Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

# Solutions to solid state physics problems, Study notes of Physics Fundamentals

Solutions with problems In Solid State Physics

Typology: Study notes

2018/2019
On special offer
30 Points

Limited-time offer

Uploaded on 09/22/2019

5

(1)

1 document

## Partial preview of the text

Download Solutions to solid state physics problems and more Study notes Physics Fundamentals in PDF only on Docsity! Foreword This collection of problems and solutions is intended to aid students taking our course in Solid State Physics. Exercises are an integral part of a course and the reader is urged to attempt most of them. The problems are selected from areas usually covered in a first course and are of a type most often assigned for class work and given on examinations. No arrangement in order of complexity has been attempted and for some problems only answers are given. Lule̊a, Sweden January 1995 Sune Marklund The layout and chapters division have been changed. Some problems have also been added or changed and some solutions are elucidated. Lule̊a, Sweden February 1996 Niklas Lehto Chapters containing some problems on superconductivity, on optical and dielectric properties, and on magnetic properties have been added. Lule̊a, Sweden February 1997 Niklas Lehto A section on X–ray diffraction has been added. The section is based on ’Elementary Solid State Physics’ by M. Ali Omar. Addison Wesley. The intention is to supplement Kittel (ed. 7) chapter 2. Lule̊a, Sweden February 2000 Hans Weber Three laboratory excersises have been added. Lule̊a, Sweden February 2001 Hans Weber Minor corrections. Lule̊a, Sweden February 2003 Hans Weber i ii FOREWORD A X-ray diffraction The text in this chapter is based on ’Elementary Solid State Physics’ by M. Ali Omar. The book is no longer available. 1 Scattering from an atom The diffraction process can be divided naturally into two stages: (1) scattering by in- dividual atoms, and (2) mutual interference between the scattered rays. Since the two stages are distinct from each other, we shall treat them independently, for convenience. Why does an atom scatter the x–ray beam? Well, any atom is surrounded by electrons which undergo acceleration under the action of the electric field associated with the beam. Since an accelerated charge emits radiation (a fact well known from electromagnetism), so do the atomic electrons. In effect the electrons absorb energy from the beam, and scatter it in all directions. But the electrons form a charge cloud surrounding the atom, so when we are considering scattering from the atom as a whole, we must take into account the phase differences between the rays scattered from the different regions of the charge cloud. We do this as follows: Consider a single electron, as shown in Fig A.1a. A plane–wave field given by u = Aei(k0·r−ωt) is incident on the electron, where A is the amplitude, k0 the wave vector (k0 = 2π/λ), and ω the angular frequency. The scattered field is an outgoing spherical wave represented by u′ = fe A D ei(kD−ωt) (A.1) where fe is the scattering length of the electron, and D is the radial distance from the electron to the point at which the field is evaluated. The quantity k is the wave number of the scattered wave, and has the same magnitude as k0. Note the the amplitude of the scattered wave decreases with distance as 1/D, a property shared by all spherical waves. Suppose now that the incident wave acts on two electrons, as in Fig A.1b. In this case, both electrons emit spherical waves, and the scattered field observed at a distant point is the sum of the two partial fields, where their phase difference has to be taken into account. Thus we have u′ = fe A D [ eikD + eikD+δ ] (A.2) 1 2 A. X-RAY DIFFRACTION (b) Incident ray Scattered ray P P N M 1 2 S S0 Incident Electron k k 0 (a) s (c) Figure A.1: Scattering from (a) a single electron, (b) two electrons. (c) The scattering vector s. Note that the vectors k0,k, and s form an isosceles triangle. where δ is the phase lag of the wave from electron 1 behind that of electron 2 1. (The time factor has been omitted for the sake of brevity, but its presence is implied.) Referring to the figure, we may write δ = (P1M − P1N) 2π/λ = (r · S− r · S0) k where r is the vector radius of electron 2 relative to electron 1, and S0 and S are the unit vectors in the incident and scattered directions, respectively. The expression for δ can be set forth in the form δ = s · r, (A.3) where the scattering vector s is defined as s = k (S− S0) = k− k0. (A.4) As seen from Fig A.1c, the magnitude of the scattering vector is given by s = 2k sin(θ) (A.5) where θ is half of the scattering angle. Substituting the expression Eq (A.3) for δ into Eq. (A.2), one finds u′ = fe A D eikD [ 1 + eis·r ] (A.6) In deriving this we have chosen the origin of our coordinates at electron 1. But it is now more convenient to choose the origin at an arbitrary point, and in this manner treat the two electrons on equal footing. The ensuing expression for the scattered field is then u′ = fe A D eikD [ eis·r1 + eis·r2 ] (A.7) where r1 and r2 are the position vectors of the two electrons relative to a new origin. Equation (A.6) is a special case of (A.7), where r1 = 0, that is, where the origin is chosen 1The distance D to the field point is assumed to be large, otherwise the denominator D in Eq. (A.2) would not be the same for the two electrons. This conditions simplifies the calculations, and is the reason why the detector is usually placed far from the crystal 1. SCATTERING FROM AN ATOM 3 at electron 1, as pointed out above. The generalisation of (A.7) to an arbitrary number of scatterers is now immediate, and the result is u′ = fe A D eikD ∑ l eis·rl where rl is the position of the l th electron, and the sum is carried out over all the electrons. By analogy with the case of the single electron, equation (A.1), the scattering length for the system as a whole is now given by the sum f = fe ∑ l eis·rl . (A.8) That is, the total scattering length is the sum of individual lengths with the phases taken properly into account. The intensity I of the scattered beam is proportional to the square of the magnitude of the field, and therefore I ∼| f |2= f 2e | ∑ l eis·rl |2 . (A.9) Results (A.8) and (A.9) are the basic equations in the treatment of scattering and diffrac- tion processes, and we shall use them time and again in the following pages. We may digress briefly to point out an important aspect of the scattering process: the coherence property involved in the scattering. This property means that the scatterers maintain definite phase relationships with each other. Consequently we can speak of interference between the partial rays. By contrast, if the scatterers were to oscillate randomly, or incoherently, the partial rays would not interfere, and the intensity at the detector would simply be the sum of the partial intensities, that is, I ∼ Nf 2e , where N is the number of scatterers. Note the marked difference between this result and that of coherent scattering in eq. (A.9). The scattering length of the electron is well known , and can be found in books on electromagnetism. Its value is fe = [( 1 + cos2(2θ) ) /2 ]1/2 re, where re, the so–called classical radius of the electron, has a value of about 10 −15m. 2 We can now apply these results to the case of a single free atom. In attempting to apply eq. (A.8), where the sum over the electrons appears, we note that the electrons do not have discrete positions, but are spread as a continuous charge cloud over the volume of the atom. It is therefore necessary to convert the discrete sum to the corresponding integral. This readily leads to f = fe ∑ l eis·rl → fe ∫ ρ(r)eis·rd3r, 2For the sake of visual thinking, consider the electron to be in the form of a sphere whose radius is roughly equal to the scattering length fe. Thus the electron ”appears” to the radius as a circular obstacle of cross section πf2 e . 6 A. X-RAY DIFFRACTION case of a simple lattice, where the unit cell contains a single atom, the factor F becomes equal to fa. The factorisation of fcr as shown in eq. (A.16) merits some emphasis: We have separated the purely structural properties of the lattice, which are contained in S, from the atomic properties contained in F . Great simplification is achieved thereby, because the two factors may now be treated independently. Since the factor F involves a sum over only a few atomic factors, it can be easily evaluated in terms of the atomic factors, as discussed in the previous section. We shall therefore not concern ourselves with this straightforward task for the moment, but press on and consider the evaluation of the lattice factor S. The lattice structure factor The lattice structure factor S, defined in eq. (A.15), is of vital importance in the discussion of x–ray scattering. Let us now investigate its dependence on the scattering vector s, and show that the values of s for which S does not vanish form a discrete set, which is found to be related to Bragg’s law. Incident ray Q 21 α α 0S 0 (a) (c)(b) Lattice Incident ray S a A B P h=0 h=1 S 2 s a N 2 0 0 2 π . Figure A.3: (a) Scattering from a one–dimensional lattice. (b) Diffraction maxima. (c) Diffraction cones for first order (h = 0) and second order (h = 1) maxima. We start with the simplest possible situation, an x–ray beam scattered from a one– dimensional monatomic lattice, as illustrated in Fig. A.3a. When we denote the basis vector of the lattice by a, the structure factor becomes S = N∑ l=1 eis·la, (A.17) where we have substituted R (c) l = la, and N is the total number of atoms. The series in eq. (A.17) is a geometric progression, the common ratio being eis·a, and can readily be evaluated. The result is S = sin [ 1 2 Ns · a ] sin [ 1 2 s · a ] e 1 2 iNs·a e 1 2 is·a Physically, it is more meaningful to examine S2 than S, since this is the quantity which 2. SCATTERING FROM AN CRYSTAL 7 enters directly into calculations of intensity. It is given by S2 = sin2 [ 1 2 Ns · a ] sin2 [ 1 2 s · a ] (A.18) We now wish to see how this function depends on the scattering vector s. As we see from eq. (A.18), S2 is the ratio of two oscillating functions having a common period s · a = 2π, but, because N is much larger than unity in any practical case, the numerator oscillates far more rapidly than the denominator. Note, however, that for the particular value s · a = 0, both the numerator and denominator vanish simultaneously, but the limiting value of S2 is equal to N2, a very large number. Similarly the value of S2 at s · a = 2π is equal to N2, as follows from the periodicity of S2, as mentioned above. The function S2 is sketched versus s · a in Fig. A.3b, for the range 0〈s · a〉2π. It has two primary maxima, at s · a = 0 and s · a = 2π, separated by a large number of intervening subsidiary maxima, the latter resulting from the rapid oscillations of the numerator in eq. (A.18). Calculations show that when the number of cells is very large, as it is in actual cases, the subsidiary maxima are negligible compared with the primary ones. For instance, the peak of the highest subsidiary maximum is only 0.04 that of a primary maximum. It is therefore a good approximation to ignore all the subsidiary maxima, and take the function S2 to be a nonvanishing only in the immediate neighbourhoods of the primary maxima. Furthermore, it can also be demonstrated that the width of each primary maximum decreases rapidly as N increases, and that this width vanishes in the limit as N → ∞. Therefore S2 is nonvanishing only at the values given exactly by s · a = 0, 2π. But because S2 is periodic, with period of 2π, it