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)
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 + B0 ) nˆ = 1 B0 nˆ
(S17.2)
(B 0 + B0 ) = 1 B0
(S17.3)
1
Substitute the value of B 0 from Eq. (17.22) in the above equation we write
( E 0 + E0 ) =
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)