Solutions Manual for Solid State Physics An Introduction to Theory 1st Edition by Joginder

Solution Manual Solid State Physics: An Introduction to Theory

Chapter 1 Problem 1.1: Solution: Operate the operator S ni R n  S mi R n  on the position vector r to write

S

ni

R n  S mi R n  r = S ni R n  (Smi r + R n )

= S ni S mi r + S ni R n + R n = Sni Smi Sni R n + R n 

(S1.1)

Problem 1.2: Solution: Operate the operator S ni R n  on the position vector r to get the transformed vector r  , i.e.,

S

ni

R n  r = r  = S ni r + R n

(S1.2)

It can be written as

S ni r = r  − R n

(S1.3)

Operate the above equation by S −ni1 from the left side to write

S −ni1 S ni r = S −ni1 r  − S −ni1 R n r = S −ni1 r  − S −ni1 R n =

S − S R  r −1 ni

−1 ni

(S1.4)

n

From the definition of the inverse transformation the above equation can also be written as r=

S

R n  r −1

ni

(S1.5)

From Eqs. (S1.4) and (S1.5) one immediately write

S



R n  = S −ni1 − S −ni1 R n −1

ni



(S1.6)

Problem 1.3: Solution: In the bcc structure there are two atoms in a cube with edge a (see Fig. 1.14).

Therefore, the volume per atom is a 3 / 2 . Alternately the volume per atom is given by

V0 = a1  a 2  a 3

(S1.7)

For a bcc structure from Eq. (1.37) one can write

a2  a3 =

=

(

)(

1 2 ˆ ˆ ˆ a − i1 + i 2 + i 3  ˆi1 − ˆi 2 + ˆi 3 4

(

1 2 ˆ ˆ a i1 + i 2 2

)

)

Perform the dot product with a1 from the left side to write

a1  a 2  a 3 =

=

(

)(

1 3 ˆ ˆ ˆ ˆ ˆ a i1 + i 2  i1 + i 2 − i 3 4

)

a3 2

Problem 1.4: .Solution: The angle between two primitive vectors a1 and a 2 can be calculated from their dot

product defined as

a1  a 2 = a1 a 2 cos cos =

a1  a 2 a1 a 2

(S1.8) (S1.9)

Here  is the angle between the two vectors a1 and a 2 .

From Eq. (1.37) it is straightforward to prove that

a1 = a 2 = 3 a/2

(S1.10)

Substitute the expressions for a1 and a 2 for bcc structure from Eq. (1.37) in Eq. (S1.9) and use Eq. (S1.10) one gets

cos = Hence

− a2 / 4 1 = − = − 0.33 2 3 3a / 4

 = cos − 1 (− 0.33) = 109.47 o or 109 o 28

Problem 1.5: Solution: In the fcc structure there are four atoms in a cube with edge a (see Fig. 1.19a).

Therefore, the volume per atom is a 3 / 4 . Alternately the volume per atom is given by

V0 = a1  a 2  a 3

(S1.11)

For fcc structure from Eq. (1.38) one can write

a 2  a3 =

(

)(

1 2 ˆ ˆ a i 2 + i 3  ˆi 3 + ˆi1 4

=

(

1 2 ˆ ˆ ˆ a i1 + i 2 − i 3 4

)

)

Perform the dot product with a1 from the left side to write

V0 = a1  a 2  a 3 =

(

)(

1 3 ˆ ˆ ˆ ˆ ˆ a i1 + i 2  i1 + i 2 − i 3 8

)

= a3 / 4 Problem 1.6: Solution: The angle between two primitive vectors a1 and a 2 can be calculated from the relation

cos =

a1  a 2 a1 a 2

(S1.12)

From Eq. (1.38) it is straightforward to prove that

a1 = a 2 = a/ 2

(S1.13)

Substitute the expressions for a1 and a 2 for fcc structure from Eq. (1.38) in Eq. (S1.12) and use Eq. (S1.13) one gets

cos = Hence

a2 / 4 1 = 2 2 a /2

 = cos − 1 (1 / 2 ) = 60 o

Problem 1.7: Solution: a) Packing fraction in bcc structure In a bcc structure there are two atoms in a cube with side a (see Fig. 1.14). Let r be the radius of an atom in a bcc structure, then in the close packing state three atoms along the diagonal of the cube should touch each other. Therefore, 4r = 3 a

(S1.14)

Hence the volume of an atom becomes V0 =

4 3  3 3 r = a 3 16

(S1.15)

The packing fraction is given by fp =

(

)

2  3 / 16 a 3  3 = = 0.68 8 a3

(S1.16)

b) Packing fraction in fcc structure In a fcc structure there are four atoms in a cube with side a (see Fig. 1.19a). Let r be the radius of an atom in a fcc structure, then in the close packing state three atoms along the diagonal of the basal plane of the cube should touch each other. Therefore, 4r = 2 a

(S1.17)

Hence the volume of an atom becomes V0 =

4 3  r = a3 3 12 2

(S1.18)

The packing fraction is given by

fp =

(

)

4  / 12 2 a 3  = = 0.74 3 a 3 2

(S1.19)

Problem 1.8: Solution: From Eq. (1.40) one can write  1 3 ˆ  ˆ a 2  a 3 =  − a ˆi1 + a i 2   c i 3 2 2   =

ac ˆ i2 + 2

(S1.20)

3ac ˆ i1 2

The volume per atom becomes ac V0 = a1  a 2  a 3 = a ˆi1   ˆi 2 +  2 =

3acˆ  i1  2 

3a2 c 2

(S1.21)

The same result can be obtained from the lattice vectors of the hexagonal structure given by Eq. (1.41).

Problem 1.9: Solution: In the hcp structure there are 6 atoms in the unit cell shown in Fig. 1.23b. The volume of the unit cell of hcp structure V (Fig.1.23b) is three times the volume of the primitive cell, i.e., V=

3 3 2 a c 2

(S1.22)

In the close packing the atoms in the basal plane of the unit cell touch each other. Therefore, twice the radius of an atom must be equal to the lattice vector in the basal plane, i.e.,

2r = a

(S1.23)

4 3 a r = 8   =  a 3 3 2 3

Volume of six atoms in the unit cell = 6

Packing fraction f p =

 a3 2

3 3 a c/2

In an ideal hcp structure

=

2 a 3 3c

(S1.24)

c = a

8 3

(S1.25)

Substitute Eq. (S1.25) in Eq. (S1.24) to get fp =

 3 2

= 0.74

(S1.26)

Problem 1.10: Solution: The planes (100), (200) and (110) in a sc structure are drawn in Fig. S1.1 below

Chapter2 Problem 2.1: Solution: Substitute for R n from Eq. (1.5) in Eq. (2.20) one writes

A L (K ) =

 e 

− K  ( n1 a1 + n 2 a 2 + n 3 a 3 )

n1 , n 2 , n 3

=  e -  n1 K  a1  e -  n 2 K  a 2  e -  n 3 K  a3 n1

n2

(S2.1)

n3

Therefore the intensity of lattice scattering of the solid becomes

I L = A L (K ) =  e 2

n1

2 -  n1 K  a1

2

2

e n2

-  n 2 K  a 2

e n3

-  n 3 K  a 3

(S2.2)

Let N 1 , N 2 and N 3 be the lattice points along the three directions. It can be easily proved that

N2

e 

- n1 K  a1

= e -  a1  K

e

-

1  N1 (a1  K ) 2 -

n1 = 1

1

e 2

 (a1  K )

1  sin  N 1 a1  K  2  1  sin  a1  K  2 

Therefore

N1

1  sin 2  N1 a1  K  2  = 1  sin 2  a1  K  2 

2

e 

- n1 a1  K

n1 = 1

(S2.3)

This is a peaked function and the maximum occurs when

a1  K = 2  N1

(S2.4)

It can further be shown that the height of the peak ( maximum value) is N 12 . Similarly one can prove that

N2

2

e 

- n 2 a 2  K

n 2 =1

N3

2

e 

n 3 =1

- n 3 a 3  K

1  sin 2  N 2 a 2  K  2  = 1  sin 2  a 2  K  2 

(S2.5)

1  sin 2  N 3 a 3  K  2  = 1  sin 2  a 3  K  2 

(S2.6)

These function exhibit maxima only when

a 2  K = 2  N 2

(S2.7)

a 3  K = 2  N 3

(S2.8)

The heights of the peaks corresponding to (S2.5) and S2.6) are N 22 and N 32 . Eqs. (S2.4), (S2.7) and (S2.8) are the Laue diffraction conditions and correspond to the maximum scattering intensity I L .

Problem 2.2: Solution: We know that a b  = 2   

(S2.9)

According to the above equation a1 is normal to both b 2 and b 3 . Therefore, one can write

a1  b 2  b 3 Or

a1 = K 1 b 2  b 3

(S2.10)

where K 1 is a constant. From Eq. (S2.9) one can write

a1 b1 = 2 

(S2.11)

Substitute a1 from Eq. (S2.10) in Eq. (S2.11) we get K1 =

2 b1  b 2  b 3

Substitute the value of K 1 in Eq. (S2.10) one gets a1 = 2 

b2  b3 b1  b 2  b 3

(S2.12)

Similarly one can obtain expressions for a 2 and a 3 as a 2 = 2

b 3  b1 b1  b 2  b 3

(S2.13)

a3 = 2

b1  b 2 b1  b 2  b 3

(S2.14)

Problem 2.3: Solution: From Eq. (1.37) one can write

a 2  a3 =

(

)(

1 2 ˆ ˆ ˆ a − i1 + i 2 + i 3  ˆi1 − ˆi 2 + ˆi 3 4 =

(

1 2 ˆ ˆ a i1 + i 2 2

)

) (S2.15)

b1 = 2 

(

a 2  a3 2 ˆ ˆ = i1 + i 2 a1  a 2  a 3 a

)

(S2.16)

Here we have used the average volume of an atom in bcc structure from Problem 1.3. Similarly one can obtain the expressions for b 2 and b 3 and are given as

b2 =

2 ˆ ˆ i2 + i3 a

(

)

(S2.17)

b3 =

2 ˆ ˆ i 3 + i1 a

(

)

(S2.18)

The above primitive vectors in the reciprocal space are shown in Fig. S2.1.

Problem 2.4: Solution: With the help of Eq. (2.82) one can write  1  3 a 2  a 3 =  − a ˆi1 + a i 2   c ˆi 3 2  2 

=

1 3 ˆ ac ˆi 2 + ac i1 2 2

Therefore the primitive reciprocal lattice vector b 1 is given as

1 ˆ 3 ˆ ac i 2 + ac i1 a2  a3 2 b1 = 2  = 2 2 a1  a 2  a 3 3 2 a c 2 b1 =

2  ˆ 1 ˆ   i1 + i 2  a  3 

(S2.19)

Similarly one can obtain

b2 =

2 2 ˆ i2 a 3

(S2.20)

b3 =

2 ˆ i3 c

(S2.21)

The primitive reciprocal lattice vectors are drawn in Fig. 2.12. The general reciprocal lattice vector is defined as G p = p1 b 1 + p 2 b 2 + p 3 b 3

=

2  ˆ 2 ˆ 1 ˆ  2 2 ˆ p1  i1 + i 2  + p2 i2 + p3 i3 a a c 3  3 

=

2 2 1 (p1 + 2 p 2 ) ˆi 2 + 2  p 3 ˆi 3 p1 ˆi1 + a a 3 c

(S2.22)

(S2.23)

Problem 2.5: Solution: From Eq. (2.85) one can write a c  a a c   a a 2  a 3 =  − ˆi1 + ˆi 2 + ˆi 3    ˆi1 − ˆi 2 + ˆi 3  2 2  2 2 2   2

=

(

ac ˆ ˆ i1 + i 2 2

)

With the help of Eqs. (2.85) and (S2.24) one can prove that

(S2.24)

a1  a 2  a 3 = a 2 c/2

(S2.25)

From Eqs. (S2.24) and (S2.25), it is straightforward to prove that b1 = 2 

(

a 2  a3 2 ˆ ˆ = i1 + i 2 a1  a 2  a 3 a

)

(S2.26)

Similarly one can prove that b2 =

2  ˆ a ˆ   i2 + i3  a  c 

(S2.27)

b3 =

2  ˆ a ˆ   i1 + i 3  a  c 

(S2.28)

The primitive lattice vectors in the reciprocal space, given by Eqs. (S2.26) – (2.28) are shown in Fig. 2.13b).