is also finite at all the values s · a = 2πh, h = any integer. (A.19) At these values S2 is equal to N2, and hence S = N . Equation (A.18) determines all the directions in which S has a nonzero value and hence the directions in which diffraction takes place. The physical interpretation of this equation is straightforward. Recalling the definition of s, Eq. (A.4), and referring to Fig A.3, we obtain s · a = 2π λ (S − S0) · a = 2π λ (AQ− PB) , which is the phase difference between the two consecutive scattered rays. Thus Eq. (A.19) is the condition for constructive interference, i.e., the lattice scattering factor survives only in these directions, which is hardly surprising. For a given h, the condition Eq. (A.19) does not actually determine a single direction, but rather an infinite number of directions forming a cone whose axis lies along the lattice line. To see this, we can write Eq. (A.19) as 2π λ (cosα− cosα0) = 2πh, (A.20) where α0 is the angle between the incident beam and the lattice line and α is the corre- sponding angle for the diffracted beam. Thus for a given h and α0, the beam diffracts along all directions for which α satisfies Eq. (A.20). These form a cone whose axis lies along the lattice , and whose half angle is equal to α. The case h = 0 is a special one; 8 A. X-RAY DIFFRACTION its cone includes the direction of forward scattering. Diffraction cones corresponding to several values of h are shown in Fig. A.3c. In treating the lattice structure factor, we have so far confined ourselves to the case of a one–dimensional lattice. Referring to Eq. (A.15) and substituting for the lattice vector, R(c) = l1a+ l2b+ l3c, where a,b and c are the basis vectors, we find for the structure factor S = ∑ l1,l2,l3 eis·(l1a+l2b+l3c), where the triple summation extends over all the unit cells in the crystal. We can separate this sum into three partial sums, S =   ∑ l1 eis·l1a     ∑ l2 eis·l2b     ∑ l3 eis·l3c   , and in this manner we factor out S into a product of one–dimensional factors, and we can therefore use the result we developed earlier. The condition for constructive interference is that each of the three factors must be finite individually, and this means that s must satisfy the following three equations simultaneously: s · a = h2π s · b = k2π s · c = l2π (A.21) where h, k, and l are any set of integers. Written in terms of the angles made by s with the basis vectors, in analogy with Eq. (A.19), these equations become respectively a (cosα− cosα0) = hλ b (cos β − cos β0) = kλ c (cos γ − cos γ0) = lλ (A.22) where α0, β0, and γ0 are the angles which the incident beam makes with the basis vectors, while α, β, and γ are the corresponding angles for the diffracted beam. Equations (A.21) and (A.22) are known as the Laue equations, after the physicist who first derived them. The question is how to determine the values of the scattering vector s which satisfy the diffraction condition Eq. (A.21). We shall show in the next section that these values form a discrete set which corresponds to Bragg’s law. 3 The reciprocal lattice and X–ray diffraction Starting with a lattice whose basis vectors are a, b and c, we can define a new set of basis vectors a∗, b∗, and c∗ according to the relations a∗ = 2π Ωc (b x c) , b∗ = 2π Ωc (c x a) , c∗ = 2π Ωc (a x b) , (A.23) 3. THE RECIPROCAL LATTICE AND X–RAY DIFFRACTION 11 b* A O a* Figure A.7: The first Brillouin zone for a rectangular lattice. falls precisely at the center, unlike the case of the direct lattice, in which the lattice points usually lie at the corners of the cell. If the first BZ is now translated by all the reciprocal vectors Gn, then the whole reciprocal lattice space is covered, as it must be, since the BZ is a true unit cell. The Brillouin zone for a 3–dimensional lattice can be constructed in a similar manner, but note that in this case the lattice vectors are bisected by perpendicular planes, and that the first BZ is now the smallest volume enclosed by these planes. In the simplest case, the sc lattice, the BZ is a cube of edge 2π/a, centered at the origin. The BZ’s for the other cubic lattices are more complicated in shape, and we shall defer discussion of these and other lattices to a later section (See Kittel chapter 2). Sometimes one also uses higher–order Brillouin zones, which correspond to vectors joining the origin to farther points in the reciprocal lattice, but we shall not discuss these here, as they will not be needed. We shall find that the concept of the Brillouin zone is very important in connection with lattice vibrations, and electron states in a crystal (See Kittel chapter 4, 5 and 7). Having defined the reciprocal lattice and discussed some of its properties, let us now proceed to demonstrate its usefulness. One important application lies in its use in the evaluation of lattice sums, and this rests on the following mathematical statement: N∑ i=1 eiA·Rl = NδA,Gn . (A.25) Here A is an arbitrary vector, the summation is over the direct lattice vectors, footnote and N is the total number of cells in the direct lattice. Because of the delta symbol, the meaning of Eq. (A.25) is that the lattice sum on the left vanishes whenever the vector A is not equal to some reciprocal lattice vector Gn. When this is equal to some Gn, however, the lattice sum becomes equal to N . To establish the validity of Eq. (A.25), we shall first treat the case A = Gn; to evaluate the exponent A ·Rl on the left of Eq. (A.25), we substitute A = Gn =n1a ∗+n2b ∗+n3c ∗ and Rl =l1a+ l2b+ l3c, and the result is A ·Rl = Gn ·Rl = (n1a∗ + n2b∗ + n3c∗) · (l1a1 + l2a2 + l3a3) = (n1l1 + n2l2 + n3l3) 2π, where in evaluating the scalar products of the basis vectors we used Eq. (A.25). For example, a∗ · a = 2π, a∗ · b = 0, etc. Each term in the sum in Eq. (A.25) is therefore of 12 A. X-RAY DIFFRACTION the form eim2π, where m is an integer and is consequently equal to unity. The total sum is then equal to N , as demanded by Eq. (A.25). In the case A 6= Gn, we can follow the same procedure employed in evaluating Eq. (A.17), and the result is the same as before, namely, that for large N the sum vanishes except for certain values of A. The exceptional values are, in fact, those singled out above, that is, A = Gn. As a final point, we shall now show that the vectors of the reciprocal lattice are related to the crystal planes of the direct lattice. In this manner, the somewhat abstract reciprocal vectors will acquire a concrete meaning. Consider the set of crystal planes whose Miller indices are (hkl) and the corresponding reciprocal lattice vector Ghkl = ha ∗ + kb∗ + lc∗, where the numbers h, k, and l are a set of integers. We shall now establish the following properties: 1 The vector Ghkl is normal to the (hkl) crystal planes. 2 The inter planar distance dhkl is related to the magnitude of Ghkl by dhkl = 2π/Ghkl. (A.26) y u v Ghkl (hkl) plane x z Figure A.8: The reciprocal lattice vector Ghkl is normal to the plane (hkl). To establish these relations, we refer to Fig. A.8, where we have drawn one of the (hkl) planes. The intercepts of the plane with the axes are x, y, z and they are related to the indices by (h, k, l) ∼ ( 1 x , 1 y , 1 z ) , (A.27) where use is made of the definition of the Miller indices (See Kittel page 12). Note also the vectors u and v which lie along the lines of intercepts of the plane with the xy and yz planes, respectively. According to the figure, these vectors are given by u = xa− yb, and v = yb − zc. In order to prove relation (1) above, we need only prove that Ghkl is orthogonal to both u and v. We have u ·Ghkl = (xa− yb) · (ha∗ + kb∗ + lc∗) = 2π (xh− yk) = 0 (A.28) where we have used Eq. (A.25) to establish the second equality; the last equality follows from Eq. (A.27). In the same manner we can also show that Ghkl is orthogonal to v, and this establishes property (1). 4. THE DIFFRACTION CONDITION AND BRAGG’S LAW 13 In order to prove Eq. (A.26), one observes that dhkl, the inter planar distance, is equal to the projection of xa along the direction normal to the (hkl) planes; this direction can be represented by the unit vector Ĝhkl = Ghkl/Ghkl, since we have already established that Ghkl is normal to the plane. Therefore dhkl = xa · Ĝhkl = (xa ·Ghkl) /Ghkl., We now note that xa ·Ghkl = 2πhx, and this is equal to 2π, because, according to Eq. (A.27), xh = 1. This completes the proof of Eq. (A.26). The connection between reciprocal vectors and crystal planes is now quite clear. The vector Ghkl is associated with the crystal planes (hkl), which are, in fact, normal to it, and the separation of these planes is 2π times the inverse of the length Ghkl in the reciprocal space. The crystallographer prefers to think in terms of the crystal planes, which have a physical reality, and their Miller indices, while the solid–state physicist prefers the reciprocal lattice, which is mathematically more elegant; the two approaches are, however, equivalent, and one can change from one to the other by using the relations connecting the two. Of the two approaches, we shall mostly use the reciprocal lattice in this book. 4 The diffraction condition and Bragg’s law We shall now employ the concept of the reciprocal lattice to evaluate the lattice–structure factor S, which is involved in the x–ray scattering process. This factor is given in Eq. (A.15). Comparing this with Eq. (A.25), we see that S vanishes for every value of s except where s = Ghkl. (A.29) The condition for diffraction is therefore that the scattering vector s is equal to a recip- rocal lattice vector. Equation (A.29) implies that s is normal to the (hkl) crystal planes [property (1) above], as shown in Fig. A.9. The equation can be rewritten in a different form. Recalling that s = 2(2π/λ) sin θ,Ghkl = 2π/dhkl, and substituting into Eq. (A.29), we find that 2dhkl sin θ = λ. (A.30) This is exactly the same form as Bragg’s law (2d sin θ = nλ), which is seen to follow from the general treatment of scattering theory. It is therefore physically meaningful to use the Bragg model, and speak of reflection from atomic planes. This manner of viewing the diffraction process is conceptually simpler than that of scattering theory. When the condition (A.29) is satisfied, the structure factor is nonzero, and its value is equal to N as seen from Eq. (A.25). Thus Shkl = N. Substituting this into Eq. (A.23), we find the crystal scattering factor fcr to be fcr,hkl = NFhkl, and the intensity I is then I ∼| fcr,hkl |2∼| Fhkl |2 . (A.31) 16 A. X-RAY DIFFRACTION B Problems in Solid State Physics 1 Crystal Structures 1.1. Given that the primitive basis vectors of a lattice are a = (a/2)(x̂+ ŷ), b = (a/2)(ŷ + ẑ), (B.1) c = (a/2)(ẑ + x̂), where x̂, ŷ, and ẑ are unit vectors in the x-, y- and z-directions of a Cartesian coordinate system. (a) Determine the Bravais lattice. (b) Calculate the volume of the primitive unit cell. 1.2. The angles between the tetrahedral bonds of diamond are the same as the larger angle between two body diagonals of a cube. Use elementary vector analysis to find the value of the angle. 1.3. Use information about the structure and the atomic weights to calculate the densities of the following solids: Al, Fe, Zn and Si. 1.4. Show that in an ideal hexagonal close-packed (hcp) structure, where the atomic spheres touch each other, the ratio c/a is given by c a = ( 8 3 )1/2 = 1.633. (B.2) 1.5. The packing ratio is defined as the fraction of the total volume of the cell that is filled by atoms. Determine the maximum values of this ratio for equal spheres located at the points of SC, BCC, FCC and diamond type crystals. 