TEST BANK FOR Introduction to Solid State Physics 8th Edition By Kittel C. (Solution manual)

1. The vectors ˆˆˆ++xyz and ˆˆˆ−−+xyz are in the directions of two body diagonals of a cube. If θ is the angle between them, their scalar product gives cos θ = –1/3, whence . 1cos1/3901928'10928'−θ==°+°=° 2. The plane (100) is normal to the x axis. It intercepts the a' axis at and the c' axis at ; therefore the indices referred to the primitive axes are (101). Similarly, the plane (001) will have indices (011) when referred to primitive axes. 2a'2c' 3. The central dot of the four is at distance cos60actn60cos303aa°=°=° from each of the other three dots, as projected onto the basal plane. If the (unprojected) dots are at the center of spheres in contact, then 222aca,23⎛⎞⎛⎞=+⎜⎟⎜⎟⎝⎠⎝⎠ or 2221c8ac;1.633.34a3==
1-1
CHAPTER 2
1. The crystal plane with Miller indices hk􀁁 is a plane defined by the points a1/h, a2/k, and . (a)
Two vectors that lie in the plane may be taken as a
3 a / 􀁁
1/h – a2/k and 1 3 a / h − a / 􀁁 . But each of these vectors
gives zero as its scalar product with 1 2 h k 3 G = a + a + 􀁁a , so that G must be perpendicular to the plane
hk􀁁 . (b) If nˆ is the unit normal to the plane, the interplanar spacing is 1 nˆ ⋅a /h . But ,
whence . (c) For a simple cubic lattice
nˆ =G/ |G|
1 d(hk􀁁) = G⋅a / h|G| = 2π/ |G| G = (2π / a)(hxˆ + kyˆ + 􀁁zˆ) ,
whence
2 2 2 2
2 2 2
1 G h k .
d 4 a
+ +
= =
π
􀁁
1 2 3
1 3a 1 a 0
2 2
2. (a) Cell volume 1 3a 1 a 0
2 2
0 0
a ⋅a ×a = −
c
1 3 a2c.
2
=
2 3
1 2
1 2 3
2 3
ˆ ˆ
(b) 2 4 1 3a 1 a 0
| | 3a c 2 2
0 0
2 ( 1 ˆ ˆ ), and similarly for , .
a 3
× π
= π = −
⋅ ×
π
= +
x ˆ
c
y z
b a a
a a a
x y b b
(c) Six vectors in the reciprocal lattice are shown as solid lines. The broken
lines are the perpendicular bisectors at the midpoints. The inscribed hexagon
forms the first Brillouin Zone.
3. By definition of the primitive reciprocal lattice vectors
3 2 3 3 1 1 2 3
3 1 2 3
1 2 3
3
C
) (a a ) (a a ) (a a ) (2 ) / | (a a a ) |
| (a a a ) |
/V .
BZ V (2
(2 )
× ⋅ × × ×
= π ⋅ ×
⋅ ×
= π
= π
For the vector identity, see G. A. Korn and T. M. Korn, Mathematical handbook for scientists and
engineers, McGraw-Hill, 1961, p. 147.
4. (a) This follows by forming
2-1
2
2 1
2
2 1
2
|F| 1 exp[ iM(a k)] 1 exp[iM(a k)]
1 exp[ i(a k)] 1 exp[i(a k)]
1 cosM(a k) sin M(a k) .
1 cos(a k) sin (a k)
− − ⋅Δ − ⋅Δ
= ⋅
− − ⋅Δ − ⋅Δ
− ⋅ Δ ⋅ Δ
= =
− ⋅Δ ⋅ Δ
(b) The first zero in
sin 1 M
2
ε occurs for ε = 2π/M. That this is the correct consideration follows from
zero, 1
as Mh is
an integer
sin M( h 1 ) sin Mh cos 1 M cos Mh sin M .
2 2 ±
π + ε =􀀈􀀋􀀉π􀀋􀀊 ε +􀀈􀀋􀀉π􀀋􀀊 ε
1
2
5. 2 i(x jv1 +yjv2 +zjv3 )
1 2 3 S (v v v ) f e
j
− π = Σ
Referred to an fcc lattice, the basis of diamond is
000; 1 1 1 .