Problem 2.6: Solution: Eq. (2.89) is written as 2 K  G p + G 2p = 0

(S2.29)

For every G p there is − G p , therefore substitute − G p for G p in Eq. (S2.29) to write

2 K  G p = G 2p

(S2.30)

In a simple cubic crystal

Gp =

(

2 p1 ˆi1 + p 2 ˆi 2 + p 3 ˆi 3 a

Therefore, G p =

2 n a

) (S2.31)

where n is an integer. We know that K = 2  /  .From Fig. S2.2 one can write

2 K G p cos (90 −  ) = G 2p Substitute for K and G p in the above equation to get

2 Or

2



sin  =

2 n a

2 a sin  = n 

(S2.32)

Chapter 4 Problem 4.1: Solution: From Eqs. (4.12) and (4.16) one can write U(R ) =

1  A B  N − m p mi + n p ni  2  R R 

(S4.1)

In writing Eq. (S4.1), R 0 has been replaced by R in Eq. (4.12). In the equilibrium state the interaction potential must be minimum, i.e.,

dU =0 dR R 0

(S4.2)

Substitute Eq. (S4.1) in Eq. (S4.2) and simplify one gets R 0n - m =

n B p ni m A p mi

(S4.3)

The interaction potential in the equilibrium state from Eq. (S4.1) can be written as

U(R 0 ) =

 1  A B N  − m p mi + n p ni  2  R0 R0 

(S4.4)

Substitute Eq. (S4.3) in Eq. (S4.4) and simplify, one gets U(R 0 ) =

1 A p mi  m  N − 1 m  2 R0  n 

(S4.5)

Problem 4.2: Solution: Form Eq. 4.19 we write m

 m p mi  n − m  m   U(R 0 ) = 2 N  p mi  − 1  n   n p ni 

Put m = 6 and n = 12 in the above equation to get

 6 p 6i   1   −  U(R 0 ) = 2 N  p 6i  12 p 2 12i    =−

(p ) 1 N  6i 2 p12i

(S4.6)

2

Substitute the values of p 6i and p 12i from Eq. (4.43) in the above equation one gets

U(R 0 ) = − 4.30 (2 N  ) Problem 4.3: Solution: The interaction potential is defined as

U(V ) =

b b12 − 62 4 V V

(S4.7)

At zero pressure, the interaction potential is minimum in the equilibrium state. So

dU =0 dV V0

(S4.8)

Here V0 is the volume in the equilibrium state. Substitute Eq. (S4.7) in Eq. (S4.8) we get  2b  V0 =  12   b6 

1/ 2

(S4.9)

The bulk modulus is given by [Eq.4.28]  d2U  B M = V 2   dV  V0

(S4.10)

Substitute Eqs. (S4.7) and (S4.9) in Eq. (S4.10) and simplify to get

BM = 2

b 5/2 6 3/2 b12

(S4.11)

Comparing Eq. (S.4.7) with Eq. (4.33) of the book, one can easily write

b12 = 2 N  P12i

and

b 6 = 2 N  P6i

From Eq. (4.34) of the book, it is easy to prove the

P6i5/2 1  3 3/2 P12i  Therefore one gets from Eq. (S4.11) BM 

 3

Problem 4.4: Solution: The interaction potential is defined as [Eq. (4.67) of the book]

 Z2 e2  B U(R ) = N  n 0 n −  M  R   R From Eq. (S4.12) one can easily prove that

(S4.12)

Z2 e2 d2U ( n −1) = N  M dR 2 R R 30

(S4.13)

0

In a cube with edge a there is one CsCl molecule. Therefore, the volume of solid with N molecules of CsCl is

V = Na3

(4.14)

The 1NN distance in the CsCl structure is given by R = 3 a/2

(S4.15)

From Eqs. (S4.14) and (S4.15) we get

V=

8NR3

(S4.16)

3 3

Now one can immediately write dV 8 N R 2 = dR 3

(S4.17)

The bulk modulus is given by [see Eq. (4.74) of the book]

d 2 U  dR  BM = V 2   dR  dV 

2

(S4.18) R0

Substitute Eqs. (S4.13), (S4.16) and (S4.17) in Eq. (S4.18) and simplify to obtain

BM =

 M Z2 e2 8 3 R 04

(n − 1)

(S4.19)

Problem 4.5: Solution: From Eq. (4.83) of the book, the total interaction potential is given by

 Z2 e2  − R/ U(R ) = N  n 0 R e − M  R   In the equilibrium state the potential energy is minimum, i.e.,

(S4.20)

dU =0 dR R 0 Substitute Eq. (S4.20) in the above equation one gets

R 02 e - R 0 /  =   M

Z2 e2 R n 0

(S4.21)

From Eq. (S4.20) the interaction energy in the equilibrium state is given by

 Z2e2  U(R 0 ) = N n 0 R e − R 0 /  −  M  R0  

(S4.22)

Substitute Eq. (S4.21) in Eq. (S4.22) and simplify to get

U(R 0 ) = − N  M

Z2e2    1 −  R0  R0 

(S4.23)

Problem 4.6: Solution: Using Eq. (S4.20) and simplifying we can get  n 0 R - R 0 /  2  M Z 2 e 2  d2U = N −  2 e  dR 2 R R 30   

(S4.24)

0

Substitute Eq. (S4.21) in the above equation we get N M Z2 e2  R 0  d2U  = − 2  2 3 dR R R0   

(S4.25)

0

Substitute Eqs. (4.75), (4.76) of the book [for volume of NaCl and its derivative] and (S4.24) in Eq. (S4.18) and simplify to get

BM =

 M Z2 e2  R 0   − 2  4 18 R 0   

Problem 4.7: Solution: The total interaction potential from Eq. (S4.1) is given by

(S4.26)

U(R ) =

1  A B  N − m p mi + n p ni  2  R R 

Substitute the value of 1/R in terms of volume from Eq. (4.32) of the book, we write U(V ) =

where

Pni =

1  B A    N  n/3 Pni − m/3 Pmi 2 V V 

N n/3 p ni 2 n/6

and

 = Pmi

N m/3 p mi 2 m/6

(S4.27)

(S4.28)

In the equilibrium state

dU = 0 dR R 0 Substitute Eq. (S4.27) in the above equation and simplify, one gets the expression for equilibrium volume  n B Pni   V0 =     m A Pmi

3 n-m

(S4.29)

Now the bulk modulus is given by  d2U  B M = V 2   dV  V0

(S4.30)

One can calculate the bulk modulus in the same way as in Problem 4.3. Finally one gets

n Pni n − m d2U 1 = N B n +2 9 dV 2 V 2 0 V03

(S4.31)

n P n − m 1 N B n ni +1 2 9 V03

(S4.32)

BM =

It can be shown that Eq. (S4.32) is the same as Eq. (4.38).

Chapter 5 Problem 5.1: Solution: Let the strain components be defined as

  =

u  r

+

u  , u 

  =

u  r

(S5.1)

According to Eq. (5.80) the force equation for the homogeneous medium is given as

m

 xy d 2 u 1  xx  xz = + + 2 x y z dt

(S5.2)

where u 1 is the displacement in the x-direction. Differentiate the expressions for  xx ,  xy and

 xz given by Eq. (5.62) and use in these Eq. (S5.1) the elements of strain tensor, one gets   2u 2   yy  zz  2u3   xx  2 u1   = C11 = C11 xx + C12  + C + +  12  x x x  x 2  x  x y x z   xy

 xy

(S5.3)

 2u  2u 2  = C 44  21 +  y x y   y

(S5.4)

  2 u1  2 u 3   xz  xz = C 44 = C 44  2 +  z z x z   z

(S5.5)

y

= C 44

Substitute Eqs. (S5.3), (S5.4) and (S5.5) in Eq. (S5.2) one can write

m

  2u 2   2 u1  2 u1   2u3   2 u1  2 u1 ( ) = C + C + + C + C +   11 44  12 44  2 t 2 x 2 z 2   y  x y z x 

(S5.6)

Eq. (S5.6) gives the equation of motion of the elastic wave produced in a cubic solid when stress is applied externally in the x-direction. In exactly the same manner one can apply the nonhomogenious stress in the y- and z-directions and can obtain the corresponding equations of motion for u 2 and u 3 which are given by



  2 u1 2u 2 2u 2   2u3   2u 2  2u 2 ( ) = C + C + + C + C +   11 44  12 44  2 t 2 y 2 z 2   x  x y y z 

(S5.7)

m

 2u3 2u3    2 u1 2u3  2u3  2u 2  ( ) = C + C + + C + C +   11 44  12 44  2 t 2 z 2 y 2   x z y z   x

(S5.8)

The Eqs. (S5.6), (S5.7) and (S5.8) can be written in a general form as

m

2u   2u  2u  2u ( ) = C + C + C + C  11 44  12 44 2 t 2 r2  (    ) r  (    ) r r

(S5.9)

Problem 5.2: Solution: Let the general wave-like solution of Eq. (S5.9) be given as

u(r ) = u 0 exp [ −  (K  r −  t )

(S5.10)

Elastic waves in [100] direction The elastic waves along the [100] direction may be longitudinal or transverse in nature. The displacement of longitudinal (L ) elastic waves in the x-direction u 1L is given by

u 1L = u 0L exp− (K r1 −  L t ) = u 0L exp− (Kx −  L t )

(S5.11)

Here u 0L and K are the amplitude and propagation wave vector along the x-direction and  L is the frequency of the L elastic wave. Substitute Eq. (S5.11) in Eq. (S5.6), the frequency of the elastic wave is given by

L =

C11

m

(S5.12)

K

It immediately gives the linear velocity as vL =

L K

=

C11

(S5.13)

m

There are two transverse (T ) elastic waves with displacements along the y- and z-directions but with K along the x-direction and are defined as

 (

)

(S5.14)

 (

)

(S5.15)

u 2T1 = u 02 exp − K r1 − T1 t

u 3T2 = u 03 exp -  K r1 − T2 t

where T1 and  T2 are the frequencies of the T elastic waves. Substitute Eq. (S5.14) in Eq. (S5.7) and Eq. (S5.15) in Eq. (S5.8), one can solve for the velocities of the waves v T1 and v T2 to get

T

v T1 =

v T2 =

1

K

T

2

K

=

C 44

=

C 44

(S5.16)

m

(S5.17)

m

Eqs. (S5.16) and (S5.17) show that the velocities of the two T elastic waves are the same, but differ from that of the L elastic wave.

Elastic waves in [110] direction Let us consider the propagation of elastic waves along the [110] symmetry direction. The propagation wave vector K in the [110] direction is given by

K = ˆi1 K 1 + ˆi 2 K 2 =

(

K ˆ ˆ i1 + i 2 2

)

(S5.18)

The components K 1 = K x and K 2 = K y have the same magnitude. From Fig. 5.8 it is evident that one of the transverse elastic waves T1 propagates in the xy-plane with particle displacement along the z-direction defined as

 K  (x + y ) −  T1 t  u 3T1 = u 03 exp −     2

(S5.19)

Substitute Eq. (S5.19) in Eq. (S5.8) one gets

T = 1

C 44

m

K

(S5.20)

Therefore, the velocity of the T1 elastic wave becomes v T1 =

C 44

m

(S5.21)

The second transverse elastic wave T2 and the L elastic wave propagate in the xy-plane with particle displacement also in the xy-plane and these are defined as

 K  (x + y ) −  t  u 1 = u 01 exp −     2

(S5.22)

 K  (x + y ) −  t  u 2 = u 02 exp −     2

(S5.23)

Substitute u 1 and u 2 in Eqs. (S5.6) and (S5.7) one gets

K2 K2   m u 1 = (C11 + C 44 ) u 1 + (C12 + C 44 ) u2 2 2

(S5.24)

K2 K2 u 2 + (C12 + C 44 ) u1 2 2

(S5.25)

2

 2  m u 2 = (C11 + C 44 )

The Eqs. (S5.24) and (S5.25) can be solved for  2 to write

12  m =

1 (C11 + C12 + 2 C 44 ) K 2 2

(S5.26)

 22  m =

1 (C11 − C12 ) K 2 2

(S5.27)

So the velocities of these elastic waves are given by

v1 =

C11 + C12 + 2 C 44 2 m

(S5.28)

v2 =

C11 − C12 2 m

(S5.29)

We have already discussed Eq. (S5.28) gives the longitudinal wave while Eq. (S5.29) gives the transverse wave Therefore,

vL =

C11 + C12 + 2 C 44 2 m

(S5.30)

v T2 =

C11 − C12 2 m

(S5.31)

Elastic waves in [111] direction

For elastic wave propagating in the [111] direction the wave vector K is defined as

K = ˆi1 K 1 + ˆi 2 K 2 + ˆi 3 K 3 = ˆi1 K x + ˆi 2 K y + ˆi 3 K z

=

ˆi + ˆi + ˆi ) ( 3

K

1

2

(S5.32)

3

Therefore, the Cartesian components of displacement are given by

 K  u 1 (r ) = u 01 exp  −  (x + y + z ) −  t     3

(S5.33)

 K  u 2 (r ) = u 02 exp  −  (x + y + z ) −  t     3

(S5.34)

 K  u 3 (r ) = u 03 exp  −  (x + y + z ) −  t     3

(S5.35)

The total displacement u(r ) becomes

(

)

 K  u(r ) = u 01 ˆi1 + u 02 ˆi 2 + u 03 ˆi 3 exp  −  (x + y + z ) −  t     3

(S5.36)

One is interested to find the velocities of the different modes of the elastic wave propagating in the [111] direction. Substitute Eqs. (S5.33), (S5.34) and (S5.35) in Eq. (S5.6) and simplify to obtain

(C11 + C 44 − 3  ) u 1 + (C12 + C 44 ) u 2 + (C12 + C 44 ) u 3 = 0

(S5.37)

where

 = m

2 K2

(S5.38)

Similarly substituting Eqs. (S5.33), (S5.34) and (S5.35) in Eqs. (S5.7) and (S5.8) one can write

(C12 + C 44 ) u 1 + (C11 + C 44 − 3  ) u 2 + (C12 + C 44 ) u 3 = 0

(S5.39)

(C12 + C 44 ) u 1 + (C12 + C 44 ) u 2 + (C11 + C 44 − 3  ) u 3 = 0

(S5.40)

Eqs. (S5.37), (S5.39) and (S5.40) can be solved in exactly the same manner as in section 5.9.3 and the final results are

vL =

L

=

K

C11 + 2 C12 + 4 C 44 3 m

v T1 = v T2 = v T =

T K

=

(S5.41)

C11 − C12 + C 44 3 m

(S5.42)

Problem 5.3: Solution: The force density in terms of the stress components is defined as

f =  

 

(S5.43)

r

According to Newton second law of force

f  =  u 

(S5.44)

where  is the mass density. If   is constant, there is no motion but leads to compression (expansion). Therefore, the gradient of   must be finite for the motion of the volume element.