1.6. Draw an FCC cubic cell. Construct a primitive cell within this larger cell, and compare the two. How many atoms are in the primitive cell and how does this compare with the number in the cubic cell. 1.7. Show that a two-dimensional lattice may not possess a 5-fold rotational symmetry. 17 18 B. PROBLEMS IN SOLID STATE PHYSICS 1.8. Demonstrate the fact that if an object has two reflection planes intersecting at π/4 it also possesses a 4-fold axis lying at their intersection. 1.9. Determine which planes in an FCC structure have the highest density of atoms and evaluate this density for Cu. a 1 2 43 b Figure B.1: Plane sets in a two-dimensional square lattice. 1.10. Determine the Miller indices of the four sets of planes in Fig.B.1. Do this in the two coordinate systems a and b. 3. DIFFRACTION IN CRYSTALS 21 3.14. The scattered intensity for a linear crystal is given by the expression I = sin2 1 2 M(a ·∆k) sin2 1 2 (a ·∆k) , (B.4) where M is the number of scattering centers and a the lattice constant. Evaluate the first subsidiary maximum and show that it is equal to 0.045M2, in the limit of large M . 3.15. The structure factor of the basis of a BCC lattice may be evaluated by assuming the cell to contain one atom at a corner and another at the center of the unit cell. Show that the same result is obtained by taking the cell to contain one eighth of an atom at each of its eight corners, plus one atom at the center. 3.16. Which of the following reflections would be missing in a BCC lattice: (100), (110), (111), (200), (210), (220) and (211)? Answer a similar question for an FCC lattice. 3.17. Evaluate the structure factor of the basis for the diamond structure. Express it in terms of the corresponding factor for the FCC Bravais lattice. Show that the allowed reflections satisfy h + k + l = 4n, where all indices are even and n any integer; or else all indices are odd. Which of the reflections in problem 3.16 are allowed? 3.18. Cesium chloride (CsCl) crystallises in a cubic structure, in which one type of atom is located at the corners and the other at the center of the cell. Draw the cubic unit cell and calculate the structure factor of the basis assuming that fCs = 3fCl. Explain why the extinction rule for the BCC Bravais lattice is violated here. 3.19. Show that if a crystal undergoes volume expansion than the reflected beam is rotated by the angle δθ = −δV 3V tan θ. (B.5) 3.20. A beam of 150 eV electrons falls on a powder nickel sample. Find the three smallest Bragg angles at which reflection takes place. 3.21. In an X-ray diffraction experiment on NaCl, diamond and CsCl the following results was determined: Bragg angle Sample A Sample B Sample C θ1 10.8 ◦ 13.7◦ 22.0◦ θ2 15.3 ◦ 15.9◦ 37.7◦ θ3 18.9 ◦ 22.8◦ 45.8◦ θ4 22.0 ◦ 27.0◦ 59.8◦ θ5 24.7 ◦ 28.3◦ 70.4◦ Determine which sample is which? 22 B. PROBLEMS IN SOLID STATE PHYSICS 4 Crystal Binding 4.1. What is the reason for the fact that the tetrahedral bond is the dominant bond in the diamond structure? 4.2. Estimate the strength of the bond between water molecules in electron volts per bond. 4.3. Show that two parallel electric dipoles attract each other. 4.4. Show that the van der Waals binding energy is ∼ 1/R6. Use two atoms and assume that they are linear harmonic oscillators separated by R and that each oscillator bears charges ±e with separations x1 and x2. 4.5. Estimate the strength of the van der Waals bond for neon in electron volts per bond. 4.6. A quantitative theory of bonding in ionic crystals was developed by Born and Meyer. The total potential energy of the system is taken to be E = N ( A Rn − αe 2 4πǫoR ) , (B.6) whereN is the number of positive-negative ion pairs and α is the Madelung constant. A and n are constants determined from experiments. (a) Show that the equilibrium inter atomic distance is given by the expression Rn−10 = 4πǫoA αe2 n. (B.7) (b) Establish that the bonding energy at equilibrium is E0 = − αNe2 4πǫoR0 ( 1− 1 n ) . (B.8) (c) Calculate the constant n for NaCl, given that the measured binding energy for this crystal is 7.95 eV/molecule and α = 1.75. 5. LATTICE VIBRATIONS 23 5 Lattice Vibrations 5.1. Plot the first three allowed vibrational modes in a continuous line under the bound- ary conditions u(0) = u(L) and u(0) = u(L) = 0. 5.2. Determine the density of states for a two-dimensional continuous medium using periodic boundary conditions. 5.3. In the Einstein model, atoms are treated as independent oscillators. The Debye model, on the other hand, treats atoms as coupled oscillators vibrating collectively, but the collective modes are regarded as independent. Explain the meaning of this independence, and contrast it with that in the Einstein model. 5.4. Would you expect to find sound waves in small molecules? If not, how do explain the propagation of sound in gaseous substances? 5.5. Explain qualitatively why the inter atomic force constant diminishes rapidly with distance. 5.6. Determine the dispersion relation of phonons in a one-dimensional crystal with one atom per primitive cell. Also plot the relation. 5.7. The dispersion relation for a one-dimensional crystal with two atoms per cell is ω4 − 2CM1 +M2 M1M2 ω2 + 2C2 M1M2 (1− cosKa) = 0, (B.9) where a is the length of the cell and M1,M2 are the masses of the two atoms. Determine the roots to the dispersion relation at the Brillouin zone center and at the zone boundary. Also plot the relation. 5.8. Show that the total number of allowed modes in the first BZ of a one-dimensional diatomic lattice is equal to 2N , the total number of degrees of freedom. 5.9. Suppose that we allow the two masses M1 and M2 in a one-dimensional diatomic lattice to become equal. What happens with the frequency gap? 5.10. Derive an expression for the specific heat of a one-dimensional diatomic lattice. Make the Debye approximation for the acoustic branch and assume that the optical branch is flat. Investigate the high- and low-temperature limits. Hint: ∫ ∞ 0 x2ex (ex − 1)2 dx = π2 3 (B.10) 5.11. According to classical theory the average energy of an atom in a crystal is given by the expression EAtom = 3 ∫+∞ −∞ ∫+∞ −∞ ( p2x 2m + 1 2 kx2 ) e − ( p2x 2m + 1 2 kx2 ) /kBT dpxdx ∫+∞ −∞ ∫+∞ −∞ e − ( p2x 2m + 1 2 kx2 ) /kBT dpxdx (B.11) Evaluation of this integral yields the value EAtom = 3kBT . 26 B. PROBLEMS IN SOLID STATE PHYSICS 6.12. Plot the Fermi-Dirac function f(E) versus the energy ratio E/EF at room temper- ature (300 K). Assume that the Fermi energy is independent of temperature. If EF = 5 eV, determine the energy values at which f(E) = 0.5, 0.7, 0.9 and 0.95. 6.13. Cyclotron resonance has been observed in Cu at a frequency of 24 GHz. Given that the effective electron mass is equal to the normal mass compute the value of the applied magnetic field. 6.14. At which temperature is the electronic contribution to the heat capacity Cv equal to the phonon contribution for Ag? 7. BAND THEORY 27 7 Band Theory 7.1. The empty lattice electron energy is obtained by cutting and displacing various segments of the free electron energy curve. Is this rearrangement justifiable for a truly free electron? How do you differentiate between an empty lattice and free space? 7.2. Band overlap is important in the conductivity of polyvalent metals. Do you expect it to take place in a one-dimensional crystal? 7.3. Convince yourself that the shapes of the first Brillouin zones for the FCC and BCC are those given in the text book. 7.4. Show that the number of allowed k-values in a band of a three-dimensional simple cubic lattice is equal to N , the number of unit cells in the crystal. 7.5. Calculate the number of electron states n in a band of a three-dimensional lattice, which contains N number of unit cells in the crystal. 7.6. Show that the first three bands in the empty lattice model in one dimension span the following energy ranges    0 ≤ E1 ≤ π 2h̄2 2mea2 π2h̄2 2mea2 ≤ E2 ≤ 2π 2h̄2 mea2 2π2h̄2 mea2 ≤ E3 ≤ 9π 2h̄2 mea2 (B.12) 7.7. Show that wave vectors, touching the centers of the hexagonal faces of the first zone of the FCC lattice, give rise to Bragg reflection from the (111) atomic planes. The shape of the zone is given in Kittel. Also show that the other faces give rise to reflection from the (200) planes. 7.8. Show similarly that the faces of the zone for the BCC lattice are associated with reflections from the (110) faces. The shape of the zones are shown in the text book. 7.9. Suppose that the crystal potential in a one-dimensional lattice is composed of a series of rectangular wells which surround the atom. Suppose that the depth of each well is V0 and its width a/5. Using the nearly free electron model, calculate the values of the first three energy gaps. 7.10. (a) Using the free electron model, and denoting the electron concentration by n, show that the radius of the Fermi sphere is given by kF = (3π 2n)1/3. (b) As the electron concentration increases, the Fermi sphere expands. Show that this sphere begins to touch the faces of the first zone in an FCC lattice when the electron to atom ratio n/na = 1.36, where na is the atom concentration. (c) Suppose that some of the atoms in a Cu crystal, which has an FCC structure, are gradually replaced by Zn atoms. Considering that Zn is divalent while Cu is monovalent, calculate the atomic ratio of Zn to Cu in a CuZn alloy (brass) 28 B. PROBLEMS IN SOLID STATE PHYSICS at which the Fermi sphere touches the zone faces. This particular mixing is interesting because the solid undergoes a structural phase change at this concentration ratio. 9. SUPERCONDUCTIVITY 31 9 Superconductivity 9.1. The superconductor tin has Tc = 3.7 K and Bc = 30.6 mT at T = 0 K. Calculate the critical current for a tin wire of diameter 1 mm at T = 2 K. What diameter of wire would be required to carry a current of 100 A? 9.2. A solenoid is used to investigate the resistivity of superconducting Nb3Zr. The solenoid is 25 cm long with a diameter of 10 cm. It has 984 turns of a Nb3Zr wire of diameter 0.47 mm. The decay of the current is governed by I(t) = I0 exp ( −Rt L ) , (B.13) where R is the resistance and L the inductance. The decay was found to be less than 1 per 109 in one hour. Deduce an upper limit for the electrical resistivity of superconducting Nb3Zr. 9.3. Estimate the strength of the electron-phonon interaction for Al and Pb. 9.4. A measurement of the critical temperature Tc for different Hg-isotopes gave the following result: M (u) 199.5 200.7 202.0 203.3 Tc (K) 4.185 4.173 4.159 4.146 Determine the relation between Tc and M . 9.5. Nb is a type II superconductor with coherence length ξ = 3.8 · 10−8 m. An external magnetic field penetrates a type II superconductor along fluxoids with an approx- imate diameter of 2ξ containing one quantum of flux. Determine the critical field Bc2 for Nb and compare with the experimental value Bc2 = 0.302 T. Assume that the critical field is reached when the fluxoids are close-packed. 32 B. PROBLEMS IN SOLID STATE PHYSICS 10 Optical and Dielectric Properties 10.1. Estimate threshold energy and wave length for optical inter band transitions in potassium, using the free electron model. 10.2. Calculate the refractive index of NaCl and CsCl, using the Clausius-Mossotti rela- tion. The electronic polarisabilities are 3.23 · 10−35 cm3 for Na+, 3.05 · 10−34 cm3 for Cs+, and 3.28 · 10−34 cm3 for Cl−. 10.3. Sketch the frequency dependence of the polarisability for a crystal with electronic, ionic, and dipolar effects contributing to it. 10.4. Consider a semiclassical model of the ground state of the hydrogen atom in an electric field normal to the plane of the orbital, as shown in the Figure. Show that α = 4πǫ0a 3 H for this model, where aH is the radius of the unperturbed orbital. Note that if the applied field is in the x-direction, then the x-component of the field of the nucleus at the displaced position of the electron orbit must be equal to the applied field. (The correct quantum mechanical result is larger than this by a factor 9 2 .) eE x Figure B.2: An electron in a circular orbit of radius aH is displaced a distance x on application of an electrical field. 