Kittel Charles-Introduction To Solid State Physics 8Th Edition-Solution Manual

Kittel Charles-Introduction To Solid State Physics 8Th Edition-Solution Manual

(Parte 1 de 8)


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

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



1. The crystal plane with Miller indices hk is a plane defined by the points aA1/h, a2/k, and . (a)

Two vectors that lie in the plane may be taken as a 3 / Aa

1/h – a2/k and 13/h/−Aaa. But each of these vectors gives zero as its scalar product with 12hk3=++AGaaa, so that G must be perpendicular to the plane

. (b) If is the unit normal to the plane, the interplanar spacing is hkAn1/h⋅na. But , whence . (c) For a simple cubic lattice

y z aab a a

xy 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

BZV( 2

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

(b) The first zero in 1sinM2 ε occurs for ε = 2π/M. That this is the correct consideration follows from

1zero, as Mh is an integer

11sin M(h)sinMh cos Mcos Mh sin M.

123S (v)fej

5. 2i(xv+yv+zv) −π=Σ

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 (2) we have S(basis) = 1 + e–i3π = 0, and this reflection is forbidden.

The integral is not difficult; it is given as Dwight 860.81. Observe that f = 1 for G = 0 and f ∝ 1/G4 for 0Ga 1.>>

7. (a) The basis has one atom A at the origin and one atom 1B 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 2 (integer)⋅∆ π×ak =

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

Thus the cohesive energy ratio bcc/fcc = 0.956, so that the fcc structure is more stable than the bcc.

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

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.


2nAqU(R)N;2 log 2 Madelung const.

In equilibrium

Un A q nN0 ; R


0 0 0 021UU(R -R ) U R R R,2R

bearing in mind that in equilibrium R0 (UR)0.∂∂=

For a unit length 2NR0 = 1, whence

6. 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 ≡ R0/ρ we have

By trial and error we find or Rx9.2 ,0 = 3.0 Å. The actual KCl structure has R0 (exp) = 3.15 Å . For the imagined structure the cohesive energy is

in units with R0 in Å. For the actual KCl structure, using the data of Table 7, we calculate 2U0.495,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.

7. The Madelung energy of Ba+ O– is –αe2/R0 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 = 2.65 eV. Thus at the specified value of R0 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.

further, also from (37), eyy = S21Xx,

9. For a longitudinal phonon with K || [1], u = v = w.

This dispersion relation follows from (57a).

10. We take u = – w; v = 0. This displacement is ⊥ to the [1] direction. Shear waves are degenerate in this direction. Use (57a).

UC ( e e ) C e

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

(Parte 1 de 8)