Substitute Eq. (5.25a) in Eq. (S5.43) and use Eq. (S5.44) one writes

 u =

 

 C r   , ,

(S5.45)



Use Eq. (5.9) for the strain component to get   2 u  2u   1  u =  C  +  2  ,  , r r   r r

In the above equation  and  are dummy variables and can be changed in the first term of the above equation. In doing so one gets

 u =  C  ,  ,

 2u  r r

(S5.46)

It is the equation of motion of the volume element of the medium. The wave-like solution of the above equation is written as

u  = u 0 exp  ( k  r −  t )

(S5.47)

Substitute Eq. (S5.47) in Eq. (S5.46) to write

 2  u 0 =  C  k  k u 0 

(S5.48)

 ,  ,

The above equation can be written as

  C k  k −     u  = 0   2

(S5.49)

0

, ,

Eq. (S5.48) or Eq. (S5.49) is a general classical wave equation for an anisotropic continuous medium and is called Christoffel equation.

Problem 5.4: Solution: The free energy of a deformed body is given by 2

1   2 F = F0 + L     +  L   2  

(S5.50)

Here  L and  L are the Lame’s coefficients. In Eq. (5.14) the strain component has been represented as the sum of the pure shear and pure hydrostatic compression. Substitute Eq. (5.15) in Eq. (S5.50) we write 2 2 2     1 1 1   F = F0 + L     +  L     −       +      2 3 3         2

  1 1   = F0 +  L   −       + K L     3 2     

where

K L = L +

2 L 3

2

(S5.51)

(S5.52)

Eq. (S5.51) expresses the free energy as the sum of pure shear strain and the hydrostatic compression. In terms of free energy the stress at constant temperature is defined as

  =

F   T

Substitute Eq. (S5.51) in Eq. (S5.53) to get

(S5.53)





1 3

  = 2  L    −       + K L      





(S5.54)



Substitute the value of K L from Eq. (S5.52) in the above expression we get

  = 2  L   + L     

(S5.55)



Note that Eq. (S5.55) for the stress can also be obtained by differentiating Eq. (S5.50) for the free energy.

Chapter 6 Problem 6.1: Solution: The total number of modes in a monatomic linear chain is given by  max

No of modes =

 g( )d 0

=

Let

2N



max

d



2  max − 2

(S6.1)

 =  max sin (x ) , then d =  max cos(x ) dx

Make the above substitution in Eq. (S6.1) we get No of modes =

=

2N



 /2

 dx 0

2N  =N  2

Problem 6.2: Solution: In a diatomic linear lattice the equations of motion are given by Eqs. (6.62) and (6.63) and are written as

M 1 u(2 n ) =  F  u (2 n + 1) + u (2 n −1) − 2 u (2 n ) 

(S6.2)

M 2 u(2 n + 1) =  F  u (2 n + 2) + u (2 n ) − 2 u (2 n + 1) 

(S6.3)

Let a be the periodicity (distance between two consecutive identical atoms) in the diatomic lattice as shown in Fig. 6.14 (see Chapter 6). Further, consider longitudinal wave in which the atoms vibrate along the direction of the wave vector K . In this case the wave-like solutions of the Eqs. (S6.2) and (S6.3) are given by

u (2 n ) = u 1 e  K n a −  t 

(S6.4)

u (2 n + 1) = u 2 e  K (2 n + 1) a/2 −  t 

(S6.5)

where u 1 and u 2 are the amplitudes of the waves for the 2nth and (2n + 1)th atoms. Substitute Eqs. (S6.4) and (S6.5) in Eqs. (S6.2) and (S6.3) one gets

(M  − 2 ) u + (2 cos Ka/2) u = 0

(S6.6)

(2  F cos Ka/2) u 1 + (M 2  2 − 2  F ) u 2 = 0

(S6.7)

2

1

F

1

F

2

The above equations can be solved simultaneously in exactly in the same way as in section 6.3.2 and the final equation is

 1 4  F2 1  2   + + 1 − cos 2 Ka/2 = 0 M1 M 2  M1 M 2 

(

 4 − 2  F 

)

(S6.8)

Eq. (S6.8) is quadratic in  2 and yields the solutions 2  1  1 1  1  4  K a     F   −  =  F  + + sin 2   M1 M 2  2   M1 M 2   M1 M 2 

1/ 2

2

(S6.9)

Here the independent values of K lie in the range from −  / a to  / a , i. e., −

 a

K

 a

(S6.10)

which defines the 1BZ for a diatomic linear lattice. One can prove in exactly the same way as in section 6.3.2 that the amplitudes of the two types of atoms are in the inverse ratio of their masses, i.e., u1 M = − 2 u2 M1

Problem 6.3:

(S6.11)

Solution: In the harmonic force approximation, the equation of motion for a lattice is given by

(n 1 , n 2 , n 3 ) = −   Fs u(n 1 , n 2 , n 3 ) − u(n 1 + s1 , n 2 + s 2 , n 3 + s 3 )  (S6.12) Mu s

where n = (n 1 , n 2 , n 3 ) represents the nth atom in a lattice and s = (s1 , s 2 , s 3 ) represents the sth nearest neighbor of the nth atom. Further, M is the mass of an atom and  Fs is the force constant for the sth nearest neighbor of the nth atom. Consider a square lattice with identical atoms and having lattice parameter a . If an atom is assumed to interact with its 1NNs then it is reasonable to assume the same force constant  F for all the 1NNs. The equation of motion (S6.12) can be written as

(n 1 , n 2 , 0) =  F u(n 1 + 1, n 2 , 0) + u(n 1 −1, n 2 , 0) + u(n 1 , n 2 + 1, 0) + u(n 1 , n 2 −1, 0) − 4 u(n 1 , n 2 , 0)  Mu (S6.13) The wave-like solution is written as  (K x n1a + K y n 2a −  t )

u(n 1 , n 2 , 0) = u 0 e (KR n −  t ) = u 0 e

(S6.14)

Substitute Eq. (S6.14) in Eq. (S6.13) and simplify one gets



M  2 = 2  F 2 − cos K x a − cos K y a



(S6.15)

The above equation can also be written as

2 =

K ya  4 F  2 K x a 2 sin + sin   M  2 2 

(S6.16)

One can calculate the vibration frequency along the symmetry directions. Along the [10] direction K x = K and K y = 0 . So the vibration frequency along the [10] direction is given by

=

4 F Ka sin M 2

(S6.17)

which is the same expression as obtained in a monatomic linear lattice [ see Eq. (6.43)]. In this direction the maximum value of wave vector K max is /a which defines the 1BZ boundary. Along the [11] direction the maximum value of wave vector which defines the 1BZ boundary is

K max =

( / a )2 + ( / a )2 =

2  / a . In the [11] direction K x = K y = K/ 2 . With this

substitution Eq. (S6.16) reduces to

=

8 F Ka sin M 2 2

(S6.18)

The dispersion relation in [11] direction is similar to that in the [10] direction but with a different proportionality constant and different K max .

Problem 6.4: Solution: In the sc lattice there are six 1NNs with coordinates a ( 1, 0, 0) , a (0,  1, 0 ) and

a (0, 0,  1) . If an atom is assumed to interact with its 1NNs then it is reasonable to assume the

same force constant  F for all the 1NNs. The equation of motion (S6.12) for the sc lattice can be written as

(n 1 , n 2 , n 3 ) =  F u(n 1 + 1, n 2 , n 3 ) + u(n 1 −1, n 2 , n 3 ) + u(n 1 , n 2 + 1, n 3 ) + u(n 1 , n 2 −1, 0 ) Mu

+  F u(n 1 , n 2 , n 3 + 1) + u(n 1 , n 2 , n 3 −1) − 6 u(n 1 , n 2 , n 3 ) 

(S6.19)

The wave-like solution can be written as  (K x n1a + K y n 2a + K z n 3a −  t )

u(n 1 , n 2 , n 3 ) = u 0 e (KR n −  t ) = u 0 e

(S6.20)

Substitute Eq. (S6.20) in Eq. (6.19) and simplify one gets



 Ky a

−  2 M =  F e K x a + e −  K x a + e

+e

−  Ky a

+ e K z a + e −  K z a − 6



 2 M = 2  F  3 − cosK x a − cosK y a − cosK z a 

2 =

K ya 4 F  2 K x a K a + sin 2 + sin 2 z   sin M  2 2 2 

(S6.21)

Eq. (S6.21) gives the dispersion relation for the sc lattice. One can study the dispersion relations in the high symmetry directions [100], [110] and [111]. In the [100] and [110] directions, one gets the same result for the vibration frequency as obtained in the square lattice [Eqs. (S6.17) and (S6.18)]. In the [111] direction all the three components of wave vector are equal and the vibration frequencies are given as

=

12 F Ka sin M 2 3

(S6.22)

The vibration frequencies along the three symmetry directions [100], [110] and [111] are shown in Fig. S6.1.

Chapter 8 Problem 8.1: Solution: Consider a two-dimensional solid with N atoms where each atom is considered to be a fixed harmonic oscillator. The energy of an atom is given by

E=

p2 1 + F r2 2M 2

(S8.1)

where p 2 = p 2x + p 2y

r2 = x 2 + y2 and

=

F M

(S8.2)

where M is the mass of an atom. The average energy of an atom is given by substituting Eq. (S8.1) in Eq. (8.18), i.e.,

 p2



1



− 0  + M r   2M 2  2  p2 1 2 2     + M  r e d r d2p − 2 M 2   E= 

e

 p2 1  − 0  + M2 r2   2M 2   

2 2

(S8.3)

d2r d2p

−

Here the Maxwell-Boltzmann distribution given by Eq. (8.19) of the book is used. Simplify the above integral we can write 0 p2



E =

p2 − 2M 2  2M e d p − 

e

−

0 p2 2M



r e 2

+

2

d p

1 − 0 M  2 r 2 2

d2r

1 M  2 − 1 − 0 M  2 r 2 2 e d2r  2

(S8.4)

−

−

With the use of the spherical coordinates the above equation reduces to 0 p2



E=



p2 − 2M 0 2 M e 2  p dp 

e

−

1 + M  2 0 2

0 p2 2M

1 − 0 M  2 r 2

2 r e 2

e

2  p dp

1 − 0 M  2 r 2 2

2  r dr (S8.5)

2  r dr

0

0

Substitute E = p 2 / 2 m and q = (1/2)  0 M  2 r 2 in the above equation and simplify to get 

E=

Ee 0 

e

− 0 E

− 0 E



dE +

dE

0

qe 1 0

0 

e

−q

−q

dq dq

0

Integrate the above equation by parts, we get E = 2 /  0 = 2 k BT

(S8.6)

Hence the total energy of N atoms becomes E = N E = 2 N k BT

The specific heat at constant volume is given by

(S8.7)

CV =

E = 2 N kB = 2R T

(S8.8)

Chapter 9: Problem 9.1: Solution: In a one-dimensional free-electron gas with N e electrons and having length L , the Schrodinger wave equation is defined as

−

2 d2  k (x ) = E k  k (x ) 2 m e dx 2

(S9.1)

where the normalized wavefunction  k (x ) is defined as

 k (x ) =

1 L

e k x = k

(S9.2)

Here k represents the normalized plane wave function.  k (x ) or k forms a complete set of wave functions. The cyclic boundary condition demands

 k (x ) =  k (x + L)

(S9.3)

Substitute Eq. (S9.2) in Eq. (S9.3) we get

e k L = 1

(S9.4)

which is satisfied if k possesses the values given by k=

2 n L

(S9.5)

with n an integer, negative, positive or zero. Eq. (S9.5) gives the discrete values of k . In the limit of large L , the k values become quasi-continuous. Substitute (S9.2) in Eq. (S9.1) the electron energy is given by

2 k2 Ek = 2 me

(S9.6)

Eq. (S9.6) gives parabolic energy bands. The linear momentum of an electron is given by

p  k (x ) = −  

d  k (x ) =  k  k (x ) dx

Or p = k

(S9.7)

N e electrons can be accommodated in different k -states until all the electrons are exhausted (see Fig. 9.2). The highest filled state is the Fermi state with wave vector k F , called the Fermi wave vector. The Fermi wave vector can be calculated from the fact that N e electrons are accommodated up to k F . According to (S9.5) there is one k -state in 2  / L , therefore the number of electrons in k F is given by

2

kF = Ne 2 /L

which gives kF = 

Ne =  n0 L

(S9.8)

Here n 0 is the number of electrons per unit length. Let N(E k ) be the density of electron states and is defined as the number of electron states per unit energy. Now the number of electron states with wave vector k is given by

n (k ) =

L k 2

(S9.9)

Therefore, the density of electron states becomes

N(E k ) = 2

dn (k ) dn (k ) dk = 2 dE k dk dE k

Use (S9.6) and (S9.9) in the above equation we get N(E k ) =

L  2 me    2   2 

1/ 2

E k−1/2

(S9.10)

The number of electron states per unit length per unit energy is given by g e (E k ) =

1  2 me    2   2 

1/ 2

E k− 1/2

(S9.11)

The total energy E T in the ground state of one-dimensional free-electron gas becomes

EF

E T =  E k N(E k ) dE k

(S9.12)

0

Substitute Eq. (S9.10) for N(E k ) in Eq. (S9.12) and integrate ET =

1 Ne EF 3

(S9.13)

Therefore, the average kinetic energy per electron is one third of the Fermi energy of the onedimensional free-electron gas.

