James Patterson Bernard Bailey. Solid-State Physics. Solutions Manual for Instructors

Transcription

1 James Patterson Bernard Bailey Solid-State Physics Solutions Manual for Instructors

2 Professor Emeritus, James Patterson, Ph.D. 0 Parkview Drive Rapid City, SD 0 USA jdp@rap.midco.net Dr. Bernard Bailey, Ph.D. 0 Taylor Ave. #C Cape Canaveral, FL 90 USA bbailey@brevard.net This file is an online supplement to the second edition of Solid-State Physics ISBN , e-isbn It contains a solution manual for instructors. Since it is meant to be available to instructors on the internet, all problem statements and their corresponding solutions are included. Only about half of the solutions are made available to the students directly in an appendix in the second edition textbook. c Springer-Verlag Berlin Heidelberg 00 This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 9, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Springer is part of Springer Science+Business Media (

3 Chapter Solutions Problem (.): Show by construction that stacked regular pentagons do not fill all twodimensional space. What do you conclude from this? Give an example of a geometrical figure that when stacked will fill all two-dimensional space. a) For example, the shaded area is not enclosed by the stacked pentagons. b) They do not form a lattice. We cannot form a lattice that maps into itself by rotation of π/. c) A square, a rectangle. Problem (.): Find the Madelung constant for a one-dimensional lattice of alternating, equally spaced positive and negative charged ions.

4 Chapter Solutions M N ( ) N N + ln( + x) x ln Problem (.): Use the Evjen counting scheme [.9] to evaluate approximately the Madelung constant for crystals with the NaCl structure. Consider for example NaCl. Place the origin on a Na + cation. Suppose the distance to the nearest Cl anions is d. We have Cl ions at a distance d, Na + ions at a distance d and Cl ions at d, and so on. Evjen s idea was to divide space, starting with the Na + ion, into neutral cubic volumes thus making the convergence relatively rapid. This requires that we divide up the charges. Thus for an ion on the face of a cube, one half of its charge is inside the cube and one half is outside. In a somewhat similar way we divide charges on the edges and corners of the cubes. For the first cube we get the following series for the Madelung constant M: / / / M +.. (ions on faces) (ions on edges) (ions on corners) If we include the second cube, the series gives M.. A more precise value of M is M. Problem (.): Show that the set of all rational numbers (without zero) forms a group under the operation of multiplication. Show that the set of all rational numbers (with zero) forms a group under the operation of addition.

5 Chapter Solutions A rational number is the ratio of two integers. Zero is excluded. To be a group they must satisfy four requirements. Let N i be a rational number ( 0), and multiplication. a. Closure N i N j is a rational number b. Associative law N N ) N N ( N N ) ( i j k i j k c. Identity ( N i ) Ni so / is the rational identity d. Inverse " " N i / N i is a well defined rational number since N i 0 The second part for addition (+) is similarly done. a. Closure x w If Ni, N j, and w, x, y, z are integers y z x w xz + wy Ni + N j + rational y z yz b. Associative law N + N ) + N N + ( N + N ) c. Identity 0 ( i j k i j k d. Inverse Ni N i Problem (.): Construct the group multiplication table of D (the group of three dimensional rotations that map a square into itself).

6 Chapter Solutions A g g g g The group D is of order with distinct elements. Let g, g, g, g be the clockwise rotations of 0, 90, 0, 0 about an axis through A and perpendicular to the page. The other four elements (g, g, g, g ) are rotations of 0 about the axes as marked by dashed lines in the figure. Using the notation of the book, we list the group elements: g g g g g g g g.

7 Chapter Solutions Then, e.g., g g g. Note that g is the identity so the first line in the table is trivial. The rest of the elements can be worked out as we did D in the text. As an example, we worked out g g (above) from the group elements. g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g g Problem (.): Show that the set of elements (,, i, i) forms a group when combined under the operation of multiplication of complex numbers. Find a geometric group that is isomorphic to this group. Find a subgroup of this group. Is the whole group cyclic? Is the subgroup cyclic? Is the whole group abelian? We first construct the group multiplication table. i i i i i i i i i i i i The table shows the group is closed, associativity is obvious, the identity is, and each element has an inverse.