10.5. Find the frequency dependence of the electronic polarisability of an electron having the resonance frequency ω0, treating the system as a simple harmonic oscillator. 11. MAGNETIC PROPERTIES 33 11 Magnetic Properties 11.1. The ground state (1s) of the hydrogen atom is ψ(r) = 1√ πa3 e−r/a, (B.14) where a is the Bohr radius. Calculate the diamagnetic molar susceptibility χm for this atom, using the Langevin formula. 11.2. Use the Hund rules to determine the ground state and the effective number of Bohr magnetons for: (a) Eu++ with the configuration 4f 75s2p6 (b) Yb3+ with the configuration 4f 135s2p6 (c) Tb3+ with the configuration 4f 85s2p6 11.3. The magnetic susceptibility is measured on a diamagnetic sample, which is suspected to be contaminated by Mn. The measurement gave the following results: T (K) 300 180 145 115 χ (SI) -8.88·10−6 -8.50·10−6 -8.24·10−6 -7.86·10−6 (a) Show that the sample contains a paramagnetic contamination. (b) Calculate the concentration of Mn under the assumption that they occur as Mn2+ ions with (3d5) configuration. (c) Determine the diamagnetic susceptibility of the substrate. 11.4. Calculate the theoretical saturation magnetisation of Fe2(SO4)3, which has a density of 3.1·103 kg/m3. 11.5. In an inhomogeneous magnetic field a body is influenced by a force F = V χH ∂B ∂z n the direction of the field gradient. This is used in an experimental setup, where a small amount of the paramagnetic salt (CrK(SO4)2 · 12H2O) is attached to a weighting-machine. The turn of the scale was 10.00 mg. Then the salt was sunken into a dewar with liquid nitrogen (77 K) and placed in the inhomogeneous magnetic field with H ∂B ∂z = −3.493 · 106 N/m3. The z axis is vertical. The new turn of the scale was 7.74 mg. The density of the paramagnetic salt is 1.83 g/cm3 and of liquid nitrogen 0.808 g/cm3. (a) Determine the number of effective Bohr magnetons in the paramagnetic salt. (b) Calculate the theoretical value of effective Bohr magnetons if Cr3+ have the configuration 3d3. 11.6. Show that the saturation magnetisation just below the Curie temperature is pro- portional to (Tc − T )−1/2, using the mean field approximation and that the spin is 1/2. 36 C. SOLUTIONS, HINTS OR ANSWERS 1.6. For a figure see the front page. The primitive cell contains one atom and the cubic cell four atoms. 1.7. Draw a vector a pointed along say the x-direction and assume it to be a vector of lattice translation. Assume further that a rotation β yielding a vector a′ is a symmetry operation. This means that a rotation of −β resulting into the vector a′′ also is a symmetry operation. The vector sum a′ + a′′ is parallel with a and is a vector of lattice translation i. e. a′ + a′′ = na with n an integer. We have thus that 2a cos β = na or that cos β = n/2. The only possible values for n are -2, -1, 0, 1 and 2 and none of these produce a lattice that have 5-fold symmetry. 1.8. This is easily seen if you draw a picture. In Fig.C.1 one triangle is reflected in the two mirror planes to generate an object of eight triangles which possesses a 4-fold axis at the intersection. 6 3 1 2 8 7 4 5 Figure C.1: One triangle reflected in two mirror planes. 1.9. For the (111)-planes every atom has six nearest neighbours and that is the highest possible density. Expressed in the cube length a this density is n = 2 √ 12 3a2 . (C.3) For Cu a = 3.61 · 10−10 m, giving n = 0.177 · 1020 atoms/m2 1.10. The Miller indices are shown in the table. Planes System a System b 1 (11) (20) 2 (12̄) (13) 3 (21̄) (1̄3) 4 (31) (42̄) 2 The Reciprocal Lattice 2.1. Let a1, a2 and a3 be the primitive translation vectors of the crystal lattice. Then the reciprocal lattice vectors are given by b1 = 2π a2 × a3 a1 · a2 × a3 ; b2 = 2π a3 × a1 a1 · a2 × a3 ; b3 = 2π a1 × a2 a1 · a2 × a3 . (C.4) 2. THE RECIPROCAL LATTICE 37 Now let c1, c2 and c3 be the reciprocal lattice vectors to the reciprocal lattice. Then we have c1 = 2π b2 × b3 b1 · b2 × b3 = 2π 2π a3×a1 a1·a2×a3 × 2π a1×a2 a1·a2×a3 2π a2×a3 a1·a2×a3 · 2π a3×a1 a1·a2×a3 × 2π a1×a2 a1·a2×a3 = V (a3 × a1)× (a1 × a2) (a2 × a3) · (a3 × a1)× (a1 × a2) , (C.5) where V = a1 · a2 × a3. Recalling the two vector identities A× (B×C) = B(A ·C)−C(A ·B) (C.6) A · (B×C) = B · (C×A) = C · (A×B), (C.7) we find that c1 = a1. In the same way we find that c2 = a2 and c3 = a3 i.e. the reciprocal lattice to the reciprocal lattice is the direct lattice. 2.2. Use the relations (C.4) to calculate the reciprocal translation vectors. Note that ai must be primitive vectors of the lattice. (a) The primitive translation vectors of the SC lattice are a1 = ax̂; a2 = aŷ; a3 = aẑ, (C.8) which gives b1 = 2π a x̂; b2 = 2π a ŷ; b3 = 2π a ẑ. (C.9) Thus, the reciprocal lattice to the SC lattice is a SC lattice with the lattice constant 2π/a. (b) The primitive translation vectors of the BCC lattice are a1 = 1 2 a(−x̂+ ŷ + ẑ); a2 = 1 2 a(x̂− ŷ + ẑ); a3 = 1 2 a(x̂+ ŷ − ẑ), (C.10) which gives b1 = 2π a (ŷ + ẑ); b2 = 2π a (x̂+ ẑ); b3 = 2π a (x̂+ ŷ); (C.11) Thus, the reciprocal lattice to the BCC lattice is an FCC lattice with the lattice constant 4π/a. (c) The primitive translation vectors of the FCC lattice are a1 = 1 2 a(ŷ + ẑ); a2 = 1 2 a(x̂+ ẑ); a3 = 1 2 a(x̂+ ŷ); (C.12) which gives b1 = 2π a (−x̂+ ŷ + ẑ); b2 = 2π a (x̂− ŷ + ẑ); b3 = 2π a (x̂+ ŷ − ẑ), (C.13) Thus, the reciprocal lattice to the FCC lattice is a BCC lattice with the lattice constant 4π/a. 38 C. SOLUTIONS, HINTS OR ANSWERS 2.3. In both cases YES. The angle is 35.26◦. 2.4. In the general case the reciprocal lattice vectors depend on all three real lattice vectors. However, for the case when the real lattice vectors are at right angles with one another (simple cubic) the answer is yes. 2.5. Just simple mathematics. Let a1, a2 and a3 be the primitive translation vectors of the crystal lattice. Then the volume is given by V = |a1 · a2 × a3|. The reciprocal lattice vectors are given by equation (C.4) and the volume of the reciprocal cell is Ω = |b1 · b2 × b3| = ∣ ∣ ∣ ∣2π a2 × a3 a1 · a2 × a3 · 2π a3 × a1 a1 · a2 × a3 × 2π a1 × a2 a1 · a2 × a3 ∣ ∣ ∣ ∣ = (2π)3 V 3 |(a2 × a3) · (a3 × a1)× (a1 × a2)| = (2π)3 V 3 |(a2 × a3) · a1(a3 · a1 × a2)| = (2π)3 V . (C.14) 2.6. The volume Ω of the Brillouin zone is equal to the volume of the primitive cell in the reciprocal lattice. The volume is Ω = (2π)3/V , where V is the volume of the primitive cell in the real lattice. Na has BCC structure with a = 4.23 Å. Thus, the volume of the cubic cell is two times V . We find Ω = (2π)3 (4.23 · 10−10)3/2 = 6.55 · 10 30 m−3. (C.15) Au has FCC structure with a = 4.08 Å. Thus, the volume of the cubic cell is four times V . We find Ω = (2π)3 (4.08 · 10−10)3/4 = 14.6 · 10 30 m−3. (C.16) 2.7. The first, second and third Brillouin zones of a square lattice are shown in Fig.C.2.                              3rd BZ                          2dn BZ 1st BZ Figure C.2: The first, second and third Brillouin zones of a square lattice. 3. DIFFRACTION IN CRYSTALS 41 3.6. The reason is that the mass of a neutron is much larger than the mass of an electron. Convince yourself of that by comparing the energies for one and the same de Broglie wave length. 3.7. Of course not. Since λlight ≈ 10−6 − 10−7 m and the distance between atoms in a crystal is ≈ 10−10 m. 3.8. The neutron is uncharged an can penetrate much deeper into the crystal. 3.9. The minimum wavelength is obtained when the electron gives all of its kinetic energy to the photon. Thus the energy E of these photons is equal to the kinetic energy of the electrons E = hc λ = 10 keV. (C.28) 3.10. First show that for a simple cubic lattice there is no extinction rule. Then, using Bragg’s law 2a√ h2 + k2 + l2 sin θ = λ, (C.29) we find Plane: (100) (110) (111) (200) (210) (211) Bragg angle: 17.1◦ 24.6◦ 30.6◦ 36.0◦ 41.1◦ 46.0◦ 3.11. (a) d(111) = 2.34 Å (b) First calculate the cube length a with the formula dhkl = a√ h2 + k2 + l2 (C.30) and recall that the cubic cell contains four atoms. We simply have that ρ = 4 · 27 NAa3 , (C.31) from which we find that NA = 6.0 · 1026 kmol−1. 3.12. (a) From Bragg’s law (C.30), we easily obtain that a = 2.91 Å. (b) Reflections from the (111)-planes are forbidden since 1 + 1 + 1 = 3 is an odd number. (c) The density is given by ρ = 2M/NA a3 = 7544 kg/m3 (C.32) 3.13. For a spherically symmetric electron distribution the atomic form factor is given by f = 4π ∫ drρ(r)r2 sinGr Gr . (C.33) Use Euler’s formula sinGr = eiGr − e−iGr 2i (C.34) and integrate by parts. 42 C. SOLUTIONS, HINTS OR ANSWERS 3.14. We know that central diffraction maxima occur for a · ∆k = h · 2π, where h is an integer. The first minimum occurs for 1 2 M(a · ∆k) = π and the second for 1 2 M(a · ∆k) = 2π , the first subsidiary maximum is then likely to be found for 1 2 M(a ·∆k) = 3π 2 . This gives an intensity of 0.045M2, in the limit of large M . 3.15. Use the relation Shkl = ∑ i fie −2πi(hxi+kyi+lzi) (C.35) where the sum is over all atoms i in the unit cell. With one atom in (0,0,0) and one in (1 2 , 1 2 , 1 2 ) we obtain Shkl = fa ( 1 + e−iπ(h+k+l) ) . (C.36) With one eight of an atom at the positions (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1) , (1,1,1) and one atom at (1 2 , 1 2 , 1 2 ) we obtain Shkl = fa 8 ( 1 + e−i2πh + e−i2πk + e−i2πl + e−i2π(h+k) + e−i2π(h+l)+ +e−i2π(k+l) + e−i2π(h+k+l) ) + fae −iπ(h+k+l). (C.37) Which after simplification is the same formula as which results when assuming the cell to contain one atom at a corner and another at the center. 3.16. Using the relation (C.35) for BCC we find the extinction rule h+ k + l = even, i.e. we have reflections from the (110)-, (200)- and (220)-planes. Using the relation (C.35) for FCC we find the extinction rule all h, k, l even or all h, k, l odd, i.e. we have reflections from the (111)-, (200)- and (220)-planes. 3.17. Diamond structure is an FCC lattice plus a basis in (0,0,0) and (1 4 , 1 4 , 1 4 ). From the relation (C.35) we get that SDiahkl = fa ∑ j ( e−i2π(hxj+kyj+lzj) + e−i2π{h(xj+ 1 4 )+k(yj+ 1 4 )+l(zj+ 1 4 )} ) = SFCChkl ( 1 + e− iπ 2 (h+k+l) ) , (C.38) where the j is the four lattice points of the cubic FCC cell. From this we see that SDiahkl 6= 0 provided that all h, k, l even and h + k + l = 4n or all h, k, l odd. Thus the (111)- and (220)-planes in problem 3.16 are allowed. 3.18. Put one atom in (0, 0, 0) (Cs) and the other in (1 2 , 1 2 , 1 2 ) (Cl), then the structure factor (C.35) is Shkl = fCs + fCle −iπ(h+k+l) ≈ fCl ( 3 + e−iπ(h+k+l) ) 6= 0 ∀ hkl. (C.39) In spite of the fact that the structure is similar to BCC it is a SC-structure with a basis of two atoms. Therefore the extinction rule for the BCC lattice is violated. 3.19. Differentiate the Bragg law 2L sin θ = λ and you will obtain L cos θ ·dθ+dL sin θ = 0 or dθ = −dL L tan θ. Also differentiate V = L3 and you will find the desired result. 4. CRYSTAL BINDING 43 3.20. All h, k, l odd or all even for allowed reflections, since Ni has FCC structure. The three smallest angles result from reflections from the (111)-, (200)- and (220)-planes. 