Baixe Solucionário Estado Sólido e outras Exercícios em PDF para Física, somente na Docsity!
Introduction to Solid State Physics, 8th Edition
Charles Kittel
CHAPTER 1: CRYSTAL STRUCTURE.
Periodic Array of Atoms. Fundamental Types of Lattices. Index System for Crystal Planes. Simple Crystal Structures. Direct Imaging of Atomic Structure. Nonideal Crystal Structures. Crystal Structure Data.
CHAPTER 2: WAVE DIFFRACTION AND THE RECIPROCAL LATTICE. Diffraction of Waves by Crystals. Scattered Wave Amplitude. Brillouin Zones. Fourier Analysis of the Basis.
CHAPTER 3: CRYSTAL BINDING AND ELASTIC CONSTANTS. Crystals of Inert Gases. Ionic Crystals. Covalent Crystals. Metals. Hydrogen Bonds. Atomic Radii. Analysis of Elastic Strains. Elastic Compliance and Stiffness Constants. Elastic Waves in Cubic Crystals.
CHAPTER 4: PHONONS I. CRYSTAL VIBRATIONS. Vibrations of Crystals with Monatomic Basis. Two Atoms per Primitive Basis. Quantization of Elastic Waves. Phonon Momentum. Inelastic Scattering by Phonons.
CHAPTER 5: PHONONS II. THERMAL PROPERTIES. Phonon Heat Capacity. Anharmonic Crystal Interactions. Thermal Conductivity.
CHAPTER 6: FREE ELECTRON FERMI GAS. Energy Levels in One Dimension. Effect of Temperature on the Fermi-Dirac Distribution. Free Electron Gas in Three Dimensions. Heat Capacity of the Electron Gas.
Electrical Conductivity and Ohm’s Law. Motion in Magnetic Fields. Thermal Conductivity of Metals.
CHAPTER 7: ENERGY BANDS. Nearly Free Electron Model. Bloch Functions. Kronig-Penney Model. Wave Equation of Electron in a Periodic Potential. Number of Orbitals in a Band.
CHAPTER 8: SEMICONDUCTOR CRYSTALS. Band Gap. Equations of Motion. Intrinsic Carrier Concentration. Impurity Conductivity. Thermoelectric Effects. Semimetals. Superlattices.
CHAPTER 9: FERMI SURFACES AND METALS. Construction of Fermi Surfaces. Electron Orbits, Hole Orbits, and Open Orbits. Calculation of Energy Bands. Experimental Methods in Fermi Surface Studies.
CHAPTER 10: SUPERCONDUCTIVITY. Experimental Survey. Theoretical Survey. High-Temperature Superconductors.
CHAPTER 11: DIAMAGNETISM AND PARAMAGNETISM. Langevin Diamagnetism Equation. Quantum Theory of Diamagnetism of Mononuclear Systems. Paramagnetism. Quantum Theory of Paramagnetism. Cooling by Isentropic Demagnetization. Paramagnetic Susceptibility of Conduction Electrons.
CHAPTER 12: FERROMAGNETISM AND ANTIFERROMAGNETISM. Ferromagnetic Order. Magnons. Neutron Magnetic Scattering. Ferrimagnetic Order. Antiferromagnetic Order. Ferromagnetic Domains.
CHAPTER 18: NANOSTRUCTURES.
Imaging Techniques for Nanostructures. Electronic Structure of 1D Systems. Electrical Transport in 1D. Electronic Structure of 0D Systems. Electrical Transport in 0D. Vibrational and Thermal Properties of Nanostructures.
CHAPTER 19: NONCRYSTALLINE SOLIDS. Diffraction Pattern. Glasses. Amorphous Ferromagnets. Amorphous Semiconductors. Low Energy Excitations in Amorphous Solids. Fiber Optics.
CHAPTER 20: POINT DEFECTS. Lattice Vacancies. Diffusion. Color Centers.