8 Chapter Solutions This group is isomorphic to the group of rotations of a square about an axis through its center and perpendicular to the square. i 0 i π i π / i π / e, e, i e, i e so θ in e i θ gives the rotation angle. A subgroup is (, ). The whole group is cyclic as i generates all elements (i i, i, i i, i i 0 ) (, ) is also cyclic. The order of multiplying the elements of the whole group is unimportant so the whole group is abelian. Problem (.): Construct the stereograms for the point groups (C ) and mm(c v ). Tell how all elements of each group are represented in the stereogram (see Table.). The lowest point group consistent with a tetragonal system is. means -fold symmetry. mm means -fold symmetry with two distinct mirror planes parallel to a -fold axis. The -fold axis is represented by the square. The dots show the lack of mirror symmetry.

9 Chapter Solutions m' m m m' mm As shown by the dots there are two mirror planes m and m. m' is generated by m by a π/ rotation. Similarly for m' and m. the equivalent dots can be generated from just one by symmetry operations. Problem (.): Draw a bcc (body-centered cubic) crystal and draw in three crystal planes that are neither parallel nor perpendicular. Name these planes by the use of Miller indices. Write down the Miller indices of three directions, which are neither parallel nor perpendicular. Draw in these directions with arrows. Use primitive unit cell with a, a, a primitive translation vectors z a a ( i j + k ) a a ( i + j + k ) y x a a ( i + j k) a length of side of conventional unit cell (cubic)

10 Chapter Solutions a) Planes could be perpendicular to a, a, a e.g. a plane perpendicular to a would have intercept (,, ) and reciprocals (, 0, 0) and have Miller indices [, 0, 0]. b) e.g. a, a, a are neither parallel nor perpendicular and [, 0, 0] are the Miller indices of a. The bcc lattice and primitive translation vectors as well as primitive and conventional unit cells are clarified in the drawing. z a a x a y a a) Primitive translation vectors of the bcc lattice a b) Conventional and primitive cells of bcc lattice

11 Chapter Solutions 9 Problem (.9): Argue that electrons should have energy of order electron volts to be diffracted by a crystal lattice. To be diffracted λ should be approximately equal to the lattice spacing. so λ h p h me h E. mλ Say λ ~ Å, m ~ m electron, so using SI and converting to ev: (. 0 ) ev E ev. 0 9 (9. 0 )( 0 ). 0 J Problem (.0): Consider lattice planes specified by Miller indices (h, k, l ) with lattice spacing determined by d (h, k, l ). Show that the reciprocal lattice vectors G (h, k, l ) are orthogonal to the lattice plane (h, k, l ) and if G (h, k, l ) is the shortest such reciprocal lattice vector then π dhkl (,, ). G ( hkl,, ) If we draw all (h, k, l) planes they pass through all lattice points. If the direct lattice is specified by the primitive vectors a, b, c then let the intercepts of the planes relative to an origin at a lattice point be denoted by a b c,,. h k l

12 0 Chapter Solutions If (h, k, l ) are the smallest set of numbers with the same ratio (i.e. Miller indices) then a/h, b/k, c/l connect adjacent planes and G (h, k, l ) is shortest. Orthogonality follows from since and Thus a b πh πk G 0 h k h k G π b b b ( h + k + l ) ai bj δij ( a a, a b, a c). G( hkl,, ) a π dhkl (,, ). G( hkl,, ) h G( hkl,, ) Problem (.): Suppose a one-dimensional crystal has atoms located at nb and αmb where n and m are integers and α is an irrational number. Show that sharp Bragg peaks are still obtained. A density function that describes the positions of the atoms is given by ρ( x) δ( r nbiˆ) + δ( r mbαi ˆ). From (.9) the scattering amplitude is n P ρ()e r m Δ i kr dv where Δ k kf k i. We assume, as usual, elastic scattering so k f k i k. Substituting the first equation in the second, and for the geometry shown, α /.