Chapter 10 Problem 10.1: Solution: Consider a two-dimensional free-electron gas in the xy-plane where magnetic field H is applied along the z-direction. The free electrons in a magnetic field will satisfy the Schrodinger wave equation 2

1  e  p − A  = E  2m e  c 

(S10.1)

where A is the vector potential and  (x, y ) is the wave function. The above equation can be written as  1  2 2e e2  p − A  p + 2 A 2   = E  2 me  c c 

(S10.2)

In order to evaluate Eq. (S10.2) consider the vector potential of the form A = ( − y H, 0, 0 )

(S10.3)

Substitute Eq. (S10.3) in Eq. (S10.2) and use the operator form of momemtum p in two dimensions one gets 2 1   2  2  e H y  e2 2 2  2    −  + − + y H  = E  2  x 2 y 2  2 me  c  x c   

(S10.4)

Rearrange the terms in the above equation to write 2 e H y  e2 H2 2  2   2  2 − + − y  = E  2 m e y 2 2 m e  x 2 c x c2  2  2

−

(S10.5)

The above equation can be written as

eH  2   2    +  − − y  = E  2 2 m e y 2 m e  x c   2

2

(S10.6)

This equation can be solved in exactly the same way as in section 10.8.3 to get the quantized energy eigenvalues Problem 10.2: Solution: The general gauze for the vector potential, which yields magnetic field in the zdirection, is given by

1  1  A =  − y H, x H, 0  2  2 

(S10.7)

Eq. (S10.2) can be written as 2 1  2 2e  p − (A x p x + A y p y ) + e 2 A 2   = E  2 me  c c 

The operator form of the above equation becomes 2 e H  e H  1  2  e2 2 2 e2 2 2  2    −  + −  y +  x + y H + 2 x H  = E  2  x 2 y 2  2 me  c  x c  y 4 c 4c   

(S10.8) 2 2 eH  eH  1  e2 e2 2  2 2 2     −  +  y − y H −  −  x − x2 H2  2 2 2 2 2 2    2 me  c  x 4 c  c  y 4 c   x   y

The above equation can be written as     eH  eH  1   −  2  + y  −  2  − x 2 me  2c  2 c    x  y  2

2

2

  = E  2

eH  eH  2   2    +   −  − y   − x  = E  2 m e  x 2c  2 m e  y 2 c  

(S10.9)

The above equation can also be solved for the energy eigenvalues which are quantized.

    = E  

Problem 10.3: Solution: Let  0 be the resistivity in the in the absence of magnetic field and  (H ) the resistivity in the presence of magnetic field. By definition the magnetoresistance is given by Magnetoresistance =

 (H ) −  (0)  (0)

(S10.10)

The conductivity in the absence of magnetic field  0 is given by

 0 = ne e  e 2

(S10.11)

me

[see Eq. (10.14)]. Here the symbols have their usual meanings. As resistivity is the reciprocal of conductivity, therefore the resistivity in the absence of magnetic field is given by

0 =

me n e e2  e

(S10.12)

Suppose both the electric and magnetic fields are applied to the conductor in the perpendicular directions (say H in the z-direction) then the electrons start rotating about the magnetic field with cyclotron frequency  c . The diagonal components of the conductivity matrix in Eq. (10.49) give conductivity in the presence of magnetic field (magnetoconductivity), i.e.,

 (H ) [ =  xx =  yy ] =

0 2 1 + ( c e )

(S10.13)

Therefore, the resistivity in the presence of magnetic field becomes

 (H ) =

1 + ( c e )

2

0



=  0 1 + ( c e )

2



(S10.14)

Reshuffling the terms we write



0

=

 (H ) −  0 2 = ( c e ) 0

(S10.15)

Substitute the value of  c from Eq. (10.24) in the above equation we finally write

Magnetoresistance =

  H2 0

Chapter 11 Problem 11.1: Solution: The mobility of charge carriers q is given by

q = − 1 q  3

f 0  q (v ) d 3 v E 3 f  0 (v, t ) d v

v2

(S11.1)

Here the denominator gives the density n q of the charge carriers, therefore

 q = − 1 q  v 2 f 0  q (v ) d 3 v 3 nq

(S11.2)

E

The above integral can be solved using the Maxwell-Boltzmann distribution give by (see Eq. 11.56)

 mq   f 0 (E, T ) = n q  2  k T B  

3/ 2

e − E / k BT

(S11.3)

Substitute Eq. (S11.3) in Eq. (S11.2) to get

4 q  m q    q = 3 k B T  2 k B T 

3/ 2



  (E ) e

− E / k BT

(S11.4)

v 4 dv

0

Substitute x = E/k B T = (1/2 ) m q v / k B T and  (E ) =  0 E in the above equation and simplify 2

q  0  p + 3/2 - x (k B T ) q = x e dx m q 0 3  4

q =

4 3 

p

(k B T )p

q 0 (p + 5/2 ) mq

p

(S11.5)

(S11.6)

Here (p + 5/2 ) is a gamma function as defined in Chapter 11.

Problem 11.2: Solution: From Eq. (11.72) we know that

 =

n e e 2  e (E F ) m e

(S11.7)

where the symbols have their usual meanings. Therefore the resistivity is given by

=

m e

(S11.8)

n e e 2  e (E F )

We are given the values of the following physical quantities:

me = 1.01m e n e = 8  5  1028 / m 3

 = 1  7  10 − 8  − m v F = 1  55  10 6 m/sec.

Substitute above values in Eq. (S11.8) we write

 e (E F ) =

1.01  9.11  10 −31

(

8  5  10 28  1  6  10 −19

= 2  4  10

−14

)  1.7  10 2

−8

sec.

The mean free path of the electron at the Fermi surface is given by

 e = v F   e (E F ) = 1  55  106  2  4  10−14 = 37  8  10−9 = 37  8 nm

Chapter 12 Problem 12.1: Solution: The atomic potential seen by an electron is given by

V a (r ) = −

Z e2 r

(S12.1)

The Fourier transform of the atomic potential is defined as (see Eq. 12.88) 

VGa p =

1  G r e p V a (r ) d 3 r  V0 0

(S12.2)

Substitute Eq. (S12.1) in Eq. (S12.2) we write VGa p =

1 V0



− 0

Z e 2  G p r cos 2 e r dr sin  d d r



=−



2

Z e2  G r cos  r dr  e p sin  d  d V0 0 0 0

Solve the angular part of the integral we get 4  sin (G p r ) Z e2 r dr  V0 0 Gpr 

VGa p = −

(S12.3)

Above integral can be easily solved to yield VGa p = −

4 Z e2 G 2p V0

(S12.4)

Problem 12.2: Solution: The screened atomic potential seen by an electron is given by

V a (r ) = −

Z e2  (r ) r

(S12.5)

Here  (r ) is the dielectric function of the metal under consideration. Eq. (S12.5) cannot be solved unless one knows the explicit form of the dielectric function but the problem becomes simple if  (r ) is assumed to be a constant, i.e.,  (r ) =  .  is generally called dielectric constant of the material. So one can write

V a (r ) = −

Z e2 r

(S12.6)

Substitute Eq. (S12.6) in Eq. (S12.2) we write VGa p =

1 V0



− 0

Z e2 =−  V0

Z e 2  G p r cos 2 e r dr sin  d d r 



0

0

 r dr  e

 G p r cos

2

sin  d  d 0

Solve the angular part of the integral we get 4  sin (G p r ) Z e2 r dr   V0 0 Gpr 

VGa p = −

(S12.7)

Above integral can be easily solved to yield VGa p = −

4 Z e2  G 2p V0

(S12.8)

Problem12.3: Solution: The free-electron energy bands as calculated in Chapter 9 are shown in Fig. S12.1 in the extended zone scheme. The energy bands in the reduced zone scheme are also shown in the second part of the diagram.

Problem 12.4: Solution: The general expression for energy of a band in the tight binding theory is

E k = E 0 −  0 −  1  e k  ( R n − R m )

(S12.9)

m

Let 2a be the size of the cube edge. In a bcc structure there are eight 1NNs with coordinates R n − R m are given as a ( 1,  1,  1) . Here we have assumed the atom at R n to be at the

origin. Substitute the values of R n − R m in Eq. (S12.9) we write

 (

Ek = E0 −  0 −  1 e

 (

 a kx + ky + kz )

 a − kx + ky + kz )

− 1 e

+e

 a (k x + k y − k z )

 a ( − kx + ky − kz )

+ e

+e

 a (k x − k y + k z )

 a (− k x − k y + k z )

+ e

 a (k x − k y − k z )

+e

 a (− k x − k y − k z )

+ e



 (S12.10)

In the above equation the various terms can be arranged so that the exponential terms are changed into trigonometric functions

 (

 a ky + kz )

E k = E 0 −  0 − 2  1 cos(k x a ) e

 a (k y − k z )

+e

 a (− k y + k z )

+e



= E 0 −  0 − 4  1 cos(k x a )cos(k y a ) e k za + e -  k za

 a (− k y − k z )

+e





= E 0 −  0 − 8  1 cos(k x a ) cos(k y a ) cos(k z a )

(S12.11)

For small values of k ( k x , k y , k z ) one can write

cos(k x a ) = 1 − cos(k y a ) = 1 −

cos(k z a ) = 1 −

(k x a )2 2

(k a )

+

(S12.12)

+

(S12.13)

+

(S12.14)

2

y

2

(k z a )2 2

Substitute Eqs. (S12.12) – (S12.14) in Eq. (S12.11) and neglect the higher powers (the terms having power more than two) one gets  k2 a2  Ek = E 0 −  0 − 8 1  1 −  2  

(S12.15)

This equation shows that near the center of the 1BZ energy bands are spherical in nature.

In fcc structure there are twelve 1NNs with coordinates R n − R m = a ( 1,  1, 0) , a ( 1, 0 ,  1) and a (0,  1,  1) . Here again we have assumed the atom at R n to be at the origin.

Substitute the values of R n − R m in Eq. (S12.9) we write

 (

+e

 (

+ e

Ek = E0 −  0 −  1 e

 a kx + ky )

 a (k x − k y )

 a ( − kx + ky )

+e

 a (− k x − k y )

+e

 a (− k y − k z )



−  1 e a ( k x + k z ) + e a ( k x − k z ) + e a ( − k x + k z ) + e a ( − k x − k z )



 a ky + kz )

− 1 e

 a ( ky − kz )

 a (− k y + k z )

+ e



Rearranging the terms in the above equation we write

+ e





E k = E 0 −  0 −  1 2 e a k x cos(a k y ) + 2 e − a k x cos(a k y )



cos(a k z )



−  1 2 e a k x cos(a k z ) + 2 e − a k x cos(a k z )



 a ky

− 1 2 e

cos(a k z ) + 2 e

− a k y





Or





E k = E 0 −  0 − 4  1 cos(a k x ) cos(a k y ) + cos(a k y ) cos(a k z ) + cos(a k z ) cos(a k x ) (S12.16) At small values of k the above equation can be written as 2 2    (  k ya) ( k xa)   E k = E 0 −  0 − 4  1  1 − . . .  1 − . . . 2 2       

 (k y a )2   (k z a )2    − 4  1 1 − . . .  1 − ... 2 2       

 (k z a )2   (k x a )2  − 4  1  1 − . . .  1 − . . .  2 2      Simplify the above equation we get

E k = E 0 −  0 − 12 1 + 4  1 k 2 a 2

(S12.17)

It shows that near the center of the 1BZ the energy bands are spherical in nature.