CHAPTER 21: DISLOCATIONS. Shear Strength of Single Crystals. Dislocations. Strength of Alloys. Dislocations and Crystal Growth. Hardness of Materials.
CHAPTER 22: ALLOYS. General Consideration. Substitutional Solid Solutions – Hume-Rotherby Rules. Order-Disorder Transformation. Phase Diagrams. Transition Metal Alloys. Kondo Effect.
CHAPTER 1
- The vectors x ˆ + y ˆ + z ˆ and − x ˆ − y ˆ + z ˆ 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
θ = cos −^1 1/ 3 = 90 ° + 19 28'° = 109 28'°.
- 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'
- The central dot of the four is at distance
cos 60 a ctn 60 cos 30 (^3)
a a
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
(^2 ) a 2 a^ c , 3 2
⎝ ⎠ ⎝^ ⎠
or
(^2) a 2 1 c ; 2 c (^8) 1.633. 3 4 a 3
2
2 1 2 2 1 2
1 exp[ iM(a k)] 1 exp[iM(a k)] |F| 1 exp[ i(a k)] 1 exp[i(a k)] 1 cos M(a k) sin^ M(a^ k) . 1 cos(a k) (^) sin (a k)
(b) The first zero in
sin 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 ) sin Mh cos M cos Mh sin M. (^2 2) ±
π + ε = (^) π ε + (^) π ε
- j^1 j^2 j^3 2 i(x v +y v +z v ) S (v v v ) 1 2 3 f (^) j e − π = Σ
Referred to an fcc lattice, the basis of diamond is
Thus in the product
S(v v v ) 1 2 3 = 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 v 1 + v 2 + v 3 = 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 22 0 0 0 2 22 0
- 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/G^4 for Ga 0 >>1.
- (a) The basis has one atom A at the origin and one atom
B at 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 ⋅ ∆ k = 2 π×(integer) A + fB e–iπn. For n odd, S = fA –
fB; for n even, S = fA + fB. (c) If fA = fB the atoms diffract identically, as if the primitive translation vector
were
a 2
and the diffraction condition
( ) 2 (integer). 2
a ⋅ ∆ k = π ×
b. (^) ( ) ( )
(^2 ) 0 0 0 2 0 0
1 U
U(R -R ) U R R R... ,
2 R
δ = + δ + ∂
bearing in mind that in equilibrium (^) R 0
( U∂ ∂R) =0.
2 2 2 n 2 3 3 3 0 0 0 0
U n(n 1)A 2 q (n 1) q 2 N N R (^) R R R R R
2
⎛ (^) ∂ ⎞ ⎛^ + α ⎞^ ⎛ + α ⎜ ⎟ =^ ⎜^ −^ ⎟^ =^ ⎜ − ⎝ ∂ ⎠ (^) ⎝ ⎠ ⎝
2
0
αq ⎞ ⎟ ⎠
For a unit length 2NR 0 = 1, whence
0
2 2 2 2 2 2 4 0 2 2 R 0 0 R^0
U q U (n 1) q log 2 (n 1) ; C R R (^) 2R R R
⎛ ∂ ⎞ α ∂ − ⎜ ⎟ =^ −^ =^ = ⎝ ∂^ ⎠ ∂
.
- For KCl, λ = 0.34 × 10 –8^ ergs and ρ = 0.326 × 10 –8Å. For the imagined modification of KCl with the ZnS structure, z = 4 and α = 1.638. Then from Eq. (23) with x ≡ R 0 /ρ we have
x e^2 −^ x^ = 8.53 ×10 .−^3
By trial and error we find x 9.2, or R 0 = 3.00 Å. The actual KCl structure has R 0 (exp) = 3.15 Å. For the imagined structure the cohesive energy is
2 2 0 0
- αq p U U= 1- , or =-0. R R q
in units with R 0 in Å. For the actual KCl structure, using the data of Table 7, we calculate (^2)
U
q
units as above. This is about 0.1% lower than calculated for the cubic ZnS structure. It is noteworthy that the difference is so slight.