150 eV electrons allows a classical evaluation of the de Broglie wave length, so λ = h/ √ 2Em = 1.00 Å. Using the lattice parameter a = 3.52 Å one finds: Plane (111) (200) (220) Angle 14.2◦ 16.5◦ 23.7◦ 3.21. The structure and extinction rules are: Substance Structure Extinction rule NaCl FCC ∀ h, k, l even ∀ h, k, l odd CsCl SC ∀ h, k, l Diamond Diamond ∀ h, k, l even, h+ k + l = 4n ∀ h, k, l odd If we take the square of Bragg’s law we find sin2 θ = λ2 4a2 (h2 + k2 + l2), (C.40) where λ2/4a2 is a constant and (h2 + k2 + l2) is an integer. Putt (h2 + k2 + l2) = X and calculate the fractions sin2 θi sin2 θ1 = Xi X1 . (C.41) Now we can identify (hkl) for the scattering planes. Sample A Sample B Sample C θ sin 2 θi sin2 θ1 Xi (hkl) θ sin2 θi sin2 θ1 Xi (hkl) θ sin2 θi sin2 θ1 Xi (hkl) 10.8◦ 1 1 100 13.7◦ 1 3 111 22.0◦ 1 3 111 15.3◦ 2 2 110 15.9◦ 4/3 4 200 37.7◦ 8/3 8 220 18.9◦ 3 3 111 22.8◦ 8/3 8 220 45.8◦ 11/3 11 311 22.0◦ 4 4 200 27.0◦ 11/3 11 311 59.8◦ 16/3 16 400 24.7◦ 5 5 210 28.3◦ 12/3 12 222 70.4◦ 19/3 19 331 When we compare the scattering planes with the extinction rules we find that sample A is CsCl, sample B is NaCl and that sample C is diamond. 4 Crystal Binding 4.1. For a silicon atom the electron configuration is (Ne)3s23p2. Thus Si has four valence electrons and lacks four electrons with respect to a filled shell structure (Ar). Such a structure can however be simulated by electron sharing. First promote one of the 46 C. SOLUTIONS, HINTS OR ANSWERS 5 Lattice Vibrations 5.1. The solutions are of the form u(x) = Aeikx. The boundary condition u(0) = u(L) gives k = n · 2π/L, n = 1, 2, 3 and the boundary condition u(0) = u(L) = 0 gives k = n · π/L, n = 1, 2, 3. Figure C.5 shows Im(u) = Im(Aeikx) in the interval x = [0, L]. 0 L (a) 0 L (b) Figure C.5: The three first modes under the boundary conditions u(0) = u(L) (a) and u(0) = u(L) = 0 (b). 5.2. Apply periodic boundary conditions over a square of side L. This leads to a k- vector density of (L/2π)2 and that 2πkdk · (L/2π)2 = D(ω)dω. After simplification D(ω) = Aω/(2πv2), where A is the area and v the velocity of sound. 5.3. According to the Einstein model the atom energy in one dimension is given by EAtom = ∑∞ n=0 nh̄ωe −nh̄ω/kBT ∑∞ n=0 e −nh̄ω = h̄ω eh̄ω/kBT − 1 , (C.57) resulting in the crystal energy U = 3NAEAtom. We see that we multiply the thermal energy for one atom with the total number of atoms i.e. we assume that the atoms vibrate independent of one another. In the Debye model the crystal energy is given by U = ∫ ωD 0 n(ω) · h̄ω ·D(ω)dω (C.58) where n(ω) = 1 eh̄ω/kBT − 1 (Bose-Einstein statistics) (C.59) We sum (or integrate) over a continuum of modes with energies h̄ω but assume that the modes do not interact with one another. 5.4. No, the wave length of sound is much larger than a molecule. The sound velocity in a gas is the velocity of a pressure pulse. 5. LATTICE VIBRATIONS 47 5.5. The interatomic attractive force results from an reorganisation of the outer electrons of the atoms. This means that they have to be close to one another for the electron clouds to overlap. This also explains why the free atom model holds so well in gases in which the average interatomic distance is much large than the corresponding distance in a crystal. 5.6. The dispersion relation is given by ω = √ 4C M ∣ ∣ ∣ ∣sin Ka 2 ∣ ∣ ∣ ∣ , (C.60) see Kittel for details. In Fig. C.6 ω is in units of √ 4C/M . 0 K π/ a0 0.5 1.0 ω Figure C.6: The dispersion relation of a one-dimensional crystal with one atom per cell. 5.7. The roots are given by ω2 = C M1 +M2 M1M2 ± C √ √ √ √ ( M1 +M2 M1M2 )2 − 4 sin 2 Ka 2 M1M2 (C.61) At the zone center we have K = 0, so we find ω2 = 2C ( 1 M1 + 1 M2 ) (optical branch) ω2 = 0 (acoustical branch). (C.62) At the zone boundary K = ±π/a, we find ω2 = 2C M1 ; ω2 = 2C M2 . (C.63) The dispersion relation is shown in Fig. C.7. 48 C. SOLUTIONS, HINTS OR ANSWERS 0 K π/ a Optical branch Acoustical branch Figure C.7: The dispersion relation of a one-dimensional crystal with two atom per cell. 5.8. Applying periodic boundary conditions over a chain of total length L containing N cells with a primitive translation vector a gives a k-vector density of L/2π = Na/2π. Multiplying this with the length of the Brillouin zone (= 2π/a) results in (Na/2π) · (2π/a) = N allowed modes. For a diatomic lattice we have two branches, the acoustical and the optical branch. Therefore there are 2N allowed modes. 5.9. The frequency gap disappears, since the gap is given by | √ 2C/M2 − √ 2C/M1|. 5.10. The density of states in one dimension is given by D(ω) = L π ( dω dk )−1 = L πν , (C.64) in the Debye approximation. The number of allowed modes in the acoustic branch is ∫ ωD 0 D(ω)dω = N, (C.65) which gives ωD = Nπν L ⇒ D(ω) = N ωD . (C.66) Now we can use the relation Cacousticv = dU dT . (C.67) After differentiation, the change to the dimensionless variable x = h̄ω/kBT and by defining the Debye temperature θD as kBθD = h̄ωD the expression for the heat capacity takes the form Cacousticv = R T θD ∫ θD/T 0 x2ex (ex − 1)2 dx. (C.68) At low temperatures, T ≪ θD, the upper limit of the integral approaches ∞, and in this limit Cacousticv = R T θD · π 2 3 . (C.69) 6. THE FREE ELECTRON MODEL 51 6 The Free Electron Model 6.1. A core, or inner shell electron has a probability distribution which is large near the nucleus but weak or nearly zero midway between the atoms. A free or delocalised electron on the other hand has a wave function very much like a plane wave. Such electrons contribute to electrical- and thermal conductivity while core electrons do not. The existence of core electrons is, however, clear from X- ray diffraction exper- iments. Remember the atomic form factor. 6.2. Electrons are charged particles while a gas usually consists of uncharged molecules or atoms. From the ideal gas equation pV = NRT we find that n = N/V ≈ 2.4·1025 molecules/m3. For Li which has BCC structure and one valence electron per atom we find that the electron density is n ≈ 4.7 · 1028 electrons/m3. Thus the electron density is much greater than the density of molecules in air. 6.3. The electric current density is given by J = Nqv. In an electrical field we realise that changing the sign of the charge is accompanied by a reversed direction of the velocity i. e. the sign of J is the same. 6.4. Classically we have Eelectron = 3kBT/2. For T = 300 K we easily find the velocity to be v = 1.2 · 105 m/s. Compare this with vF = 1.57 · 106 m/s. 6.5. For an external electric AC - field, with a frequency coinciding with the cyclotron frequency, strong absorption of the signal can be observed and the sample absorbs energy and the velocity of the excited electrons increases. In the steady state this velocity reaches a constant value limited by collisions with phonons and impurities and a balance between absorbed energy and energy leaving the sample through radiation exists. 6.6. (a) The electron density, n, is given by n = Zv NA V , (C.82) where Zv is the valence of the metal and V is the volume of 1 kmol. But V =M/ρ, thus we obtain n = Zv ρNA M = 8.48 · 1028 m−3. (C.83) (b) τ = 2.7 · 10−14 s (c) EF = 7.0 eV (d) vF = 1.57 · 106 m/s (e) lF = vF τ ≈ 400 Å 6.7. EF = h̄2 2m (3π2N/V )2/3 and when β = 1 V ( ∂V ∂T ) P we realise that the Fermi energy decreases when the temperature increases. Differentiate the expression for EF and you will find that dEF EF = −2 3 dV V = −2 3 β∆T. (C.84) Here we have assumed that β is constant. Calculation shows a decrease of 3 %. 52 C. SOLUTIONS, HINTS OR ANSWERS 6.8. Identical calculations as above shows a decrease of 3.7 % . 6.9. See definition of the Fermi temperature TF in the text book. We conclude that room temperature is much lower than TF . 6.10. The fraction of electrons excited above the Fermi level is given by the integral η = 1 N ∫ ∞ EF D(E)f(E, T )dE, (C.85) where f(E, T ) = 1 e(E−EF )/kBT + 1 , (C.86) i. e. we replace the chemical potential µ with the constant EF (see argument in the text book). Further we have D(E) = V 2π2 ( 2m h̄2 )3/2 E1/2 (C.87) EF = h̄2 2m ( 3π2N V )2/3 . (C.88) From the expressions for D(E) and EF we eliminate h̄2 2m to obtain D(E) = 3N 2E 3/2 F E1/2 (C.89) and as the main contribution to the integral results for energies very near EF we can make the approximation D(E) = D(EF ) ≈ 3N 2EF . (C.90) It follows then that η ≈ 3 2EF ∫ ∞ EF dE e(E−EF )/kBT − 1 . (C.91) Introducing the dimensionless variable x = (E − EF )/kBT we obtain the simpler expression η = 3kBT 2EF ∫ ∞ 0 dx ex + 1 , (C.92) which can be solved analytically. Putting y = ex we get ∫ ∞ 0 dx ex + 1 = ∫ ∞ 1 dy y(y + 1) = ∫ ∞ 1 ( 1 y − 1 y + 1 ) dy = ln 2. (C.93) Thus η = 3 ln 2 2 kBT EF ≈ kBT EF (C.94) and for T = 300 K we find ηCu ≈ 3.7 · 10−3 and ηNa ≈ 8.0 · 10−3. 6. THE FREE ELECTRON MODEL 53 6.11. The phonon contribution to the heat capacity given by Einstein is Cphv = 3R ( θE T )2 eθE/T (eθE/T − 1)2 (C.95) The other expressions can be found in the text book. The ratios asked for are given below. T(K) 0.3 4 20 77 300 Ratio 117 0.65 0.026 0.0033 0.0064 From this it is clear that the electronic contribution to the heat capacity is important only at very low temperatures. Why does the ratio increase somewhat from 77 to 300 K? 6.12. The energies are f(E, T ) 0.5 0.7 0.9 0.95 E (eV) 5 4.98 4.94 4.92 Note that f(E, T ) rapidly approaches 1 for energies below the Fermi level. 6.13. The cyclotron frequency, ωc, is given by the formula, ωc = eB/m, and ωc = 2πfc thus B = 2πfcm e ≈ 0.86 Tesla. 6.14. The electron contribution to the heat capacity is Celv = π2 2 NkB T TF . (C.96) From the Debye model we find that the phonon contribution to the heat capacity is Cphv = 9NkB ( T θD )3 ∫ xD 0 dx x4ex (ex − 1)2 →    3NkB d̊a T ≫ θD 12π4 5 NkB ( T θD )3 d̊a T ≪ θD (C.97) TF = 6.36 · 104 K and θD = 225 K for Ag. There are three intersections. The intersection at T = 0 K is trivial. The intersection at high temperature is unphysical, since Celv = C ph v when π2 2 NkB T TF = 3NkB → T = 6 π2 TF , (C.98) which gives T = 3, 87 · 104 K. This temperature is fare above the boiling point 2466 K of Ag. For the third intersection Celv = C ph v when π2 2 NkB T TF = 12π4 5 NkB ( T θD )3 → T = √ 5θ3D 24π2TF , (C.99) which gives T = 1, 94 K. Thus the approximation is correct, since T ≪ θD. 56 C. SOLUTIONS, HINTS OR ANSWERS the cubic cell of an FCC structure, with one atom per primitive cell, contains four atoms we have ( 3π2 4z a3 )1/3 = π √ 3 a ⇒ z = π √ 3 4 ≈ 1.36 (C.105) (c) Let N denote the total number of atoms, out of which N1 is the number of Cu atoms. The average valence is then N1 · 1 + (N −N1) · 2 N = z (C.106) from which we find N1 N = 2− z = 0.64. Thus 64 % Cu and 36 % Zn. 8 Semiconductors 8.1. The so-called bond model has its roots in quantum chemistry for analyses of elec- tron structures of molecules. One simple example is the molecule CH4 for which the tetrahedral symmetry is associated with the formation of covalent bonds. Trans- ferred to covalent solids, like Si, a reasoning like that for answering problem 4.1 is generally attempted. The strength of this model is that it is easily grasped, gives some understanding of the bonding mechanism and easily illustrates the known fact that the electron density is largest in the region midway between nearest neighbours. However, for quantitative purposes this model is unsatisfactory. The electrons, are not actually localised to bonds. Like free electrons in a metal the wave functions extend throughout the crystal. This is clear from the statement of the Bloch theorem ψk(r) = uk(r)e ik·r, (C.107) which is central in the band model. Having solved the Schrödinger equation Ĥψnk(r) = Enkψnk(r) for a dense mesh of wave vectors within the 1st BZ, the electron density can be calculated from the formula ρ(r) = ∑ n ∫ BZ ψ∗nk(r)ψnk(r)d 3k, (C.108) where n is a sum over all bands occupied with electrons and the integration is over the 1st BZ. In agreement with the bond model this density is largest between nearest neighbours. However, much more quantitative information can be obtained from the solutions; such as effective masses, band gaps etc. In the bond model there is no way for instance to explain the difference between an indirect band gap and a direct one. 8.2. The bond orbitals correspond to the valence band. Usually the wave functions of the valence band are called ”bonding” while the wave functions corresponding to the conduction band are called ”antibonding”. 