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Copyright © 2005 by John Wiley & Sons, Inc.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-
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which the textbook has been adopted. Any other reproduction or translation of this work
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further information should be addressed to the Permissions Department, John Wiley &
Sons, Inc., 111 River St. Hoboken, NJ 07030-
cube. If θ is the angle between them, their scalar product gives cos θ = –1/3, whence
1 cos 1/ 3 90 19 28' 109 28'
− θ = = ° + ° = °
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'
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 ) 2 a^ c a ,
3 2
or
2 2 1 2 c 8 a c ; 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
2 i(x v +y v +z v )j 1 j 2 j 3 S (v v v ) 1 2 3 f e j
− 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
1 i (v v v ). 2 S (basis) 1 e
− 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
(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.
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 ×
2 2 0 0 0 2 0 0
δ = + δ + ∂
bearing in mind that in equilibrium 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 (^0) R 0 0
R
U q U (n 1) q log 2 (n 1) ; C R R (^) 2R R R
⎛ ∂ ⎞ α ∂ −
⎜ ⎟ =^ −^ =^ = ⎝ ∂^ ⎠ ∂
.
ergs and ρ = 0.326 × 10 - Å. 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
2 x 3 x e 8.53 10.
− − = ×
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
in units with R 0 in Å. For the actual KCl structure, using the data of Table 7, we calculate 2
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.
O
2 /R 0 per ion pair, or –14.61 × 10
with –4(9.12) = –36.48 eV for Ba
++ O
--
. To form Ba
and O
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
++ O
-- is 13.83 eV; the latter is indeed the stable form.
Thus
2 2 Y = (C 11 + C C 12 11 − 2C 12 ) (C 11 +C 12 );
further, also from (37), eyy = S 21 Xx,
whence (^) yy 21 11 12 11 12 xx
σ = e e = S S = − C (C + C ).
2 2 11 44 12 44
1 2 11 12 44
ω ρ = + + +
= ω = + +
This dispersion relation follows from (57a).
this direction. Use (57a).
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 –
δ ij(λ^ + p – q) as aij = p –^ λ′^
δ ij, where^ λ′^ =^ λ^ + p – q, and^
δ ij is the
Kronecker delta function. With λ′ the matrix is in the “standard” form. The root λ′ = Rp gives λ = (R – 1)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 ,
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.
i(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
2 S
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
(^2 2 ) S
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 sin ; cos sin 0. 2
Thus the square of u{} above is
u [1 2cos Ka cos Ka sin Ka] u (1 cos Ka). 2
The potential energy per bond is
Cu (1 cos Ka), 2
− and by the dispersion relation ω
2 = (2C/M) (1 –
cos Ka)
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.
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 2 p 0
u u M ( p a C ) t > x
which is of the form of the continuum elastic wave equation with
2 1 2 2 p p 0
v M p a C
−
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
sin pk a sin pKa K M
(cos (k K) pa cos (k K) pa) 2
ω = Σ −
∂ω = Σ ∂
When K = k 0 ,
2
0 p 0
(1 cos 2k pa) , K M >
∂ω = Σ − ∂
which in general will diverge because p
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
−
a
= − + + − ω
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 =
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
| sin Ka|. 2
ω = ω We solve this for K to obtain
, whence and, from (15),
1 K (2/a) sin ( / (^) m)
− = ω ω
(^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 , and the density of orbitals near ω
3 Ω = 4 K / 3π = (4 π / 3)[( ω − ω 0 ) / A]
3/
1/ 2
0 is
, provided ω < ω
3 3 3/ D( ω)= (L/2 ) | d π Ω/d ω =| (L/2 ) (2π π / A )( ω − ω 0 ) 0. It is apparent that
D(ω) vanishes for ω above the minimum ω 0.
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 and
2 47 24 ( V) 1.5 10 ;( V) (^) rms 4.7 10 cm ;
− − < ∆ > = × ∆ = ×
3 ( ∆V) (^) rms / V = 0.125. Now
3 a/a∆ ≈ ∆V/V , whence ( ∆a) (^) rms / a =0.04.
2 R (h/2 V)
− < > = / ρ Σω
1 − 1 Σω
, whence
1 2 3 d D( ) 3V (^) D / 4 v
− = ∫ ω ω ω = ω π
3 2 3 v
2 2 < R > = 3h/ ωD / 8π ρ. (b) In one dimension from
(15) we have D( ω =) L/ vπ , whence
1 d D( )
− ∫ ω ω ω diverges at the lower limit. The mean square
strain in one dimension is
2 2 2 0
( R/ x) K u (h/2MNv) K 2
(^2 2 ) = (h/2MNv) (K/ (^) D / 2) = h/ ωD / 4MNv.
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,
2 2 2 N = (A/4 π ) ( K )π = A ω / 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
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
ω = ω
=ω >> τ, we have
3 2
2 2 0 x
2A x U dx. 2 v e 1
τ^ ∞ ≅ π −
Thus the heat capacity.
2 C = k (^) B∂U/ ∂τ ∝T
(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
τ << =ω (^) D ≈=vN (^) layer/ L, where Nlayer/L is the number of
layers per unit length.
(^1 1) x x 1 n (e 1) /(e 1) coth (x/2) 2 2 2
< > + = + − = , where
x = h/ ω/k T (^) B. The partition function
x/2 sx x/2 x 1 Z e e e /(1 e ) [2sinh (x/2)]
− − − − = Σ = − =
− 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
(2 10 ) 3 10 cm
π × ≈ × 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 ≈ ⋅ ≈ ≈ (b) The value of kF is not
affected by relativity and is ≈ n
1/ , where n is the electron concentration. Thus
3 εF hck/ (^) F hc/ √n. (c) A
change of radius to 10 km = 10
6 cm makes the volume ≈ 4 × 10
18 cm
3 and the concentration ≈ 3 × 10
38 cm
3
. Thus (The energy is relativistic.)
27 10 13 4 8 F 10 (3.10^ ) (10^ )^ 2.10^ erg^10 eV.
− − ε ≈ ≈ ≈
3 is 81 × 10
21 atoms cm
3
. The mass of an atom of He
3 is (3.017) (1.661) × 10
− − ε × × ≈ ×
− erg, or TF ≈ 5K.
–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
2 ne 1 i j n( e)v E. m
2
τ + ωτ = − = ⋅ 1 + (ωτ)
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
2 2 2 c E E t
2 ∇ = ε∂ ∂ ,
we have, for solutions of the form e
i(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 −
Here
2 xx yy 1 P and^ xy yx i^ c p.
2 ε = ε = − ω ω ε = −ε = ω ω ω
(^2 ) The determinantal equation gives the
dispersion relation.
r (^0 2 )
0 0
e ρ r 4 r drπ = −3e 2
r ,
where the electron charge density is –e(3/4πr 0
3 ). (b) The electron self-energy is
2 r^03 2 1 0 0
dr 4 r 3 4 r r 3e 5r.
− ρ π π =
The average Fermi energy per electron is 3εF/5, from Problem 6.1; because
3 N V = 3 4 rπ 0 , the average
2 3 (^22) 3 9 π 4 h/ 10mr 0. The sum of the Coulomb and kinetic contributions is
2 s (^) s
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.
c y yx x 2 0 x
c
j E E
1
ω τ = σ = σ
2 σyx ≅ σ 0 ω τ =c ne τ m mc eB τ = neB c.
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