- The Madelung energy of Ba+^ O–^ is –αe^2 /R 0 per ion pair, or –14.61 × 10 –12^ erg = –9.12 eV, as compared with –4(9.12) = –36.48 eV for Ba++^ O--. To form Ba+^ and O–^ from Ba and O requires 5.19 – 1.5 = 3.7 eV; to form Ba++^ and O--^ requires 5.19 + 9.96 – 1.5 + 9.0 = 22.65 eV. Thus at the specified value of R 0 the binding of Ba+^ O–^ is 5.42 eV and the binding of Ba++^ O--^ is 13.83 eV; the latter is indeed the stable form.
- From (37) we have eXX = S 11 XX, because all other stress components are zero. By (51), 3S 11 = 2 (C 11 − C 12 ) + 1 (C 11 +C 12 ).
Thus Y = (C 11 2 + C C 12 11 − 2C 12 2 ) (C 11 +C 12 );
further, also from (37), eyy = S 21 Xx,
whence σ = e (^) yy e (^) xx= S 21 S 11 = − C 12 (C 11 + C 12 ).
- For a longitudinal phonon with K || [111], u = v = w.
2 2 11 44 12 44 1 2 11 12 44
[C 2C 2(C C )]K 3,
or v K [(C 2C 4C 3 ρ)]
ω ρ = + + + = ω = + +
This dispersion relation follows from (57a).
- We take u = – w; v = 0. This displacement is ⊥ to the [111] direction. Shear waves are degenerate in this direction. Use (57a).
- Let exx = − eyy= 12 e in (43). Then
1 1 2 1 2 1 2 11 4 4 4 12 1 1 2 2 2 11 12
U C ( e e ) C e [ (C C )]e
2
so that
2 2 2 n 2 3 3 3 0 0 0 0
U n(n 1)A 2 q (n 1) q 2 N N R (^) R R R R R
2
⎛ (^) ∂ ⎞ ⎛^ + α ⎞^ ⎛ + α ⎜ ⎟ =^ ⎜^ −^ ⎟^ =^ ⎜ − ⎝ ∂ ⎠ ⎝ ⎠ ⎝
2
0
αq ⎞ ⎟ ⎠
is the effective shear
constant.
12a. We rewrite the element aij = p – δij(λ + p – q) as aij = p – λ′ δij, where λ′ = λ + p – q, and δij is the Kronecker delta function. With λ′ the matrix is in the “standard” form. The root λ′ = Rp gives λ = (R – 1)p
- q, and the R – 1 roots λ′ = 0 give λ = q – p.
b. Set
i[(K 3) (x y z) t] 0 i[.... .] 0 i[.... .] 0
u (r, t) u e ; v(r, t) v e ; w(r, t) w e ,
=^ +^ +^ −ω
=
as the displacements for waves in the [111] direction. On substitution in (57) we obtain the desired equation. Then, by (a), one root is
2 2 ω ρ = 2p + q = K (C 11 + 2C 12 +4C 44 ) / 3,
and the other two roots (shear waves) are
2 2 ω ρ = K (C 11 − C 12 +C 44 ) / 3.
- Set u(r,t) = u 0 ei(K·r – t)^ and similarly for v and w. Then (57a) becomes
2 2 2 2 0 11 y 44 y z 12 44 x y 0 x z 0
u [C K C (K K )]u (C C ) (K K v K K w )
ω ρ = + +
0
and similarly for (57b), (57c). The elements of the determinantal equation are
C HAPTER 4
1a. The kinetic energy is the sum of the individual kinetic energies each of the form (^) S^2
Mu. 2
The force
between atoms s and s+1 is –C(us – us+1); the potential energy associated with the stretching of this bond is
2 s 1
C(u u ) 2
− (^) s + , and we sum over all bonds to obtain the total potential energy.
b. The time average of (^) S^2 2
Mu is M u. 2 4
ω In the potential energy we have
u s 1 u cos[ t (s 1)Ka] u{cos( t sKa) cos Ka sin ( t sKa) sin Ka}.
