1000 Solved Problems in Modern Physics

300 5 Solid State Physics

5.2 Volume of the unit cell=a^3. Since there are two atoms per unit cell, 8× 1 / 8
for the corner atoms and 1×1 for the centre atom,

Volume= 2 ×

4

3

πr^3

Since the body diagonal atoms touch one another,

4 r=a

√

3

Volume of atoms in terms of a is

2 ×

4

3

πr^3 = 2 ×

4

3

π[a

√

3 /4]^3

=

√

3 πa^3 / 8

Or the fraction of the volume occupied by the body-centred cubic structure is√

3 π/8.

5.3

2 dsinθ=nλ

d 1 =

1 .λ

2sinθ

=

0. 1

2sin4◦

= 0 .717 nm

d 2 =

0. 1

2sin8◦

= 0 .359 nm

5.4 nλ=

2 a

(h^2 +k^2 +l^2 )^1 /^2

sinθ=

2 × 0. 4

(1^2 + 12 + 12 )^1 /^2

sinθ

sinθ=

0. 3

√

3

0. 8

= 0. 6495

θ= 40. 5 ◦

5.5 2 dsinθ=nλ

d=

1 .λ

2sinθ

=

0. 16

2sin30◦

= 0 .136 nm

Forn=2,

sinθ=

2 × 0. 16

2 × 0. 136

= 1. 176

a value which is not possible. Thus higher order reflections are not possible.

5.6 The de Broglie wavelength for electrons is calculated from

λ=

√

150

V

=

√

150

54

= 1. 66 A ̊

Bragg’s equation will be satisfied for neutrons of the same wavelength.

λ=

0. 286

√

E

A ̊