Problem 12.5: Solution: The pseudowave equation is given by (Eq. 12.205)  2   2 + W (r )  s (r ) = E  s (r ) −  2 me 

where

W(r ) = V(r ) +  (E − E c ) c (r ) c (r ) c

(S12.18)

(

)

(S12.19)

W (r ) is called pseudopotential which is nonlocal in nature and

 s (r ) as the pseudowave

 = V(r ) + E − H P

function given by Eq. (12.203). Add a linear combination of core states c to  s (r ) to obtain a new wave function  s (r ) written as

 s (r ) =  s (r ) +  a c c

(S12.20)

c

where a c is a constant. Substitute the new peudowave function given by Eq. (S12.20) in Eq. (S12.18) to write   2   2 + V(r ) +  (E − E c ) c (r )  c (r )    s (r ) +  a c  c (r )  − c c   2 me 

  = E   s (r ) +  a c  c (r )  c  

(S12.21)

Open the brackets we get   2 −  2 + V(r ) +  (E − E c ) c (r )  c (r )   s (r )  c  2 me  

+ − 

 2  2 + V(r ) +  (E − E c ) c (r )  c (r )   a c  c (r ) 2 me c  c

= E  s (r ) + E  a c  c (r )

(S12.22)

c

The eigenvalue equation for the core states defined as (see Eq. 12.196)  2   2 + V(r )  c (r ) = E c  c (r ) −  2 me 

(S12.23)

Use Eq. (S12.23) in Eq. (S12.22) along with the orthogonality property of core states one can write   2 −  2 + V(r ) +  (E − E c ) c (r )  c (r )   s (r ) +  c  2 me 

 E a  (r ) c

c

c

c

+  a c (E − E c )  c (r ) = E  s (r ) + E  a c  c (r ) c

c

The above equation becomes   2  2 + V(r ) +  (E − E c ) c (r )  c (r )   s (r ) = E  s (r ) − c  2 me 

(S12.24)

which is nothing but the usual pseudowave equation (see Eq. S12.18). Let us the what happens to the actual wave function defined as (see Eq. 12.202 )





 (r ) = ( 1 − P )  s (r ) =  1 −   c (r )  c (r )   s (r ) 

c

(S12.25)



The new wave function after the addition of a linear combination of core states becomes







 (r ) =  1 −   c (r )  c (r )    s (r ) + +  a c  c (r )  



c

c



Use the orthogonality relations of the core states one obtains





 (r ) =  1 −   c (r )  c (r )   s (r ) 



c

=  (r )

(S12.26)

Problem 12.6: Solution: The pseudowave equation is defined as  2   2 + W (r )  s (r ) = E  s (r ) −  2 me 

where

(S12.27)

W(r ) = V(r ) +  (E − E c ) c (r ) c (r ) c

The orthogonalized plane wave (OPW) and is defined as

k OPW = k −  c (r ) c (r ) k c

(S12.28)

Also one can write

k OPW = k −  k  c (r )  c (r )

(S12.29)

c

Operate k OPW from the left side on Eq. (S12.27) one can write   2 k OPW  −  2 + W (r )  s (r ) = E k OPW  s (r )  2 me 

(S12.30)

  2 k OPW  −  2 + W (r )  s (r ) = E k OPW  a k k k  2 me 

(S12.31)

Here we have used the pseudowave function defined in Eq. (12.203). Substitute the value of

k OPW from Eq. (S12.29) in the above equation and simplify to get

ak k

2k 2 2k 2  kk  +  a k k  W(r ) k −  a k k   c (r )  c (r ) k 2 me 2 me k k,c

= E  a k  kk  − E  a k k   c (r )  c (r ) k k

(S12.32)

k,c

Here we have used the orthogonality relations of the plane waves and the core states. Grouping the terms we get

  2 k2    a k k   2 m − E    k k − c k  c (r ) c (r ) k  + k  W(r ) k  = 0 e   

(S12.33)

Chapter 13 Problem 13.1: Solution: In a sc structure the reciprocal lattice has also the cubic symmetry. The 1BZ is a cube with boundary at a distance of d = /a where a is the lattice parameter. Here d is half the distance of the 1NN in the reciprocal space. For a monovalent metal, the Fermi wave vector from Eq. (9.19) of the book is given by

 3 2   k F =  V  0 

1/ 3

(S13.1)

In a sc structure V0 = a , therefore 3

(3 ) k =

2 1/ 3

F

(S13.2)

a

Now for a sc structure one can write

(

kF 3 2 = d a

)

1/ 3

3 =     a

1/ 3

(S13.3)

From Eq. (S13.3), it is evident that k F  d. Hence, a metal with sc structure when treated as free-electron metal, the whole of the Fermi sphere will be well within the 1BZ as in Fig. 13.2 of the book. But if the metal with sc structure is treated in the nearly free-electron approximation, there are zone boundary effects, i.e. the bands bulge in the outward direction near the zone boundary and intersect it perpendicularly.

Chapter 14 Problem 14.1: Solution: At low temperatures the electrons from the valence band cannot be excited to the conduction band. On the other hand the electrons from the valence band get excited to the accepter band and fill it. In the p-type semiconductor at low temperatures the conductivity is mainly due to the holes and is given as  h = n h e h

(S14.1)

Substitute the value of hole density n h from Eq. (14.66) in the above equation to write

 h = e  h N a N h e − E / k T av

B

(S14.2)

Substitute the value of N h from Eq.(14.43) in the above equation to get

 h =  0h e − E / k T av

B

(S14.3)

where

 0h = e  h

 m k T  2 N a  h B2   2  

3/ 4

(S14.4)

At high temperatures a large number of electrons from the valence band will be excited to the conduction band. At such temperatures, the p-type semiconductor will behave like an intrinsic semiconductor with n e  n h .

Problem 14.2: Solution: The current density and the electric fields are given as

Jx =

Ix 2.4  10 − 3 = = 10 −1 A/cm 2 W d 0.12  0.2

Ex =

Vx 1.0 = = 1.0 V/cm L 1.0

VH 10  10 − 3 Ey = = = 0.833 V/cm W 0.12 The conductivity of the Ge crystal is given by

Jx 10 −1 −1 = = 10 −1 (ohm - cm) Ex 1.0

 =

The density of electrons is given by

ne = −

1 Ix H 1 2.5  10 − 3  3  10 9  5000 = − 1 e c VH d 4.8  10 − 10  3  1010 10  10 − 3   10 − 2  0.12 3

= 6.51  10

14

cm − 3 = 6.51  10 20 m − 3

The mobility of the electrons is given by



10 −1 e = = = 0.96  10 − 3 m 2 / V - sec. − 19 20 n e e 1.6  10  6.5  10 Problem 14.3: Solution: The ellipsoidal energy bands for the conduction and valence bands are defined as

k 2y k 2z   2  k 2x E = Ec + +  +  2  m x my m z 

(S14.5)

k 2y k 2z   2  k 2x E = Ev − +  +  2  m x my m z 

(S14.6)

where E c is the energy values at the bottom of the conduction band and E v is the energy at the top of the valence band. Here m x , m y and m z are the effective component masses along the x-, y- and z-directions. It is convenient to define

x =

m kx , y = m x

m ky, z = m y

m kz m z

(S14.7)

Make the substitution (S14.7) in Eqs. (S14.5) and (S14.6) we write

where

E = Ec +

 2 2 2 m

(S14.8)

E = Ev −

 2 2 2 m

(S14.9)

 2 =  x2 +  y2 +  z2

(S14.10)

The density of electron states in the conduction band is given by N(E k ) dE k =

V

d k (2  )  3

(S14.11)

3

Substitute the values of k x , k y and k z from Eq. (S14.7) in the above expression we get

 m x m y m z   N(E k ) dE k =   m 3   

( )

1/ 2

V

d (2  )  3

(S14.12)

3

The above equation can be solved using Eqs. (S14.8) and (S14.9) to get

V  2  N(E k ) =   2 2   2 

3/ 2

(m m m ) E − E   x

 y

 1/ 2 z

1/ 2

k

c

(S14.13)

Similarly one can find the density of hole states in the valence band and is given as

N(E k ) =

V  2    2 2   2 

3/ 2

(m m m ) E − E   x

 y

 1/ 2 z

1/ 2

v

k

(S14.14)

Comparison of Eq. (S14.14) with Eq. (14.29) shows that the effective mass can be defined as

(

m  = m x m x m x

)

1/ 3

(S14.15)

Problem 14.4: Solution: From Eq. (14.52) the Fermi energy is given by

 m h  3 EF = + k B T ln    2 4  me  Eg

(S14.16)

In defining the above equation the zero reference of energy is taken at the top of the valence band, i.e., E v = 0 . The band gap is given be 1eV , i.e., E g = 1 eV . At absolute zero, Eq. (S14.16) gives the Fermi energy as E F = E g / 2 = 0.5 eV

(S14.17)

According to the above equation the Fermi energy lies in the middle of the band gap as expected.

If the Fermi energy shifts by 5% from the middle of the band gap then E F = 0.55 eV . Substitute the new value of Fermi energy in Eq. (S14.16) we write

0.55 eV = 0.50 eV + We know that

 m  3 k B T ln  h  4  me 

(S14.18)

m h = 4 m e

k B = 1.38  10 − 16 erg K − 1 1 eV = 1.602  10 − 12 erg  m h  3 k B T ln    = 0.05  1.602  10 − 12 erg 4  me  Therefore, the temperature is given by

T=

4  0.05  1.602  10 −12 3  1.38  10 −16  ln (4)

= 558 K

(S14.19)

SOLUTION MANUAL

Chapter 15 Problem 15.1: Solution: When a dielectric material in the form of a sphere with radius a is placed in the presence of externally applied electric field, it is polarized as shown in the Fig. S15.1.

The electric dipoles inside the material cancels the effect of each other but the finite charge density appears on the surface of the sphere as shown in the figure. The surface charge density of nˆ  P appears on half of the spherical surface while on the other half it is − nˆ  P . To evaluate the polarization field consider a small elemental surface dS . The charge on the surface element becomes − nˆ  P dS . The electric field at the center of the sphere due to the element dS is given by dE P = −

nˆ  P dS a2

(S15.1)

It is clear from the symmetry of the spherical surface that the components of polarization field perpendicular to the applied electric field E cancel each other. The finite polarization field is parallel to the applied electric field and is given as

nˆ  P cos dS a2

dE P = −

(S15.2)

Total polarization field is obtained by integrating over whole of the spherical surface, therefore 2



P cos EP = −  cos a 2 sin  d  d 2 a 0 0 =−

4 P 3

(S15.3)

In the vector notation one can write EP = −

4 P 3

(S15.4)

Problem 15.2: Solution: The expression for the electric susceptibility is given by

 E (K ) = 

f (E k , T ) − f (E k + K , T ) Ek − Ek +K

k

(S15.5)

Expand the above expression to write

 E (K ) =  k

f (E k , T ) f (E k , T ) + Ek − Ek +K k Ek − Ek −K

At absolute zero the Fermi-distribution function is unity, i.e., f (E k , T ) = 1 , therefore

 E (K ) =  k

1 1 + Ek − Ek + K k Ek − Ek −K

(S15.6)

Let us evaluate the first term of Eq. (S15.6)

1

E −E k

k

= k +K

2 me 1  2 2 2  k k − k + K

Here we have used the expression for energy E k =  2 k 2 / 2 m e .

1

E −E k

k

= k +K

2 me 1  2 2  k − K − 2 k K cos

(S15.7)

Change the summation into integration we write 2



2m 2V F sin  d 1 k E − E = −  2 Ke (2  )3  dk k 2  d  K + 2 k cos k k +K 0 0 0 k

(S15.8)

Solve the angular part of the integral to get

 K + 2k  2 me V F 1  = − dk k ln  k E − E 2 2   K (2  ) 0  K − 2k  k k +K k

(S15.9)

Similarly one can prove that

 K + 2k  2 me V F 1  = − dk k ln  k E − E 2 2  K − 2 k  K ( ) 2    k k −K 0 k

(S15.10)

Substitute Eqs. (S15.9) and (S15.10) in Eq. (S15.6) we get

 K + 2k  4m V F   E (K ) = − 2 e dk k ln  2   K (2  ) 0  K − 2k  k

(S15.11)

Solve the above integral by parts to get

 E (K ) = −

kF m e V  k 2F K + 2kF k 2 dk  ln − 2 K  0 K 2 − 4 k 2  K − 2kF  2  2 K  2 

The integral in the last term of the above equation can be easily solved to get

 E (K ) = −

m e V  k F K  k 2F K 2  K + 2 k F   ln +  −   8  K − 2 k F   2  2 K  2  2

(S15.12)

The above equation can be written as me k F V  4 k 2F − K 2 1 + K/2 k F   E (K ) = − ln 1 +  4kF K 1 − K/2 k F  2 2  2 

(S15.13)

Eq. (S15.13) gives the expression for the electric susceptibility of a metallic solid. The dielectric function can be obtained by substituting Eq. (S15.13) in the following expression

 (K ) = 1 − v(K )  E (K ) one finally obtains

(S15.14)

 (K ) = 1 +

 me e2 1 − 2 1+  1 + ln  2 2  2 1−  2 k F   

 = K/2 k F

where

(S15.15) (S15.16)

Eq. (S15.15) is called the Lindhard dielectric function for simple metals.