- =^ ω −^ +^ =^ ω −^ ⋅
- ω − ⋅
Then u s u (^) s 1 u {cos( t sKa) (1 cos Ka) sin ( t sKa) sin Ka}.
− (^) + = ω − ⋅ − − ω − ⋅
We square and use the mean values over time:
cos^2 sin^2 1 ; cos sin 0. 2
Thus the square of u{} above is
(^1) u [1 (^2) 2cos Ka cos Ka (^2) sin Ka] (^2) u (1 (^2) cos Ka). 2
The potential energy per bond is 2
Cu (1 cos Ka), 2
− and by the dispersion relation ω^2 = (2C/M) (1 –
cos Ka) 2 2
this is equal to M u. 4
ω Just as for a simple harmonic oscillator, the time average potential
energy is equal to the time-average kinetic energy.
- We expand in a Taylor series
2 2 2 2 s (^) s
u 1 u u(s p) u(s) pa p a ; x 2 x
⎛ ∂ ⎞^ ⎛^ ∂ ⎞
⎝ ∂^ ⎠ ⎝ ∂ ⎠
On substitution in the equation of motion (16a) we have
2 2 2 2 (^2) p 0 p 2
u u M ( p a C ) t > x
which is of the form of the continuum elastic wave equation with
2 1 2 2 v M (^) p 0 p a C p −
- From Eq. (20) evaluated at K = π/a, the zone boundary, we have
2 1 2 2
M u 2Cu ; M v 2Cv.
−ω = − −ω = −
Thus the two lattices are decoupled from one another; each moves independently. At ω^2 = 2C/M 2 the motion is in the lattice described by the displacement v; at ω^2 = 2C/M 1 the u lattice moves.
(^2 )
2 0
0 0
p 0
p 0
2 sin pk a
- A (1 cos pKa) ; M pa 2A sin pk a sin pKa K M 1 (cos (k K) pa cos (k K) pa) 2
ω = Σ −
∂ω = Σ ∂
− − +
When K = k 0 ,
2 p 0^0
A
(1 cos 2k pa) , K M >
∂ω = Σ − ∂
which in general will diverge because p
- By analogy with Eq. (18),
2 2 s 1 s s 2 s 1 s 2 2 s 1 s s 2 s 1 s 2 iKa 1 2 2 iKa 1 2
Md u dt C (v u ) C (v u ); Md v dt C (u v ) C (u v ), whence Mu C (v u) C (ve u); Mv C (u v) C (ue v) , and
−
−
−ω = − + − −ω = − + −
2 iK 1 2 1 2 iKa 2 1 2 1 2
(C C ) M (C C e ) 0 (C C e ) (C C ) M
2 1 2 2 1 2
For Ka 0, 0 and 2(C C ) M. For Ka , 2C M and 2C M.
= ω = + = π ω =
- (a) The Coulomb force on an ion displaced a distance r from the center of a sphere of static or rigid conduction electron sea is – e^2 n(r)/r^2 , where the number of electrons within a sphere of radius r is (3/4 πR^3 ) (4πr^3 /3). Thus the force is –e^2 r/R^2 , and the
C HAPTER 5
- (a) The dispersion relation is (^) m
| sin Ka|. 2
ω = ω We solve this for K to obtain
K = (2/a) sin −^1 ( ω / ωm ), whence and, from (15), (^2 2) 1/ 2 dK/d (2 / a)( (^) m ) ω = ω − ω − D( ω)
. This is singular at ω = ω (^2 2) 1/ 2 (2L/ a)( (^) m ) = π ω − ω − m. (b) The volume of a sphere of radius K in Fourier space is Ω = 4 K / 3π 3 = (4 π / 3)[( ω − ω 0 ) / A] 3/2, and the density of orbitals near ω 1/ 2
0 is D( ω)= (L/2 ) | d π 3 Ω/d ω =| (L/2 ) (2π 3 π / A3/2 )( ω − ω 0 ) , provided ω < ω 0. It is apparent that D(ω) vanishes for ω above the minimum ω 0.