8.3. If the wave functions are calculated for the conduction band one finds that the prob- ability distribution is rather flat. Thus these electrons are more like free electrons than like electrons in covalent bonds. 8. SEMICONDUCTORS 57 8.4. A broken bond in the bond model corresponds to an excitation of an electron from the valence band to the conduction band. 8.5. Hall effect- and cyclotron resonance experiments show that the holes in the valence band behave as charge carriers. For this to be possible the electrons must be mobile. 8.6. No. Even a doped sample exhibits intrinsic behaviour if the temperature is high enough. 8.7. The Hall coefficient for a sample with holes and electrons i. e. two types of carriers is given by RH = pµ2h − nµ2e e(nµe + pµh)2 . (C.109) Putting RH = 0 one finds that p n = ( µe µh )2 . Taking Si as an example µe = 1300 cm2/Vs and µh = 500 cm 2/Vs. This gives p n ≈ 6.8, meaning that the sample is weakly p - doped. 8.8. (a) Using T = 300 K, Eg = 1.10 eV and the relation ni = pi = 2 ( kBT 2πh̄2 )3/2 (memh) 3/4e−Eg/kBT , (C.110) we find that ni = pi = 1.1 · 1016 m−3. (b) From the relation µ = Ec + 1 2 Eg + 3 4 kBT ln ( mh me ) (C.111) we get µ = Ec + Eg/2 + 0.007 eV. 8.9. (a) The atoms may be considered as hydrogen atoms in a dielectric surrounding. The energy levels of a hydrogen atom in vacuum are given by En = − me4 8ǫ20h 2 Z2 n2 = 13.6 Z2 n2 eV, (C.112) where Z is the effective charge of the atomic nucleus. In the dielectric surround- ing ǫ0 is changed to ǫrǫ0 and the mass m is changed to m ∗m. The ionisation energy is equal to |E1|, so we find Ed = 13.6 · 0.07/(13.13)2 = 5.5 meV, and Ea = 13.6 · 0.09/(13.13)2 = 7.1 meV. (b) The radius of a hydrogen atom in its ground state is given by r = h2ǫ0 πme2 1 Z = a0 1 Z , (C.113) where a0 = 0.53 Å. This gives for the donors and acceptors that r = ǫrma0/m ∗, so we find ad = 99 Å, aa = 77 Å (c) At sufficiently low temperatures the donors and acceptors are not ionised i. e. the holes and electrons are ”frozen” at their impurity sites. The temperature at which freezing takes place can be estimated from the equation Ed ≈ kBT which for the donor gives that T = 64 K. We can therefore always be sure that all donors and acceptors are ionised at room temperature. 58 C. SOLUTIONS, HINTS OR ANSWERS 8.10. (a) From problem 8.8 we have that ni = pi = 1.1 · 1016 m−3. We see directly that this number is much smaller than the concentration supplied by the donors provided that they are ionised which always can be assumed at room temper- ature. (b) From the law of mass action we have, np = n2i , but n = Nd+p so p 2+pNd−n2i = 0. Solving for p we obtain (note that p > 0) p = −Nd 2 + √ ( Nd 2 )2 + n2i . (C.114) Numerically we might find that p = 0, if this is the case we have to blame our pocket calculator. Instead of purchasing a new one we solve the equation pNd − n2i = 0, (C.115) i. e. neglecting p2 as it must be small. We then find that p = 1.21 · 109 m−3 and n = 1.0 · 1023 m−3. From the expressions in the text book for n and p an expression for the Fermi level can be obtained and be given by µ = Ec + Eg 2 + kBT 2 ln ( n p ) + 3 4 kBT ln ( mh me ) (C.116) Numerically we find µ = Ec + Eg/2 + 0.421 eV. (c) If we introduce acceptors with a concentration of i Na = 6.0 · 1021 m−3 we note that the sample above is still effectively n-doped withNd = 1.0·1023−6.0·1021 = 9.4 · 1022 m−3. Identical calculations as above shows a slightly lower value for the Fermi level. 8.11. (a) From the expressions for electron- and hole mobilities in the text book we obtain τe = meµe e , τh = mhµh e (C.117) and numerically τe = 5.4 · 10−13 s, τh = 2.7 · 10−13 s. (b) From problem 8.8 we have that ni = pi = 1.1 · 1016 m−3, so we have that σi = nie(µe + µh) = 3.2 · 10−4 (Ωm)−1. 8.12. From the text book we know that the Hall coefficient is given by RH = − 1 ne = − Ey BzJx , (C.118) from which we obtain that n = BzJx eEy . (C.119) With reference to the Figure C.9 below we have n = BzI/bh eVH/h = BzI ebVH = 4.7 · 1021 m−3. (C.120) 11. MAGNETIC PROPERTIES 61 where ni is the concentration and αi is the polarisability of atom i. The refractive index is given by n2 = ǫ. NaCl has FCC structure with a = 5.63 Å, so ∑ niαi/3ǫ0 ≈ 0.303, which give n = 1.52. CsCl has SC structure with a = 4.11 Å, so ∑ niαi/3ǫ0 ≈ 0.343, which give n = 1.60. 10.3. See the figure in chapter 13 in Kittel. 10.4. The force on the electron due to the nucleus is F = −e2 4πǫ0r2 r̂, (C.131) if the origin is at the nucleus. The polarisability α is defined by p = αElocal, where p is the electric dipole moment p = ex. The requrement that the x-component of the force from the nucleus and the force from the electrical field must be equal in magnitude give e2 4πǫ0r2 x r = eE. (C.132) If x≪ aH we get p = e2E4πǫ0a 3 H e2 , (C.133) so the polarisability can be identified to be α = 4πǫ0a 3 H . 10.5. Let the local electrical field be given by Elocal = E0 sinωt. Then the equation of motion is given by m d2x dt2 +mω20x = −eE0 sinωt, (C.134) so that for x = x0 sinωt, one find m(−ω2 + ω20)x0 = −eE0. (C.135) The dipole moment has the amplitude p0 = −ex0 = e2E0 m(ω20 − ω2) ⇒ α = e 2/m ω20 − ω2 . (C.136) 11 Magnetic Properties 11.1. The Langevin formula is χm = − µ0NZe 2 6nV < r2 >, (C.137) where N V = NAρ M . The molar susceptibility is χmolarm = χmM ρ and the expectation value < r2 > is < r2 >= ∫ ψ∗r2ψd3r = − 1 πa3 ∫ ∞ 0 r2e−2r/a4πr2dr = 4a2 ∫ ∞ 0 e−2xx4dx = 3a2, (C.138) so χmolarm = − NAµ0e 2a2 2m = −3.0 · 10−8 m3/kmol (C.139) 62 C. SOLUTIONS, HINTS OR ANSWERS 11.2. The Hund rules affirm that electrons will occupy orbitals in such a way that the ground state is characterised by • The maximum value of the total spin S allowed by the exclusion principle. • The maximum value of the orbital angular momentum L consistent with this value of S • The value of the total angular momentum J is – equal to |L− S| if the shell is less than half full – equal to L+ S when the shell is more than half full – equal to S when the shell is half full, since then the first rule give L = 0 From this one find ml −3 −2 −1 0 1 2 3 S L Eu++ (4f 7) ↑ ↑ ↑ ↑ ↑ ↑ ↑ S = 7/2 L = 0 Yb3+ (4f 13) ↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ S = 1/2 L = 3 Tb3+ (4f 8) ↑ ↑ ↑ ↑ ↑ ↑ ↑↓ S = 3 L = 3 and if one uses the symbolic notation 2S+1LJ for the atomic state one find that the ground states are given by: • Eu++: 4f 75s2p6 8S7/2 • Yb3+: 4f 135s2p6 2F7/2 • Tb3+: 4f 85s2p6 7F6 and calculating the effective number of Bohr magnetons by p = g √ J(J + 1), where g is given by the Lande’ equation g = 1 + J(J + 1) + S(S + 1)− L(L+ 1) 2J(J + 1) , (C.140) give (a) p = 7.94, (b) p = 4.54, and (c) p = 9.72. 11.3. (a) If the diamagnetic sample contain a paramagnetic contamination there will be a temperature dependent contribution to the total magnetic susceptibility. This paramagnetic is determined by the Curie law χpara = C/T . Thus the total magnetic susceptibility must be given by χ = χdia + χpara = χdia + C T . (C.141) If χ is plotted vs. 1/T , it is confirmed that there is a paramagnetic contribution to the total magnetic susceptibility. (b) The Curie constant is given by C = µ0np 2µ2B 3kB , (C.142) 11. MAGNETIC PROPERTIES 63 where p is the effective number of Bohr magnetons. From the Hund rules one find that Mn2+, with the configuration 3d5, has S = 5 2 , L = 0, and J = S = 5 2 , so the Lande’ g-factor is g = 2. From the plot one find that the slope, which is equal to the Curie constant is C = 1.90 · 10−4 K. This give n = 2.1 · 1024 m−3. (c) The intercept in the plot is equal to the diamagnetic susceptibility and is found to be χdia = −9.53 · 10−6. 11.4. The concentration of Fe3+ ions is n = 9.3 · 1027 m−3. Fe3+ has the configuration 3d5 so the Hund rules give the ground state 6S5/2, with L = 0 and J = S = 5 2 . If all the spins are parallel, then Sz = 5/2, which give the maximal magnetisation as Mmax = ngµBSz = 4.32 · 105 A/m. 11.5. (a) In the first turn the force on the weighting-machine is given by F1 = gV ρsalt. In the second turn there will be two additional forces. One, Fm, from the magnetic field as Fm = V χH ∂B ∂z and one due to the Archimedes’ principle, FA = gV ρN , where ρN is the density of liquid nitrogen. Thus the total force, F2, in the second case is F2 = F1 − FA + V χH ∂B ∂z , (C.143) where C is the Curie constant C = µ0np2µ2B 3kB . This give p = 3.85. (b) Theoretically Cr3+ with the configuration 3d3 has S = 3/2. Since Cr is a iron group ion, the effective number of Bohr magnetons is given by p = g √ S(S + 1) = 3.87. 11.6. The spontaneous magnetisation is determined by M = nµB tanh ( µ0µBλM kBT ) . (C.144) Introduce the reduced magnetisation m = M/nµB and the reduced temperature t = T/Tc, where Tc = nµ0µ 2 Bλ/kB, so that m = tanh(m/t) or t = m tanh−1(m) . An expansion of tanh−1(m) as m+m3/3 + ..., when m→ 0 at the Curie point give t = m tanh−1(m) ≈ m m+m3/3 ≈ 1− m 2 3 (C.145) so m2 = 3(1− t) = 3Tc − T Tc ⇒ m ∼ (Tc − T )1/2. (C.146) 66 D. LAB: X – RAY CRYSTALLOGRAPHY parts of a crystal as they are not strongly absorbed. Electrons interact strongly with the electrons of the crystal, and hence they are more suitable for investigations of surface effects and thin films (see CK page 560 figure 5). The physical process that gives rise to the diffraction of the X–rays is the interaction of the X–ray with the electrons of the crystal. The electromagnetic field of the incident X– ray photons accelerate the electrons, and they will oscillate with the field. The oscillating electrons emit electromagnetic radiation with the same frequency as the incident X–rays. The emitted radiation from different atoms can give rise to positive interference. This we see as a reflected beam. The geometrical diffraction condition (see CK page 29), Bragg’s law 2d sin(θ) = nλ (n: 1,2,3,... gives the order of the reflection ) (D.2) is simple to deduce (figure D.1). The relation eq. (D.2) shows that we will only have a reflected beam for certain values of θ, for an ordinary mirror any θ will do. The reflected θ d θ Figure D.1: Reflection according to Braggs law beam consists of reflections from many partially reflecting planes that in certain directions perform a constructive interference. We also see that the wave length must satisfy the condition λ ≤ 2d which implies that ordinary light cannot be used for the study of crystal structures, as the separation d between planes is in the range of a few Å. As can be seen (310) a b (010) (110) (130) (120) Figure D.2: Examples of some reflecting planes with Miller indices. from figure D.2 there are many ways to form a set of parallel planes suitable for reflections. 