cubic (hcp and ccpStructures)

Let ‘r’ be the radius of sphere and ‘a’ be the edge length of the cube
As there are 4 sphere in fcc unit cell
∴Volume of four spheres = 4 (4/3πr3)
In fcc, the corner spheres are in touch with the face centred sphere. Therefore, face diagonal AD is equal to four times the radius of sphere
AC= 4rBut from the right angled triangle ACDAC = √AD2+ DC2=√a2+a2=√2a
4r = √2a
or a = 4/√2 r
∴volume of cube = (2/√2 r)3
As sphere are touching each other
Therefore a = 2r
No. of spheres per unit cell = 1/8 × 8 = 1
Volume of the sphere = 4/3 πr3
Volume of the cube = a3= (2r)3= 8r3
∴Fraction of the space occupied =1/3πr3/ 8r3= 0.524
∴% occupied = 52.4 %
Body-centred cubic unit cell

In body centred cubic unit cell
In Triangle EFD
Let DF= b
and we know that
ED=EF= a (edge length)
Now,
b2= a2+ a2= 2a2
In [\Delta AFD]
Let, AF = c
We know that
FD = b
& AD = a (edge length)
Now,
c2= a2+ b2=a2+ 2a2= 2a2
or c =√3 a
we know that c is body diagonal.As the sphere at the centre touches the sphere at the corner. Therefore body diagonalc = 4r
i.e.√3 a = 4r
or r = (√3/4)a
or a = 4r /√3
∴Volume of the unit cell = a3= (4r / √3)3= 64r3/ 3√3
No. of spheres in bcc = 2
∴volume of 2 spheres = 2×4/3πr3

Face-Centred Cubic (hcp and ccpStructures)

Let ‘r’ be the radius of sphere and ‘a’ be the edge length of the cube
As there are 4 sphere in fcc unit cell
∴Volume of four spheres = 4 (4/3πr3)
In fcc, the corner spheres are in touch with the face centred sphere. Therefore, face diagonal AD is equal to four times the radius of sphere
AC= 4rBut from the right angled triangle ACDAC = √AD2+ DC2=√a2+a2=√2a
4r = √2a
or a = 4/√2 r
∴volume of cube = (2/√2 r)3
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