13 Chapter Solutions k f b x z k i bα y ˆ 0 i k i b bα P e e e N ik fx n 0e ik fxbn ik fxb k i fx b( N ) i nb M k fxmbα + m 0e ik fxbαm e + ik fxbα e k fxbn sin k i + e k fxb sin fx bα ( M ) k fxbαm sin. k fxbα sin The scattering intensity is proportional to P. We assume a limit in which N, M very large. Following the usual arguments found in optics for interference from many slits we know the principal maxima occur for Since () () k fx b k fx bα π integer ( l) and π integer ( p). (a) () k fx b () k fx b α l p or () k fx b l α, () k b p fx () () k fx k fx since α is irrational, we need not be concerned with cross terms in the square P, and the maxima of k fx () and k fx () are different. If we look only in the z 0 plane since k fy k k fx, then k fy is also determined. We thus get sharp Bragg peaks as determined by (a). Similar comments can be made for other observation geometries.

14 Chapter Solutions Problem (.): Find the Bragg peaks for a grating with a modulated spacing. Assume the grating has a spacing d n nb + εbsin( π knb) where ε is small and kb is irrational. Carry your results to first order in ε and assume that all scattered waves have the same geometry. You can use the geometry shown in the figure of this problem. The phase φ n of scattered wave n at angle θ is π ϕ λ n d n sinθ where λ is the wavelength. The scattered intensity is proportional to the square of the scattered amplitude, which in turn is proportional to N E 0 exp( iϕ ) for N+ scattered wavelets of equal amplitude. n θ d d sinθ Let so π K λ ϕn Kd n sinθ.

15 Chapter Solutions We need to evaluate Thus we need to evaluate N E 0 exp( iϕ ). n Define So But Thus N E 0 exp[ i( nbk sinθ + K sinθ ε bsin(π knb))]. Z ε sin(πknb) N E 0 B K sinθ, Z exp( ibb). Z n Z ε sin( πknb) exp[(ln Z) ε sin(π knb)] m 0 [ ε ln Z sin(π knb)] m! + ε ln Z sin(π knb) to first order. N n N n E 0 Z 0 ) + Z( ε ln Z) Z sin(π knb.. For Z we can perform both sums for N and obtain sin(π kb) E + ε Z ln Z. Z Z cos(π kb) + Z Notice if ε 0 we retain only the first term that diverges when Z, that is when Bb πm where m is an integer or sinθ mλ/b, which is like the familiar multi slit interference rule. The second term must also be considered if ε 0. It blows up when (using G πkb) Z Z cosg + 0, which can be shown to occur when Z exp(±ig). But since Z exp(ibb) Bb ± G + π m, where m is an integer. This can be rewritten as λ sin θ [ m ± kb], b a condition for (small subsidiary) maximum in multi-scatterer interference. I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, Academic Press, San Diego, 90, p. 0 (.).

16 Chapter Solutions Problem (.): Find all Bragg angles less than 0 degrees for diffraction of x-rays with wavelength. angstroms from the (00) planes in potassium. Use a conventional unit cell with structure factor. Solution : a K.Å Using Braggs law nλ dsinθ d 00 a.å. d lattice planes. ( K bcc) λ. sinθ θ. d. (.) sinθ θ. 0.. Solution : Using structure factor and conventional unit cell n λ d 00 sinθ λ. iπ sin θ, θ 0. S00 f( + e ) 0 doesn t happen d00 0. λ (.) iπ sin θ, θ. S00 f( + e ) f as above d00 0. λ (.) iπ sin θ, θ. S00 f( + e ) 0 doesn t happen d00 0. λ (.) iπ sin θ, θ.0 S00 f( + e ) f as above d00 0.

17 Chapter Solutions Problem (.): Find the normal modes and normal-mode frequencies for a three-atom lattice (assume the atoms are of equal mass). Use periodic boundary conditions. Lattice vibrations in a three atom monatomic lattice with periodic boundary conditions: m m m γ γ γ γ x x m is the atomic mass, x i is the displacement from equilibrium, and γ is the spring constant. From Newton s second law and using periodic boundary conditions, we have the equations of motion. m x γ ( x x) γ ( x x) (a) m x γ ( x x) γ ( x x) (b) m x γ ( x x) γ ( x x ) (c) We seek normal mode solutions of the form x n u n e iωt. Substituting and canceling e iωt we have x or in matrix form γ ω m γ γ γ γ ω m γ γ u γ 0 u γ ω m u () MU 0. ()