- The potential energy associated with the dilation is (^2 3) B
B( V/V) a k T 2 2
∆ ≈. This is (^) B
k T 2
and not
B
k T 2
, because the other degrees of freedom are to be associated with shear distortions of the lattice cell.
Thus < ∆( V) 2 > = 1.5 × 10 −^47 ;( ∆V) (^) rms= 4.7 × 10 −^24 cm ;^3 and ( ∆V) (^) rms / V = 0.125. Now
3 a/a∆ ≈ ∆V/V , whence ( ∆a) (^) rms / a =0.04.
- (a) < R 2 > = (h/2 V)/ ρ Σω− 1 , where from (20) for a Debye spectrum Σω−^1
= ∫ d ω D( ω ω) − 1 = 3V ωD 2 / 4 π^3 v^3 , whence< R 2 > = 3h/ ωD 2 / 8π ρ^2 v 3. (b) In one dimension from
(15) we have D( ω =) L/ vπ , whence ∫ d ω D( ω) ω− 1 diverges at the lower limit. The mean square
strain in one dimension is 2 2 02
( R/ x) K u (h/2MNv) K 2
(^2 2 ) = (h/2MNv) (K/ (^) D / 2) = h/ ωD / 4MNv.
- (a) The motion is constrained to each layer and is therefore essentially two-dimensional. Consider one plane of area A. There is one allowed value of K per area (2π/L)^2 in K space, or (L/2π)^2 = A/4π^2 allowed values of K per unit area of K space. The total number of modes with wavevector less than K is, with ω = vK,
N = (A/4 π^2 ) ( K )π 2 = A ω^2 / 4 v .π 2
The density of modes of each polarization type is D(ω) = dN/dω = Aω/2πv^2. The thermal average phonon energy for the two polarization types is, for each layer,
D D 0 0 2
A
U 2 D( ) n( , ) d 2 d , 2 v exp(h / ) 1
ω ω (^) ω ω = ω ω τ ω ω = ∫ ∫ (^) π ω τ −
= ω
where ωD is defined by dω. In the regime D D
N D( )
ω = (^) ∫ ω =ω (^) D >> τ, we have
3 2 2 2 0 x
2A x U dx. 2 v e 1
τ^ ∞ ≅ π = ∫ −
Thus the heat capacity C = k (^) B∂U/ ∂τ ∝T^2.
(b) If the layers are weakly bound together, the system behaves as a linear structure with each plane as a vibrating unit. By induction from the results for 2 and 3 dimensions, we expect C. But this only holds at extremely low temperatures such that
∝ T
τ << =ω (^) D ≈=vN (^) layer/ L, where Nlayer/L is the number of
layers per unit length.
- (a) From the Planck distribution x^ x
n (e 1) /(e 1) coth (x/2) 2 2 2
< > + = + − = , where
x = h/ ω/k T (^) B. The partition function Z = e −^ x/2^ Σ e −^ sx^ = e −x/2^ /(1 − e −x^ ) = [2sinh (x/2)]−^1 and the free energy is F = kBT log Z = kBT log[2 sinh(x/2)]. (b) With ω(∆) = ω(0) (1 – γ∆), the condition
∂F/ ∂∆ = 0 becomes B
B h coth (h /2k T) 2
∆ = γΣ / ω /ω on direct differentiation. The energy
< n > h/ω is just the term to the right of the summation symbol, so that B∆ = γU (T ). (c) By definition
of γ, we have δω ω = −γδ/ V /V , or d log ω = −δ d log V. But , whence
.