Problem 15.3: Solution: The bare-ion potential is given as

V b (K ) = −

4 Z e2 V0 K 2

(S15.17)

When the wave vector K approaches zero, the logarithmic term in Eq. (S15.15) reduces to unity and therefore  (K ) reduces to

 (K ) →

4 me e2 k F  2 K2

(S15.18)

The screened potential is given by

4 Z e 2   2 K 2 V b (K ) V(K ) = = −  (K ) V0 K 2 4 m e e 2 k F Z  2 2 =− V0 m e k F

(S15.19)

From Eq. (9.19) the value of k F is given by  Z  k F =  3 2 V0  

which gives

1/ 3

k3 Z = F2 V0 3 

(S15.20)

Substitute Eq. (S15.20) in Eq. (S15.19) V(K ) = −

2 EF 3

(S15.20)

Problem 15.4: Solution: The dielectric function in the Thomas-Fermi approximation is given by 2  TF (K ) = 1 + K TF / K2

(S15.21)

where K TF is the Thomas-Fermi wave vector. If the bare-ion potential as seen by an electron is given as

4 Z e V (K ) = − 2

2

b

(S15.22)

V0 K

Then the expression for the screened potential in the crystal space becomes

V(r ) =

V0

d 3 K e K r 3 

(2  )

V b (K )  TF (K )

(S15.23)

Substitute the values of V b (K ) and  TH (K ) in the above equation

V(r ) = −

4 Z e 2

d Ke (2  )  3

3

 K r

1 2 K + K TF 2

(S15.24)

The angular part of the integral can be easily solved to give V(r ) = −

   Z e2  1 1 Kr dK K e − dK K 2 e − K r   2 2 2   r  0 K + K TF K + K TF 0 

Change K to − K in the second term of the above equation we write V(r ) = −

 0  Z e2  1 1 Kr Kr dK K e + dK K e   2 2    r  0 K 2 + K TF K 2 + K TF  − 

=−

Z e2 1 dK K 2 e K r 2    r − K + K TF

(S15.25)

The integral in Eq. (S15.25) can be solved by the method of residues. The integrand of the integral in Eq. (S15.25) has two poles at K =   K TF . These poles on the z-axis are shown in Fig. S15.2.

Consider the upper half of the complex plane for the contour integration which contains only one pole at K =  K TF . The residue R at K =  K TF is given by R = (K −  K TF )

=

K e K r 2 K + K TF K = K 2

TF

1 − KTF r e 2

(S15.26)

According to Cauchy’s residue theorem one can write 

1

 dK K K + K 2

−

2 TF

e K r = 2   (sum of the residues )

= 2 

1 − KTF r e =   e − KTF r 2

(S15.27)

Substitute the value of the integral from Eq. (S15.27) in Eq. (S15.25), one finally gets the screened potential as

V(r ) = −

Z e 2 − KTF r e r

Hence the screened potential decreases exponentially with the increase in distance.

Problem 15.5: Solution: Consider a spherical atom with electron cloud having radius R at the center of which is situated the nucleus. The density of the electron charge is given by

ne = −

Ze 3  Ze   = −  3 4   R 3  (4 / 3) R

(S15.28)

Here Z is the number of electrons in the atom. With the application of the external electric field E 0 the nucleus shifts its position in the direction of the field while the electron cloud in the opposite direction (see Fiq. 5.10). Therefore, the centers of the positive and negative charges separate from each other inducing finite electric dipole moment due to the polarization of electron charge p e in the atom. The magnitude of the electric dipole moment depends on the strength of the electric field, i.e., p e =  Ea E 0

(S15.29)

The Lorentz force acting on the atom is given by 1   FL = − Z e  E 0 + v  B  = − Z e E 0 c  

(S15.30)

Due to the separation of the centers of negative and positive charges the Coulomb attractive force comes into play which opposes the displacement of charges. In the equilibrium state the two opposing forces (Lorentz and Coulomb forces) should be equal. To evaluate the Coulomb force consider a sphere of radius r . The charge inside this sphere is given by

Q=

 r3  4 3 r n e = − Z e  3  3 R 

(S15.31)

Hence the Coulomb force acting on the atom becomes FC =

Ze r Q = − Z2 e2 3 2 r R

In the equilibrium state the two forces must be equal and opposite, i.e., − Z2 e2

r = − Ze E0 R3

(S15.32)

which gives

E0 R 3 r= Ze

(S15.33)

The induced dipole moment becomes pe = Z e r = E0 R 3

(S15.34)

Hence the electronic polarizability is given by

 Ea = R 3

(S15.35)

which is temperature independent. Polarizability can also be expressed in terms of the dielectric function. If there are N atoms per unit volume, the polarization is given by Pe = N p e = N  Ea E 0

(S15.36)

From Eq. (15.32) one can write

D =  E 0 = E 0 + 4  Pe

(S15.37)

Therefore the polarization is given by Pe =

 −1 E0 4

(S15.38)

Substitute the value of Pe from Eq. (S15.38) in Eq. (S15.36) the polarizability becomes

 Ea =

1  −1 N 4

(S15.39)

Therefore, the electronic polarizability is independent of temperature.

Problem 15.6: Solution: In the presence of the external electric field E 0 unequal electric dipole moments are induced on the ions which give rise to finite polarization. If the motion of the ions in the positive direction is taken to be positive, then

p I1 = Z e x 1

(S15.40)

p I2 = − Z e x 2

(S15.41)

Here x 1 and x 2 are the distances moved by the positively and negatively charged ions having masses M1 and M 2 , respectively. Z is the valency of both the negatively and positively charged ions (assumed to be the same). The net dipole moment per unit cell becomes

p I = p I1 − p I2 = Z e (x 1 + x 2 )

(S15.42)

Let  F1 and  F2 are the force constants for the positively and negatively charged ions then the forces acting on the ions are given by Z e E 0 = F1 =  F1 x 1 = M1 02 x 1

(S15.43)

− Z e E 0 = F2 = −  F2 x 2 = − M 2 02 x 2

(S15.44)

Here  0 is the natural frequency of vibration. From the above equations x1 =

Ze E0 M1  02

(S15.45)

x2 =

Ze E0 M 2  02

(S15.46)

Substitute Eqs. (S15.45) and (S15.46) in Eq. (S15.42) we write

pI =

Z2e2 E 0

  02

(S15.47)

where  is the reduced mass defined as 1



=

1 1 + M1 M 2

(S15.48)

The ionic polarizability is given by

 Ia =

pI Z2 e2 = E0   02

(S15.49)

Chapter 17 Problem 17.1: Solution: Fig. 17.4 shows the propagation of a plane wave from one medium into the other where the polarization of the wave is parallel to the interface. The four boundary conditions are defined by Eqs. (17.37) to (17.40) or by Eqs. (17.43) to (17.46). From the Fig. 17.4 it is evident that the magnetic field is parallel to the interface separating the two media. Therefore, there is no component of magnetic field B perpendicular to the interface and hence the boundary condition (17.42) becomes redundant. The boundary condition (17.45) can be written as

E 0 sin (/2 − i ) − E 0 sin (/2 − i ) = E 0 sin (/2 − r ) which can be written as

(E 0 − E 0 ) cos i = E 0 cos r

(S17.1)

The boundary conditions (17.37) and (17.40) become the same. Eq. (17.40) can be written as

1

(B 0 + B0 )  nˆ = 1 B0  nˆ

(S17.2)

(B 0 + B0 ) = 1 B0

(S17.3)



 1





Substitute the value of B 0 from Eq. (17.22) in the above equation we write

 ( E 0 + E0 ) = 

 E 0 

(S17.4)

Eqs. (S17.1) and (S17.4) can be solved for E 0 / E 0 and E 0 / E 0 to yield

E 0 =2 E0

 sin 2i    sin 2r +  sin 2i 

 sin 2i − sin 2r E 0  =  E0 sin 2r + sin 2i  For  =   the above equations reduce to

(S17.5)

(S17.6)

E 0 2 cos i sin r → E0 sin (i + r ) + cos (i − r )

(S17.7)

E 0 tan (i - r ) → E0 tan (i + r )

(S17.8)

Problem 17.2: Solution: Eqs. (17.84) and (17.85) of the text can be written here for completeness as

1.

n 2 − n 22 = n 12 − n 22 =  1

(S17.9)

2 n n 2 = 2 n1 n 2 =  2

(S17.10)

(n + n ) = (n − n ) + 4 n n

n 2 + n 22 =

2

2 2 2

2

2 2 2

2

2 2

=  12 +  22

(S17.11)

2. Add Eqs.(S17.9) and (S17.11) we get

2 n 2 =  12 +  22 +  1





(S17.12)



(S17.13)

1  12 +  22 +  1 2 3. Subtract Eq. (S17.9) from Eq. (S17.11) we get n2 =

2 n 22 =  12 +  22 −  1



1  12 +  22 −  1 2 4. One can write n 22 =

(

2 n = 4 n 2 = 2 n 2 + n 22 + n 2 − n 22 =

(

2  12 +  22 +  1

)

) (S17.14)

Problem 17.3: Solution: According to Eq. (17.99) the reflection coefficient is given by

R=

(n − 1)2 + n 22 (n + 1)2 + n 22

(S17.15)

n 2 + n 22 + 1 − 2 n R= 2 n + n 22 + 1 + 2 n

(S17.16)

Substitute for n 2 + n 22 and 2 n from Eqs. (S17.11) and (S17.14) in the above equation one gets

) ( 1 +  +  + 2(  +  +  ) 1 +  12 +  22 −

R=

2 1

2 2

2  12 +  22 +  1 2 1

2 2

(S17.17)

1

Problem 17.4: Solution: The equation of motion of the electron is given by Eq. (17.154) and is written as

me

d2x dx + = e E = e E 0 e  t 2 dt dt

(S17.18)

The electrons collide with the atoms and ultimately acquire a small constant velocity called drift velocity under the influence of a steady and slowly varying electric field. The constant drift velocity yield zero acceleration, i.e.

d2x =0 dt 2 Therefore, the equation of motion given by Eq. (S17.18) reduces to



dx = e E = e E 0 e  t dt

(S17.19)

The above equation gives

 =

eE vd

(S17.20)

where the drift velocity v d = dx/dt . The current density j is defined as

J = n ee vd Or

vd =

 J = 0 E n ee n ee

Substitute Eq. (S17.22) for v d in Eq. (S17.20) we get

(S17.21) (S17.22)

 =

n ee2

(S17.23)

0

Problem 17.5: Eq. (17.177) is written as

 Pi2  ( ) =  ( ) − 2  −  02

(S17.24)

From the above equation for  = 0 we get

 (0) −  ( ) =

 Pi2  02

(S17.25)

Now Eq. (S17.24) can be written as

 ( ) −  ( ) =

 Pi2 1 2  0 1 −  2 /  02

(S17.26)

With the help of Eq. (S17.25) one gets

 ( ) −  ( ) =  (0 ) −  ( )

 ( ) −  ( ) =  (0) −  ( )

1 1 −  2 /  02

 02  02 −  2

(S17.27)

Problem 17.6: Eq. (S17.27) can be written as

 ( ) =  ( ) +  (0) −  ( )

 ( ) =

 ( ) =

 02  02 −  2

( −  )  () +   (0) −  () 2 0

2

2 0 2

 02 − 

 02  (0) −  2  ( )  02 −  2   (0 ) − 2  ( )   2 2 0 − 



 ( ) =

 ( )  02

Using Eq. (17.186) the above equation becomes

(S17.28)

 L2 −  2  ( ) =  ( ) 2 T −  2 Or

2 − 2  ( ) = L2  ( )  T −  2

(S17.29)

Problem 17.7: The equation of motion of an electron in the presence of finite mean free path is given by

me

m dx d2x + e = − eE 2  e dt dt

(S17.30)

In terms of velocity the above equation can be written as me

m dv + e v = − eE dt e

(S17.31)

The current density due to the flow of electrons is defined as

J = − ne e v

(S17.32)

Multiply Eq. (S17.31) by − n e e , we write me

m d (− n e e v) + e (− n e e v) = n e e 2 E dt e

n e2 dJ 1 + J= e E dt  e me Rearrange the terms we write

 me dJ n e e 2   E − = J  2 dt me  ne e  e  n e e2  dJ J   E −  = dt me   0 

(S17.33)

Chapter 18 Problem 18.1: Solution: The diamagnetic susceptibility is given by

d = −

 a Z e2 6 me c

2

r2

(S18.1)

Here we are given



r 2 = a 02 = 0.53  10− 8

 cm 2

2

 a = 2.7  10 24 cm − 3 We know that

e2 = 2.8  10 −13 cm 2 mec

Substitute the values in Eq. (S18.1) to write





2 2.7  10 24  2  2.8  10 −13 d = − 0.53  10 − 8 6

= − 0.71  10 − 5 in emu This value is in reasonable agreement with the experimental value − 0.86  10 − 5 emu for He. Problem 18.2: Let the paramagnetic susceptibility be given as

M =

 a  B2 3 k BT

(S18.2)

where  B is the Bohr magnetron. If the density of atoms  a = 2  10 22 atoms/cm3 find the paramagnetic susceptibility at room temperature taken as T = 300 K .

