Gaurav
8 years ago

# Simple cubic unit cell

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