θ ∝ ω D d log θ = −γd log V
4 π (^) (2 × 10 ) (^9 3) ≈ 3 × 1028 cm 3 3
the electron concentration is
57 28 3 28
3 10 cm. 3 10
≈ ≈ ×^ −
×
Thus
2 2 3 27 20 7 4 F
h 1 1 (3 n) 10 10 10 ergs, or 3.10 eV. 2m 2 2
ε = / π^2 ≈ −^ ⋅ ≈ − ≈ (b) The value of k F is not
affected by relativity and is ≈ n1/3, where n is the electron concentration. Thus εF hck/ (^) F hc/^3 √n. (c) A change of radius to 10 km = 10^6 cm makes the volume ≈ 4 × 1018 cm^3 and the concentration ≈ 3 × 1038 cm – (^3). Thus 27 10 13 4 8 (The energy is relativistic.) F 10 (3.10^ ) (10^ )^ 2.10^ erg^10 eV. ε ≈ −^ ≈ − ≈
- The number of moles per cm^3 is 81 × 10 –3/3 = 27 × 10 –3, so that the concentration is 16 × 1021 atoms cm– (^3). The mass of an atom of He (^3) is (3.017) (1.661) × 10 –24 (^) = 5.01 × 10 –24 (^) g. Thus 54 23 21 2 3 16 F [(1.1^10 ) 10^ ][(30)(16)^10 ]^7 ε × −^ − × ≈ × −^ erg, or T F ≈^ 5K.
- Let E, v vary as e–iwt. Then
eE m e E 1 i v , i m 2
τ + ωτ = − = − ⋅ − ω + (1 τ) 1+ (ωτ)
and the electric current density is
ne^2 1 i j n( e)v E. m 2
τ + ωτ = − = ⋅ 1 + (ωτ)
- (a) From the drift velocity equation
i ωv (^) x = (e m)E (^) x + ωc v (^) y ; i ωv (^) y = (e m)E (^) y− ωcv. x
We solve for vx, vy to find
2 c x x c 2 c y y c
( )v i e m E e m E
( )v i e m E e m E
2
2
ω − ω = ω( ) + ω ( )
ω − ω = ω( ) + ω ( )
y
x
We neglect the terms in ωc^2. Because j = n(–e)v = σE, the components of σ come out directly. (b) From the electromagnetic wave equation
c^2 ∇^2 E = ε∂ 2 E ∂t^2 ,
we have, for solutions of the form ei(kz –^ ωt), the determinantal equation
2 2 xx xy 2 2 yx yy
c k
c k
2 2 2 2
ε ω − ε ω
ε ω ε ω −
Here ε (^) xx = ε (^) yy = 1 − ωP 2 ω^2 and εxy = −εyx = i ω ωc p 2 ω^3. The determinantal equation gives the
dispersion relation.
- The energy of interaction with the ion is
r (^0 2 )
e ∫ 0 ρ r 4 r drπ = −3e 2 r , 0
where the electron charge density is –e(3/4πr 03 ). (b) The electron self-energy is
2 r^03 2 1 0 dr 4 r^3 4 r^ r^ 3e^ 5r.^0 ρ π π − =
The average Fermi energy per electron is 3εF/5, from Problem 6.1; because N V = 3 4 rπ 03 , the average
is ( )
2 3 (^22) 3 9 π 4 h/ 10mr 0. The sum of the Coulomb and kinetic contributions is
2 s (^) s
U
r (^) r
which is a minimum at
2 3 s s s
, or r 4.42 1.80 2.45. r r
The binding energy at this value of rs is less than 1 Ry; therefore separated H atoms are more stable.
- From the magnetoconductivity matrix we have
c y yx x 2 0 x c
j E E 1
ω τ = σ = σ
For ωcτ >> 1, we have σyx ≅ σ 0 ω τ =c ( ne^2 τ m )( mc eB τ =) neB c.
- For a monatomic metal sheet one atom in thickness, n ≈ 1/d^3 , so that
2 2 2 R (^) sq ≈ mv (^) F nd e ≈ mv d eF.
If the electron wavelength is d, then mv dF ≈ h/ by the de Broglie relation and
2 R (^) sq≈ h e/ =137 c
in Gaussian units. Now