We know that

 B = 0.927  10 − 20 k B = 1.3807  10 − 16

k B T = 1.3807  10 − 16  300 = 4.142  10 − 14 erg

Therefore, the magnetic susceptibility is given by

(

)

2  10 22  0.927  10 − 20 M = = 1.38  10 − 5 − 14 3  4.142  10 Problem 18.3: Solution: We have seen that the magnetization in a paramagnetic material is given by

 H   M =  a  L  k BT 

(S18.3)

where L(x ) is the Langevin function and is defined as L(x ) = coth x −

1 x

For very small values of x, coth x function can be expanded in the form of a series as

coth x =

1 x x3 + − +. . . x 3 45

From the above two equations we get

x x3 − 3 45

L (x ) =

(S18.4)

Therefore, at high temperatures Langevin function in Eq. (S18.3) can be expanded and is given as 3  H 1  H      M=   − 45  k B T    3 k B T  a

(S18.5)

The paramagnetic susceptibility can be defined as 2 a 2  M 1  H      1 − M = = H 3 k BT  15  k B T    

One can also get differential susceptibility as follows

(S18.6)

2 a 2  dM 1  H      1 −  M = = dH 3 k BT  5  k B T    

(S18.7)

We see that Eqs. (S18.6) and (S18.7) differ in the second term which itself is quite small.

Problem 18.4: Solution: In the ion the partially filled shell contribute towards paramagnetic susceptibility while the filled shells contribute towards the diamagnetic susceptibility. In general also when magnetic field is applied, ions of all the elements exhibit finite diamagnetic susceptibility.

The diamagnetic susceptibility is given by

d =

 a Z e2 6 me c

2

r2

(S18.8)

The paramagnetic susceptibility at high temperatures is given as

M =

 a g 2J  B2 J (J + 1)

=  a g 2J

3 k BT e 2  2 J (J + 1) 4 m e2 c 2 3 k B T

(S18.9)

Divide Eq. (S18.9) by Eq. (S18.8) we write

 M g 2J 2 J (J + 1)  2 = d 4 Z k BT me r 2

(S18.10)

If we take Lande’s splitting factor g J as 2 then the above equation gives

 M 2 J (J + 1)  2 = d Z k BT me r 2

(S18.11)

Problem 18.5: Solution: There are a number of metals in which the valence electrons are free to move and these have no orbital motion but only spin. For example consider paramagnetic Na metal with electron configuration:

Na : 1s 2 2s 2 2p 6 3s1

In Na metal there is only one unpaired electron in the outer most orbit, i.e., 3s1 . In this case orbital quantum number is zero, i.e., L = 0 but the spin is finite. Therefore, J = S = 1/2 and from Eq. (18.25) the Lande’s splitting factor g J = g S = 2 . The magnetic moment of the free  electron  becomes

  = − gS B S = − 2 B S

(S18.12)

In the presence of magnetic field H , applied in the z-direction, the potential energy of the electron becomes

 E = −  H = 2 B S  H = 2 B H ms =  B H

(S18.13)

Here m s is the eigenvalue of spin quantum number S and has two values as  1 / 2 . Therefore, the energy state in the presence of the magnetic field splits up into two states with energy values   B H (see Fig. S18.1).

Let the population of atomic magnetic moments be governed by the Maxwell-Boltzmann distribution in which the number of atomic magnetic moments N in a state with energy E is given by defined as

N  e − E/k BT = e −  0 E

(S18.14)

Let N 1 and N 2 be the number of atomic magnetic moments in the states with energies −  B H and  B H . The total number of atomic magnetic moments in the two states is given by

N = N1 + N 2 From Eq. (S18.14) we can write

(S18.15)

N1  e 0 BH N 2  e − 0 BH Dividing the above two equations we get

N1 e  0 BH = −  0 BH N2 e

(S18.16)

Add one to both sides of the above equation to write

e  0 BH + e −  0 BH N = N2 e − 0 BH

(S18.17)

Subtract one from both sides of Eq. (S18.16) to write

N1 − N 2 e  0 BH − e − 0 BH = N2 e −  0  BH

(S18.18)

Divide Eq. (S18.18) by Eq. (S18.17) one gets

N1 − N 2 e  0 BH − e −  0 BH =  0 BH = tanh ( 0  B H ) N e + e − 0 BH

(S18.19)

Hence the magnetization in the metal is given by M =  B (N1 − N 2 ) = N  B tanh

BH k BT

(S18.20)

Let us study the particular cases with respect to T and H . When H is very small and the temperature T is large, the tangent hyperbolic can be expanded to write

 H  M = N  B  B + . . .   k BT  or

M = MH =

N  B2 H k BT

(S18.21)

The magnetic susceptibility becomes

M =

N  B2 k BT

(S18.22)

Hence when the temperature is high the magnetic susceptibility shows the Curie behavior. On the other hand if H is very large and the temperature T is very small, tanh

BH k BT

1

Therefore, one obtains the saturation magnetization with all the magnetic moments aligned along the direction of the magnetic field and is given by

M = N B

(S18.23)

Fig. S18.2 shows the variation of magnetization with the applied magnetic field at constant temperature. From Eq. (S18.17)

e −  0 BH N 2 = N 0 BH e + e − 0 BH

(S18.24)

For high values of H and low values of T ,  0  B H  1 . In this limit Eq. (S18.24) gives

N2 = 0 In other words, N1 = N Therefore, in the limit of high values of magnetic field all the atoms lie in the lower energy state and the spins of all the electrons are aligned in the direction of the field yielding saturation magnetization.

Chapter 19 Problem 19.1: Solution: The magnon dispersion relation for a crystalline solid is given by (see Eq. 19.109)

(

)

   K = 2 J s R  n −  exp  K       

(S19.1)

Let a be the size of the cube edge. In a bcc structure there are eight 1NNs with coordinates are   given as  = (a/2)(1,  1, 1) . Substitute the values of  in Eq. (S19.1) we write



 (K x + K y + K z )a/2

 K = 2 J s 8 − e

 (

 − K x + K y + K z )a/2

− 2Js e

 (K x + K y − K z )a/2

−e

 ( − K x + K y − K z )a/2

+ e

 (K x − K y + K z )a/2

−e

 (− K x − K y + K z )a/2

+ e

 (K x − K y − K z )a/2

−e

 (− K x − K y − K z )a/2

+ e





(S19.2) Here we have assumed s R = s as all the atoms are identical in a crystal. In the above equation the various terms can be arranged so that the exponential terms are changed into trigonometric functions



 (

 K y + K z )a/2

 K = 2 J s 8 − 2 cos(K x a/2) e



 (K y − K z )a/2

+e



= 2 J s 8 − 4 cos(K x a/2) cos(K y a/2) e K za/2 + e -  K za/2



 K = 16 J s 1− cos(K x a/2) cos(K y a/2) cos(K z a/2)

For small values of K ( K x , K y , K z ) one can write



 (− K y + K z )a/2

+e

 (− K y − K z )a/2

+e

 (S19.3)



cos(K x a/2 ) = 1 −

cos(K y a/2) = 1 − cos(K z a/2 ) = 1 −

(K x a/2)2 2

(K a/2)

+

(S19.4)

+

(S19.5)

+

(S19.6)

2

y

2

(K z a/2)2 2

Substitute Eqs. (S19.4) – (S19.6) in Eq. (S19.3) and neglect the higher powers (the terms having power more than two) one gets

 K = 2 J s K 2 a 2 Therefore

 K = K 2a 2 2Js

(S19.7)

 In fcc structure there are twelve 1NNs with coordinates  = (a/2)( 1,  1, 0) ,  (a/2)(1, 0 , 1) and (a/2)(0, 1, 1) . Substitute the values of  in Eq. (S19.1) we write

 ( − 2 J s e

 (K x + K y )a/2

 K = 2 J s 12 − e

 K y + K z )a/2

 (K x − K y )a/2

−e

 (K y − K z )a/2

+ e

 ( − K x + K y )a/2

−e

 (− K y + K z )a/2

+ e

 (− K x − K y )a/2

−e

 (− K y − K z )a/2

+ e







− 2 J s e ( K x + K z )a/2 + e (K x − K z )a/2 + e (− K x + K z )a/2 + e (− K x − K z )a/2



Rearranging the terms in the above equation we write



K = 2 J s 12 − 2 e K x a/2 cos(K y a/2) + 2 e − K x a/2 cos(K y a/2)



cos(K z a/2)



− 2 J s 2 e K x a/2 cos(K z a/2) + 2 e −  K x a/2 cos(K z a/2)



 K y a/2

− 2Js 2 e

cos(K z a/2) + 2 e

−  K y a/2





Or  K = 8 J s 3 − cos(K x a/2) cos(K y a/2) + cos(K y a/2) cos(K z a/2) + cos(K z a/2) cos(K x a/2)



(S19.8) At small values of K the above equation can be written as



2 2     (   K ya) ( K xa)      K = 8 J s  3 − 1 − . . . 1 − . . .    2 2      

 (K y a )2   (K z a )2  − 8 J s 1 − . . .  1 − ... 2 2      (K z a )2   (K x a )2  − 8 J s 1 − . . .  1 − . . . 2 2    

Simplify the above equation we get  K = K 2a 2 2Js

(S19.9)

Chapter 20 Problem 20.1: Solution: The spin wave frequency in an antiferromagnetic material is given by [see Eq. (20.58)] 2   1 2  2   K     =  ex  n −   e    4      2

(S20.1)

In a sc structure there are six 1NNs with coordinates a ( 1, 0, 0 ) , a (0,  1, 0 ) and a (0, 0,  1) .

Therefore, for a sc structure Eq. (S20.1) becomes

(

)

2 =

2 1 2  2 K a − K a  ex 6 − e K x a + e − K x a + e y + e y + e K z a + e − K z a    4

2 =

1 2  ex 36 − 2 cos K x a + 2 cos K y a + 2 cos K z a2 4





Further simplify the above equation we write

2 = −





1 2  ex 36 − 4 cos 2 K x a + 4 cos 2 K y a + 4 cos 2 K z a 4





1 2  ex 8 cos K x a cos K y a + 8 cos K y a cos K z a + 8 cos K z a cos K x a 4



2 = −

2 = −



(

)

(

)

(

1 2  ex 24 + 4 1 − cos 2 K x a + 4 1 − cos 2 K y a + 4 1 − cos 2 K z a 4

)



1 2  ex 8 cos K x a cos K y a + 8 cos K y a cos K z a + 8 cos K z a cos K x a 4



1 2  ex 24 + 4 sin 2 K x a + 4 sin 2 K y a + 4 sin 2 K z a 4







1 2  ex 8 cos K x a cos K y a + 8 cos K y a cos K z a + 8 cos K z a cos K x a 4



 2 = ex2 6 + sin 2 K x a + sin 2 K y a + sin 2 K z a 



− ex2 2 cos K x a cos K y a + 2 cos K y a cos K z a + 2 cos K z a cos K x a



(S20.2) This is the dispersion relation for the magnons in an antiferromagnetic material with sc structure. At small values of K

sin Ka = Ka And

(S20.3)

2 ( Ka ) cos Ka = 1 − ...

2

(S20.4)

Substitute Eqs. (S20.3) and (S20.4) in Eq. (S20.2) and simplify we get

 2 = ex2  6 + K 2 a 2 − 2 ( 3 − k 2 a 2 )

 2 = 3ex2 K 2 a 2 Hence the magnon frequency becomes

 = 3 ex Ka

(S20.5)

Problem 20.2: Solution: Let a be the size of the cube edge. In a bcc structure there are eight 1NNs with   coordinates are given as  = (a/2)(1,  1, 1) . Substitute the values of  in Eq. (S20.1) and simplify as done in Problem 19.1 we get



 2 =  ex2 8 2 − (8 cos K x a/2 cos K y a/2 cos K z a/2)2 1 4







= 16 ex2 1 − cos 2 K x a/2 cos 2 K y a/2 cos 2 K z a/2

(S20.6)

We know that for small values of K ( K x , K y , K z ), cos K x a/2 , cos K x a/2 and cos K x a/2 are given by Eqs. (S19.4)-(S19.6). Take the square of Eq. (S19.4) to write  (K x a/2)2  cos K x a/2 =  1 −  2  

2

2

Retaining the lowest order terms we get



2



(S20.7)

cos 2 K y a/2 = 1 − (K y a/2)



2



(S20.8)



2



(S20.9)

cos2 K x a/2 = 1 − (K x a/2)

Similarly we can get

cos2 K z a/2 = 1 − (K z a/2)

From Eqs. (20.7)-(S20.9) we can write



cos 2 K x a/2 cos 2 K y a/2 cos 2 K z a/2 = 1 − (K x a/2)

= 1 − K 2a 2 / 4

2

 1 − (K a/2)  1 − (K a/2)  2

y

2

z

(S20.10)

Substitute Eq. (S20.10) in Eq. (S20.6) we write

 2 = 4 ex2 K 2 a 2

(S20.11)

Therefore, the magnon frequency can be written as

 = Ka 2  ex

(S20.12)

Chapter 21 Problem 21.1: Solution: The magnetic moment of a nucleus with angular momentum I is give by



I =  I  I

(S21.1)

If magnetic field H is applied in the z-direction the magnetic moment in the direction of the field is given by 

ˆ =  I  Iz =  I  H I z

(S21.2)

Here Ĥ is a unit vector in the direction of magnetic field. The value of the magnetic moment becomes

 Iz =  I  m I

(S21.3)

Let the population of the nuclear state be given as

N mI  e

− E m I / k BT

(S21.4)

where the energy of the nuclear state is given by the Zeeman energy as E mI = −  I  H m I

(S21.5)

The average magnetic moment is defined as

  m e  0

I

 I =  Iz =

mI

I  H mI

I

 e 0  I  H mI

(S21.6)

mI

where

 0 = 1/ k BT

(S21.7)

m e

y mI

I

I =  I 

mI

e

y mI

(S21.8)

mI

where

y = 0  I  H =

I H k BT

(S21.9)

Eq. (S21.8) can be written as

I =  I 

 d  ln   e y m I  dy  m I 

The summation in the above equation can be easily performed to write

(S21.10)

2 I +1 − y   2 I2+ 1 y e − e 2  d = I  ln   dy  e y/2 − e − y/2   

I

This equation can further be simplified to get 2 I +1 1 y  2 I +1 coth y − coth  2 2 2  2

I =  I  

(S21.11)

Eq. (S21.11) gives the expression for the average nuclear magnetic moment.

