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