3. X–RAY DIFFRACTION, SOME GENERAL PRINCIPLES 67 The distance d(hkl) between parallel planes is a function of the Miller – indices (hkl) (see section below) that characterise different sets of parallel planes. In the example in figure 2 we have a simple crystal structure consisting of a simple square lattice with a basis of just one atom. With a more complicated basis we will get reflections from all atoms in the basis. So what will we see? As every atom in the basis is a member of a point lattice with the same properties as the crystal lattice, the atoms in the basis will not change (=add more) the allowed Bragg reflections of the crystal lattice. One can regard the crystal as consisting of a set of point lattices, one for each atom in the basis. All these point lattices have the same structure as the crystal lattice and they are offset to each other according to positions of the atoms in the basis. We see that all the point lattices have the same Bragg condition for reflection as the crystal lattice. As the different point lattices are offset from each other a reflection from one point lattice will be out of phase with the same reflection for another point lattice. It might even be so that the reflection from one point lattice is one half wave length out phase with that from another point lattice. In this case they will interfere destructively and fully extinguish each other and one of the Bragg reflections of the crystal lattice will be missing (if there are more point lattices the situation becomes more complicated). Concluding. For a crystal structure the possible directions are given by the crystal lattice. The atoms of the basis only affect the intensity of the different reflections, and some reflections might even disappear. A consequence of this is that a determination of the diffraction angles θ will only alow us to determine the crystal lattice but will not help us to determine the basis. To do anything on the basis like determine the positions of the atoms, we need to study the intensities of the Bragg reflections as well. So it is quite a difficult task to determine the structure of the unit cell, both theoretically and experimentally, as there are no simple and general rules. 3.2 Miller indices The orientation of a plane in a lattice is specified by giving its Miller indices, which are defined as follows: To determine the indices of the plane P in figure D.3, we find its intercepts with the axes along the basis vectors a, b, and c. Let these intercepts be x, y, normal x x y z P (b) y z (a) (122) plane P γ α β Figure D.3: (a) The (122) plane. Solid circles show the lattice points and z. Usually x is a fractional multiple of a, y a fractional multiple of b, and so forth. 68 D. LAB: X – RAY CRYSTALLOGRAPHY We form the triplet ( x a , y b , z c ) , (D.3) invert it to obtain the triplet ( a x , b y , c z ) , (D.4) and then reduce this set to a similar one having the smallest integers by multiplying by a common factor. This last set is called the Miller indices of the plane and is indicated by (hkl). Let us make an example: Suppose that the intercepts are x = 2a, y = 3 2 b, and z = 1c. We first form the set [ x a , y b , z c ] = (2, 3 2 , 1), (D.5) then invert it 1 2 , 2 3 , 1 and finally multiply by the common denominator, which is 6, to obtain the Miller indices (346) (pronounced as ”three four six”). We note that the Miller Indices are so defined that all equivalent, parallel planes are represented by the same set of indices. Thus the planes whose intercepts are x, y, z; 2x, 2y, 2z;−3x,−3y,−3z; etc., are all represented by the same set of Miller in- dices. Therefore a set of Miller indices specifies not just one plane, but an infinite set of equivalent planes, as indicated in figure D.2. There is a good reason for using such notation, as will be evident from the study of X–ray diffraction from crystal lattices. A diffracted beam is the result of scattering from large numbers of equivalent parallel planes, which act collectively to diffract the beam. The reason for inverting the intercepts in defining the the Miller indices is more subtle, and has to do with the fact that the most concise, and mathematically convenient, method of representing lattice planes is by using the so–called reciprocal lattice. 3.3 Spacing between planes of the same Miller indices. In connection with x–ray diffraction from a crystal, one needs to know the inter–planar distance between planes labelled by the same Miller indices, say (hkl). Let us call this distance dhkl. The actual formula depends on the crystal structure, and we confine our- selves to the case in which the axes are orthogonal. We can calculate this by referring to figure D.3 visualising another plane parallel to the one shown and passing through the origin. The distance between these planes ,dhkl, is simply the length of the normal line drawn from the origin to the plane shown. Suppose that the angles which the normal line makes with the axes are α, β, and γ, and that the intercepts of the plane (hkl) with the axes are x, y, and z. Then it is evident from the figure that dhkl = x cosα = y cos β = z cos γ. (D.6) But there is a relation between the directional cosines cosα, cos β, and cos γ. That is cos2 α + cos2 β + cos2 γ = 1. If we solve for cosα, cos β, and cos γ from the previous equation, substitute into the one immediately above, and solve for dhkl in terms of x, y, and z, we find that dhkl = 1 √ 1 x2 + 1 y2 + 1 z2 (D.7) 5. EQUIPMENT AND SAFETY 71 5 Equipment and safety 5.1 X–ray equipment and safety At the lab site there is a more thorough description of the equipment. You can find this on a lead attached to the X–ray equipment. As we all know X–rays are dangerous. Never try to run the X–ray generator without the lid in it’s proper position ie. closed. In figure D.4 you see an overall picture of the equipment a Tel–X–Ometer made by TELTRON. Start Figure D.4: The Tel–X–Ometer. by having a look at the Tel–X–Ometer. A prominent feature is the plastic lid (radiation cover). It covers the spectrometer table. In this area the experiment is made. In the centre of the spectrometer table the sample under investigations is mounted. From the centre an arm carrying the Geiger–Müller tube figure D.5 reaches out under the radiation Figure D.5: The Geiger–Müller tube. cover. As you move the arm along the perimeter of the spectrometer table you see that 72 D. LAB: X – RAY CRYSTALLOGRAPHY the sample holder in the centre also rotates but at half the angular rotation of the arm. The X–rays are generated by a miniature X–ray tube (see figure D.6), that is the small (size of medium sized apple) glass dome ’sitting’ on the spectrometer table. You open the Figure D.6: The miniature X–ray tube. lid (radiation cover), in the following way: Let the metal part with the international sign for radiation point towards you. The lid can be moved sideways either to the left or to the right the appropriate directions is towards the side of the Geiger–Müller tube. If moved in the opposite direction the lid will jam the rotating arm carrying the Geiger–Müller tube, see figure D.5. The lid seems to be made out of some simple plastic but when closed hardly no X–rays can escape from the Tel–X–Ometer. The plastic it is made of has a high content of Chlorine and absorbs scattered radiation. The beam in itself (the non scattered part) is stopped by the metal part of the lid. The metal part bears the international symbol for radiation. The outside diameter of the three foil represents the maximum diameter of the un collimated X–ray beam. The lid also operates a safety switch to prevent the X–ray generator from operating while the lid is open. The plastic lid cover the upper area of the equipment this part is called the spectrometer table. In figure D.7 you see the two switches to operate the Figure D.7: Operating controls of the Tel–X–Ometer. power. The right one turns the power on but X–rays re not generated yet. This you do 5. EQUIPMENT AND SAFETY 73 by turning the left switch on. In between the two switches you see a timer. You have to set the timer to some appropriate time if it runs out of time the X–ray source will be turned of. This is not only a safety precaution but also saves the X–ray tube as it only has a limited lifetime. Your operations of the switches are confirmed by the two lamps (see figure D.8) on top of the spectrometer table one white to indicate the power on switch Figure D.8: Indication lamps on the Tel–X–Ometer. and one red to indicate that the X–ray tube is operating. 5.2 The Geiger–Müller counter Several years ago one used film to record the X–ray diffraction pattern in a Debye–Scherrer camera. In the experiment you will do we will not use film but we will use a Geiger–Müller counter. It makes operation of the experiment simpler as we do not have to handle film anymore. As you are familiar with the operation of the physical processes in a G–M tube I will not mention them here. I will now describe how to operate the counter figure D.9. Turn it on by pushing the power button. It first initiates itself displaying ’r 1.0’, after a while a zero appears and the apparatus is ready. The lamp Manual is on continuously and the lamp Memory is on intermittently. In principle the apparatus is now ready for manual measurement. But lets be a bit more sophisticated. Now we have to set it for our purpose. 1. Firstly you have to adjust the G–M voltage. Push the ’select’ button several times until the lamp G–M voltage lights up intermittently, now push ’enter’, the G–M voltage lamp is on continuously. In the display window you see the voltage, it should be 405 – 415 Volts, if not adjust with the knob to the far left. When satisfied (or if correct) with adjustments you push ’enter’ again and the G–M voltage lamp lights up intermittently again. 2. Sound Push the ’select’ button several times til the sound lamp lights up intermit- tently, push ’enter’, now the sound lamp is on continuously. Pushing the ’select’ 76 D. LAB: X – RAY CRYSTALLOGRAPHY With an ordinary pocket calculator you can quickly establish the number x to multiply onto all the 1/d2 values to make them into a series of integers. d 2.06 1.26 1.08 0.89 0.82 1/d2 0.2356 0.6299 0.8573 1.2625 1.4872 x=2/0.2356 gives: 2 5.35 7.28 10.72 12.62 x=3/0.2356 gives: 3 8.02 10.92 16.08 18.94 Integers: 3 8 11 16 19 We have now to compare these ’experimental’ integers with the theoretical ones given below: h2 + k2 + l2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sc x x x x x x x x x x x x x x x x x x fcc x x x x x x x x bcc x x x x x x x x x x diamond x x x x x Values of h2 + k2 + l2 for which S 6= 0, for some cubical structures. . By studying our experimental integers and comparing it to the table above we find that the only cubical structure that fits is the diamond structure. From the structure factor for the diamond structure you have calculated earlier you can obtain the following table for (hkl) and h2 + k2 + l2. (hkl) 111 220 311 400 331 422 511 333 h2 + k2 + l2 3 8 11 16 19 24 27 27 To determine the crystal structure of an unknown crystal we have to use the structure factor to determine the possible (hkl) that might occur. Problem 3 Calculate a similar table as the above one for the diamond structure for the sc, fcc and bcc structures. A determination of the NaCl– and CsCl–structures is done by a careful study of the intensities of the lines as well. As the structure factor S usually dominates the intensities we can study it approximately from the intensities. Problem 4 Present the analysis of your results (see experimental task above) in the form of a table representing the values for θ, d, h2 + k2 + l2, (hkl). Also give the result of your analysis regarding the structure and lattice constant a. Lab–report Your lab–report should contain answers to all problems 1–4. 8. CREDITS 77 8 Credits This material has evolved from a similar lab session developed by Günter Grossman, Department of Solid State Physics, Lund University. 