4 4 4
Thus in the product
1 2 3 S(v v v ) = S(fcc lattice)× S (basis) ,
we take the lattice structure factor from (48), and for the basis
1 2 3
i 1 (v v v ).
S (basis) 1 e 2 − π + + = +
Now S(fcc) = 0 only if all indices are even or all indices are odd. If all indices are even the structure factor
of the basis vanishes unless v1 + v2 + v3 = 4n, where n is an integer. For example, for the reflection (222)
we have S(basis) = 1 + e–i3π = 0, and this reflection is forbidden.
2 3 1
G 0 0
3 3
0 0
3 3 2 2 2
0 0 0
2 2 2
0
6. f 4 r ( a Gr) sin Gr exp ( 2r a ) dr
(4 G a ) dx x sin x exp ( 2x Ga )
(4 G a ) (4 Ga ) (1 r G a )
16 (4 G a ) .
= ∞ π π − −
= −
= +
+
∫
∫
0
The integral is not difficult; it is given as Dwight 860.81. Observe that f = 1 for G = 0 and f ∝ 1/G4 for
0 Ga >>1.
7. (a) The basis has one atom A at the origin and one atom
B at 1 a.
2
The single Laue equation
defines a set of parallel planes in Fourier space. Intersections with a sphere are
a set of circles, so that the diffracted beams lie on a set of cones. (b) S(n) = f
a ⋅Δk = 2π× (integer)
A + fB e–iπn. For n odd, S = fA –
2-2
fB; for n even, S = fA + fB. (c) If fA = fB the atoms diffract identically, as if the primitive translation vector were 1a2 and the diffraction condition 1()2(integer).2⋅Δ=π×ak
2-3
CHAPTER 3
1. E = (h/ 2 2M) (2π λ)2 = (h/ 2 2M) (π L)2 , with λ = 2L.
2. bcc: U(R) = 2Nε[9.114(σ R )12 −12.253(σ R)6 ]. At equilibrium and
6 6
0 R =1.488σ ,
0 U(R ) = 2Nε( − 2.816).
fcc: U(R) = 2Nε[12.132(σ R )12 −14.454(σ R)6 ]. At equilibrium and
Thus the cohesive energy ratio bcc/fcc = 0.956, so that the fcc structure is
more stable than the bcc.
6 6
0 R =1.679σ ,
0 U(R ) = 2Nε( − 4.305).
23 16 9
3. | U| 8.60 N
(8.60) (6.02 10 ) (50 10 ) 25.9 10 erg mol
2.59 kJ mol.
−
= ε
= × × = ×
=
This will be decreased significantly by quantum corrections, so that it is quite reasonable to find the same
melting points for H2 and Ne.
4. We have Na → Na+ + e – 5.14 eV; Na + e → Na– + 0.78 eV. The Madelung energy in the NaCl
structure, with Na+ at the Na+ sites and Na– at the Cl– sites, is
2 10 2
12
8
e (1.75) (4.80 10 ) 11.0 10 erg,
R 3.6610
−
−
−
α ×
= = ×
×
or 6.89 eV. Here R is taken as the value for metallic Na. The total cohesive energy of a Na+ Na– pair in the
hypothetical crystal is 2.52 eV referred to two separated Na atoms, or 1.26 eV per atom. This is larger than
the observed cohesive energy 1.13 eV of the metal. We have neglected the repulsive energy of the Na+ Na–
structure, and this must be significant in reducing the cohesion of the hypothetical crystal.
5a.
2
n
U(R) N A q ; 2 log 2 Madelung const.
R R
⎛ α ⎞
= ⎜ − ⎟ α = =
⎝ ⎠
In equilibrium
2
n
n 1 2 0 2
0 0
U N nA q 0 ; R n
R R R +
∂ ⎛ α ⎞
= ⎜ − + ⎟ = =
∂ ⎝ ⎠ α
A,q
and
2
0
0
U(R ) N q (1 1).
R n
α
= − −
3