Problem 21.2: Solution: From Eq. (S21.9) it is evident that y is very small at very high temperatures. For small y , we can expand cotangent hyperbolic function as coth y =

1 y y3 + − ... y 3 45

(S21.12)

So Eq. (S21.11) for small y can be expanded as  2 I + 1  2 1 2 I + 1 y  1  2 1 y   −  +  + 2 3  2  y 3 2   2  (2 I + 1) y

I =  I  

 1 (2 I + 1)2 y 1 1 y  = I   + − −  4 3 y 6 2 y

=  I  I ( I + 1)

y  I2  2 I (I + 1) = H 3 3 k BT

(S21.13)

Let  nuc be the number of nuclei per unit volume. The magnetization M becomes

M =

 nuc  I2  2 I (I + 1) 3 k BT

H

The nuclear magnetic susceptibility is given by

(S21.14)

M =

 nuc  I2  2 I (I + 1) 3 k BT

(S21.15)

Problem 21.3: Solution: The quadrupole moment of a nucleus is defined as

(

)

Q =  3 z 2 − r 2  N (r ) d 3 r

(S21.16)

If the nucleus is spherical in shape and has spherically symmetric charge density  N (r ) , then all the three Cartesian directions are equivalent, i.e., x 2 = y2 = z2 =

1 2 r 3

(S21.17)

The use of Eq. (S21.17) in Eq. (S21.16) yields

Q=0

(S21.18)

One should note that if condition (S21.17) is satisfied then Q xx and Q yy are also zero. The value of quadrupole moment measures the deviation of the shape of a nucleus from spherical symmetry. Further, the quadrupole moment of a nucleus with ellipsoidal shape is positive if its major axis is along the z-direction and negative if its minor axis is along the z-axis. Another property is that the quadrupole moment tensor is symmetric and traceless, i.e., Q = Q 

(S21.19)

Q  = 0  

(S21.20)

Chapter 22 Problem 22.1: Solution: The London penetration depth is given by

 me c2   d L =  2  4  n e s  

1/ 2

We know that

c = 3.0  1010 cm/sec.

(S22.1)

m e = 9.1 10− 28 gm.

n s = 6.0  1022 electrons/cm3

e = 4.8  10 − 10 e.s.u.

Substitute these values in Eq. (S22.1) we get

(

)

2   9.1  10 − 28  3.0  1010   dL = 2  4  3.141  6.0  10 22  4.8  10 −10   

(

1/ 2

)

= 217.2  10 − 8 cm = 217.2 A In simple metals the London penetration depth is of the order of few hundred angstroms.

Problem 22.2: Solution:

Fig. S22.1 shows the geometry of the plate with respect to the Cartesian coordinates. The magnetic field outside the plate is equal to the applied magnetic field H and decreases exponentially as it enters the plate. The London equation is written as

d 2 H(x ) 1 = 2 H(x ) 2 dx dL

(S22.2)

Here d L is the London penetration depth defined in Eq. (S22.1). The general solution of Eq. (S22.2) is given as

 x   x  H(x ) = A cosh  + B sinh    dL   dL 

(S22.3)

The boundary conditions are

H(− d/2 ) = H( d/2 ) = H

(S22.4)

Applying the boundary conditions we can easily get

H(x ) = H

cosh (x/d L ) cosh (d/2 d L )

(S22.5)

The magnetic field inside the plate is given as

H(x ) = H + 4  M(x )

(S22.6)

From Eqs. (S22.5) and (S22.6) one can solve for magnetization

M(x ) =

H cosh (x/d L ) − cosh (d/2 d L ) 4 cosh (d/2 d L )

(S22.7)

For d  d L the trigonometric functions can be expanded in terms of series. Retaining the lowest order terms one can get

M(x ) =

H  2 1 2 x − d  4  8  d 2L 

(S22.8)

Problem 22.3: Solution: We know that the variation of magnetic field with temperature is given by (see Eq. 22.43)

2  T  H c (T ) = H c (0 )  1 −      Tc  

(S22.9)

We know that

Tc = 3.7 K

H c (0) = 310 gauss T = 2K Substitute these values in Eq. (S22.2) we get 2   2   H c (2) = 310  1 −     3.7   

= 219.4 gauss Problem 22.4: Solution: From Eq. (22.86) we write Ss − S n =

1 d 2 Hc 8  dT

=

dH c 1 Hc 4 dT

(S22.10)

Differentiating Eq. (S22.9) we write dH c T = − 2 H c (0) 2 dT Tc

(S22.11)

Substitute Eqs. (S22.9) and (S22.11) in Eq. (S22.10) we get 2  T  1 2 H c (0) T  1 −    Ss − S n = − 2  Tc2   Tc  

(S22.12)

which is the desired result. The negative sign indicates that the entropy decreases as the material undergoes a transition from the normal to the superconducting state. Substitute T = Tc / 2 in Eq. (S22.12) the change in entropy is given by

Ss − S n = −

H c (0)2

3 16  Tc

(S22.13)

Problem 22.5: From Eq. (22.184) the temperature dependence of the London penetration depth is given by 4  T  d L (T ) = d L (0)  1 −      Tc  

−1 / 2

(S22.14)

Therefore, the penetration depths at temperatures T1 and T2 are given by

T4 − T4  d L (T1 ) = d L (0)  c 4 1   Tc 

−1 / 2

 Tc4 − T24  d L (T2 ) = d L (0)   4  Tc 

−1 / 2

(S22.15)

(S22.16)

Divide Eq. (S22.16) by Eq. (S22.15) to write

d L (T2 )  Tc4 − T14   = d L (T1 )  Tc4 − T24 

1/ 2

(S22.17)

The given quantities are as under

d L (T1 ) = 396 A , T1 = 3 K , d L (T2 ) = 1730 A , T2 = 7.1 K From Eq. (S22.17) the value of Tc is given by

 d (T )  T  L 2  − T14  d L (T1 )  T4 = 2

4 2

c

 d L (T2 )    − 1  d L (T1 )  2

Substitute the given values in the above equation, we write 2

(7.1)4  1730  − (3)4 Tc4 =

 396  2  1730    −1  396 

(S22.18)

Solve the above equation we get

Tc = 7.19 K Chapter 23 Problem 23.1: Solution: The energy of formation of a vacancy E v is given to us E v = 1.1eV = 1.1 1.6  10−12 = 1.76  10−12 ergs

The Boltzmann constant has the following standard value

k B = 1.38  10 − 16 ergs/K From Eq. (23.14) the number of vacancies at temperature is given as

Nv = NL e

_

Ev k BT

Therefore the fraction of vacancies present in the crystal at temperature T is Ev

_ Nv = e k BT NL

(S23.1)

At 1000K the above equation gives 1.76 10 - 12

_ Nv - 16 = e 1.3810 1000 = e − 12.75 = 2.90  10 − 6 NL

(S23.2)

At 500K one can write 1.76 10 - 12

− Nv − 16 = e 1.3810 500 = e − 25.5 = 8.42  10 − 12 NL

(S23.3)

Problem 23.2: Solution: The Gibb’s free energy is defined as

G(P, V, T) = H − TS where H is the enthalpy given by

(S23.4)

H = E + PV

(S23.5)

P , V and T are pressure, volume and temperature of the crystal. E is the energy of the

crystal. Let N P point defects are introduced in the lattice with N L lattice points. The change in the Gibb’s free energy due to the introduction of the point defects at constant temperature is given by

where

G = N P H − T S

(S23.6)

H = E + P V + V P

(S23.7)

Here E is the energy of formation of the point defect. The change in entropy S consists of two contributions: the configurational entropy S c arising from the distribution of point defects on the lattice sites and second is the vibrational entropy S v which arises due to the change in vibrational frequency of the atoms in the presence if the point defects. Here we neglect the vibrational entropy and therefore Eq. (S23.6) reduces to

G = N P H − T S c

(S23.8)

In the equilibrium state G must be minimum with respect to the distribution of the point defects which gives  (G ) = H − T  (Sc ) = 0 N P N P

(S23.9)

Eq. (S23.9) describes the equilibrium state of the lattice with point defects.

Let there be N s substitutional point defects in the crystal with N L lattice points. Now eq.

(S23.9) becomes H s − T

 (Sc ) = 0 N s

(S23.10)

H s is the change in enthalpy required to form a substitutional point defect. S c is given as

S c = k B ln (W )

(S23.11)

The number of ways in which N s substitutional point defects can be distributed on the lattice is given by Eq. (23.9) of the book. Substitute the value of S c obtained from Eqs. (S23.11) and (23.9) in Eq. (S23.10) we write

H s − k B T

  N L!  =0 ln  N s  (N L − N s )! N s ! 

(S23.12)

One can solve the above equation in the same way as done in section 23.1.4.1 to get

 H s   N s = N L exp  −  k BT 

(S23.13)

Similarly one can derive the expressions for the distribution of other point defects such as interstitial point defects and the Frenkel defects etc.

Chapter 25 Problem 25.1 The energy of the electron states in a quantum well is given by [see Eq. (25.12)]

E nz =

n 2z h 2 8 m e L2z

(S25.1)

The first excited state corresponds to n z = 2 and the ground state to n z = 1. So the difference in energy is obtained as

E =

(2 − 1 ) h 2

2

8 m e L2z

2

=

3 h2 8 m e L2z

(S25.2)

Substitute the values we get

3  (6.626  10 − 34 J s) 2 E = 0.067  8  (9.1  10 −31 kg)  (10  10 −9 m) 2

= 2.679  10 −20 J The wavelength of the radiations emitted can be obtained from the equation E = h =

Or

=

hc



(

) (

hc 6.626  10 − 34 J s  3  10 8 m s −1 = E 2.697  10 − 20 J

= 7.375  10 − 6 m

)

The wave length of the radiations emitted lies in the infrared region.

Problem 25.2: From Eq. (25.36) one can write

N(E k ) dE k = N(k ) dk Or

N (E k ) =

N (k ) dE k / dk

(S25.3)

The energy eigenvalues in the quantum ring are given by (see Eq. 25.64)

 2 k 2n E kn = 2 me

(S25.4)

Here k n is a quantized wave vector and has values kn =

n R

(S25.5)

with n = 0,  1 ,  2 ,  3 , . . . From Eq. (S25.5), there is one state in 1/R distance in the k space. Hence the density of states become

N(k n ) = R

(S25.6)

Differentiate Eq. (S25.4) to write

dE k n dk n

=

2k n 2 n = me me R

(S25.7)

Eq. (S25.3) for a quantum ring can be written as

( )

N E kn =

N(k n ) dE k n / dk n

(S25.8)

Substitute Eqs. (S25.6) and (S25.7) in Eq. (S25.8) we get

( )

N E kn

me R 2 = 2  n

which is the desired result.

Problem 25.3:

(S25.9)

Probability of tunneling of an electron through a potential barrier having width b = 10 − 6 cm and height V0 = 0.3 eV is given by

P=e

−2b

2 me 2

(V0 − E )

(S25.10)

The known quantities are m e = 9.1 10− 28 gm

b = 10 − 6 cm  = 1.0546  10 − 27 erg − sec.

Substitute the values of m e , b and  in the Eq. (S25.10) we write

P=e

=e

2 9.110 − 28

- − 210 − 6

 1.054610 − 27   

2

( V0 − E )

( V0 − E )

− 8.092 10 7

(S25.11)

If the electron has energy E = 0.1 eV then

P = e − 8.092 10  (0.3 − 0.1)1.6022 10 7

− 12

= e − 45.8 = 1.286  10 − 20 If the energy of the electron is E = 0.2 eV, then

P = e − 8.092 10  (0.3 − 0.2 )1.6022 10 7

− 12

= e − 32.4 = 8.489  10 − 15 If the electron has energy is E = 0.28 eV, then

P = e − 8.092 10  (0.3 − 0.28 )1.6022 10 7

P = e − 14.48 = 5.145  10 − 7

− 12

From the above example we see that as the energy of the electron approaches the potential energy of the barrier the probability of tunneling increases exponentially and when E = Vo the probability becomes one.