78 D. LAB: X – RAY CRYSTALLOGRAPHY 4. ELECTRON IN A PERIODIC POTENTIAL 81 L, M,.. shells. Electrons outside filled shells are loosely bound, whereas electrons in filled inner shells are tightly bound. As we put the atoms together into a crystal, electrons in the closed shells will still be bound to their respective atoms. These electrons are localised. The outer electrons in the non filled shell (shells) are more loosely bound to the free atom and will de localise and can no longer be associated with a specific atom. The de localised electrons will form the conduction/valence bands and the localised electrons are called core–electrons. We will use VN + Veff = U as a known potential for the conduction electrons. U(r) is periodic, U(r+R) = U(r), where R is a lattice vector. In many cases (like metals) U(r) is a weak potential (ie. the electrons are nearly free). This somewhat surprising property can be explained by two effects: 1. the core–electrons screen the nuclear potential for the conducting electrons. 2. a conducting–electron cannot come close to the nucleus due to the Pauli–principle. The volume around the nuclei is occupied by the core–electrons. In the 1970:s it was a difficult task to perform a self consistent electron–band cal- culation. It was at the frontiers of research. Today’s computers together with powerful software like MATLAB have rendered the calculation to a relatively simple one. Today results from band calculations are used as in data for other phenomena in crystals (photo emission, absorption, Auger–spectra, etc.). 4 Electron in a periodic potential We will now turn to the task of actually performing the self consistent calculation. Firstly we will rewrite { − h̄ 2 2m ∇2 + U(r) } ψ = Eψ (E.5) into a dimensionless form. Set the following h̄2 2ma20 = 1 , a0 − Bohr radius = 0.53 Å (E.6) and hence we calculate energies in units of Rydbergs (Ry) 1 Ry = h̄2 2ma20 = 2.17 · 10−18 J = 13.6 eV (E.7) Now we have all energies in Ry and all lengths are measured in units of the Bohr radius. The S–E in these units will be { −∇2 + U(r) } ψ = Eψ (E.8) From U(r+R) = U(r) we have that the potential can be written as U(r) = ∑ G UGe iG·r (E.9) 82 E. LAB: ELECTRONS IN PERIODIC POTENTIALS, BAND STRUCTURE OF CRYSTALS We will also express the wave function ψ(r) in a Fourier–series (ie in free–electron wave functions). The k–vectors are determined from the periodic boundary conditions ψ(r) = ∑ k cke ik·r (E.10) Putting these two series (eq. (E.9) and (E.10)) into the S–E (eq. (E.8)) gives ∑ k k2cke ik·r + ∑ G UGe iG·r∑ k cke ik·r = E ∑ k cke ik·r or ( k2 − E ) ck + ∑ G UGck−G = 0 (E.11) The wave vector can take all values consistent with the periodic boundary conditions, the equation system is more or less infinite. To solve this eigenvalue problem (eq.(E.11)) we choose a wave vector k in the first Brillouin zone (1BZ). The equation system eq.(E.11) relates the corresponding coefficient ck only to coefficients ck−G with a wave vector differing by a reciprocal lattice vector G. A set of coefficients {ck−G,k fixed in BZ, and all possible G} that satisfies eq.(E.11) and all all the other ck′ vannish in the series expansion eq.(E.10) gives us a solution ψk(r). (As this solution depends on the particular choice of k, we have to keep track of the k.) The wave function can now be written as ψk(r) = ∑ G ck−Ge i(k−G)·r = eik·r · uk(r) (E.12) where uk(r) = ∑ G ck−Ge −iG·r (E.13) is a lattice periodic function. From the fact that the potential U(r) and therefore the Hamiltonian is lattice periodic, we have deduced that the wave function can be written (Blochs theorem) as ψk(r) = e ik·ruk(r) , uk(r+R) = uk(r) (E.14) We can no longer characterise a solution with only the wave vector k as for the free electron. For each k we can in principle assign an infinite number of solutions. We will keep track of the different solutions by ordering them according to increasing energy ψnk(r). The Bloch–vector k, the vector that characterises the wave–function, can always be chosen to be within the first Brillouin zone. Suppose k is not within the 1BZ we can always subtract a reciprocal lattice vector G so that k′ = k−G is within the 1BZ. This operation does not change any of the results. This can easily be seen from eq. (E.12) and (E.13) uk(r) = ∑ G′ uG′e iG′·r = e−iG·r ∑ G′ uG′e i(G′+G)·r (E.15) ψk(r) = e ik·re−iG·r · ∑ G′ uG′e i(G′+G)·r = eik ′·r · uk′(r) = ψk′(r) (E.16) As we see it is sufficient to study only the Bloch–vectors in the 1BZ only. The wave– function is normalised if the coefficients satisfy ∑ G | ck−G |2= 1 V (E.17) 4. ELECTRON IN A PERIODIC POTENTIAL 83 where V is the volume of the crystal. In the equation system ( k2 − E ) ck + ∑ G UGck−G = 0 (E.18) the k was a wave–vector that appeared in the series expansion of the wave–function. According to eq. (E.12) we can write these wave–vector like k−G, where k is a Bloch– vector (in the 1BZ) and G is a reciprocal lattice–vector. Hence it is appropriate to rewrite the equation system into ( (k−G)2 − Ek ) ck−G + ∑ G′ UG′−Gck−G′ = 0 (E.19) where k is the chosen wave–vector (Bloch–vector) that characterises our solution. The Bloch–vector k is a ’good’ quantum number and together with n it uniquely specifies the solution ψnk(r). Conclusively the S–E eq. (E.8), with series expansions eq. (E.9) and (E.10), has been rewritten into a homogeneous equation system for the coefficients ck−G. From it we can determine the energy as an eigenvalue Ek and it’s associated eigenvector ck Hck = Ekck (E.20) and the wave–function is given by ψk(r) = ∑ G ck−Ge i(k−G)·r (E.21) Equations (E.8) and (E.19) are fully equivalent, and no approximations have been made so far. Most real systems can only be solved approximately. The equation system (E.19) is of infinite dimension and can only be solved approximately. The accuracy of the solution is determined from the number of terms we have in the expansion eq. (E.21), the number of terms sets the dimension of the matrix for H in eq. (E.20) or (E.19). A analysis based on a perturbation calculation shows that if the potential U(r) is ’smooth’ our approximation improves as we increase the dimension of H. We can also from a perturbation calculation estimate the error we make. We see that eq. (E.19) reduces to the simple free–electron model if we set U = 0. This limiting case is called the empty–lattice approximation. For small U we expect to get good results. The general idea is that if we know the solution to a related problem – the free electron model – and regard the difference between the simple problem and our real problem as a perturbation and we investigate what changes this perturbation invokes. Depending on the strength of the perturbation U we have two cases. 1. The perturbation is small compared to the separation between energy levels – non– degenerate perturbation calculation. In this case the free–electron solution is a good first approximation to the final solution. 2. The perturbation is large compared to the separation between energy levels – degen- erate perturbation calculation. This is the case if several energy levels are degenerate or nearly degenerate, which happens if free–electron bands cross. In this case also a perturbation calculation calls upon solving an equation system. Problem H2c deals with a typical case with two degenerate energy levels. 86 E. LAB: ELECTRONS IN PERIODIC POTENTIALS, BAND STRUCTURE OF CRYSTALS b) Determine the lowest energy for k = 0 with second order perturbation calcu- lation. According to the Appendix a non–degenerate perturbation calculation gives Ej = E 0 j + 〈j | H′ | j〉+ ∑ n 6=j | 〈j | H′ | n〉 |2 E0j − E0n (E.27) where E0j is the energy of the unperturbed j–th energy level, 〈j | H′ | j〉 is the matrix element of the perturbation H′ between the unperturbed wave– functions | j > and | n >. For the case studied here the unperturbed free– electron wave function is 1√ L eikx, normalised over the one dimensional crystal ”volume” L. Is the result consistent with the result received in a)? c) Do the same calculation as in a) for k = π/a, but make an ansatz for ψ consisting of only two plane waves (ck and ck−G), be careful to choose the correct G. What is the band gap ∆1 between band 1 and 2 ? Note this calculation corresponds to a degenerate perturbation calculation. d) Why did we choose a wave function with three terms in a) but only two terms in c)? e) A necessity for an approximation with just a few plane waves to give any decent results is that the amplitude A of the perturbation is small. Assume that the two lowest bands and the the two lowest band gaps ∆1 and ∆2 are to be determined such that the error in energy is less than δE. (An error in energy arises because we have neglected higher free–electron bands in the expansion of the wave–function. If we neglect, as we have done here, the fourth free–electron band we might not expect that energies in the third band are that correct especially not in the upper part.) Estimate for what A the calculation in a) and c) give an acceptable approximation of the energy, in the range from the bottom of the band up to ∆2. Estimate an upper boundary Amax for A that makes the error in energy less than an acceptable δE if A < Amax. (δE you choose yourself, you may take 0.1Ry.) Make a choice of the lattice constant a (typical lattice constants are in the range 5–10 Bohr radius). Draw the three energies determined in a) at k = 0 and the two energies determined in c) at k = π/a as a function of A. Use both values the are under and above Amax. C3 Now we start to use MATLAB, on cosine potential. a) Choose an A << Amax from the figure you made previously. Use three (k = 0) plane waves and calculate the energies numerically with MATLAB. Repeat the same for k = π/a. Do the numerical results corroborate your previous results? b) Choose an A ∼ Amax and increase the number of plane waves ie increase the dimension NDIM of the matrix. What is the lowest value of NDIM where a further increase will not change the results for the energy? This solution corresponds to the exact solution. c) Calculate with MATLAB, for the same A as in P2e, the exact energies and draw these in the same figure as in P2e. When do the ”exact” energies start to deviate from your earlier results in P2e? Was your estimate of Amax you made in P2e correct? 6. WORK 87 C4 Continuation on MATLAB, on rectangular potential. a) Choose a barrier width ω 6= a/2e in the rectangular potential and use the same a as you did in P2 and C3. Calculate, with MATLAB, for some small A the band gaps ∆1 and ∆2. Compare your results between the two potentials: how do the band gaps depend on A, are they linear in A or are they like A2? What may cause the difference between the two cases? C5 Continuation on rectangular potential. a) Choose a decently large value for A, such that only a few coefficients ck are clearly nonzero and rest are small or zero. Study the wave–functions (coeffi- cients of the series expansion) at k = 0 and k = π/a and sketch the electron probability distribution | ψk(x) |2. How can we understand that a band gap occurs at k = π/a (see CK p 177f / HH 4.1)? b) Show for a k (close to the BZ–edge) that the expectation value of the velocity < v > (∼ group velocity vg) is given by 〈v〉 = 1 h̄ ∂E ∂k (E.28) by first: i) Calculate the quantum–mechanical expectation value of the momentum operator p with the numerically determined coefficients 〈ψk | p | ψk〉 = ∑ G ∑ G′ c∗k−Gck−G′ ∫ e−i(k−G)x(−ih̄)i(k −G′)ei(k−G′)x dx = ∑ G h̄(k −G) | ck−G |2 L(E.29) ii) and after this calculate the derivative ∂E ∂k numerically. (Calculate 〈v〉 in m/s.) Note units! MATLAB normalises according to ∑ G | ck−G |2= 1 (E.30) Equation (E.29) assumes normalising according to eq. (E.17). c) Draw a graph of 〈v〉 as a function of k for the two lowest bands. What happens at the BZ–edge? d) Calculate m∗ = { 1 h̄2 ∂2ǫ ∂k2 }−1 (E.31) by an explicit calculation of the second derivative and plot m∗/m as a function of k for the two lowest bands. Now we let A become larger and larger. The dimension NDIM of the matrix should be adjusted in such a way that errors in the energy do not exceed ∼1%. 88 E. LAB: ELECTRONS IN PERIODIC POTENTIALS, BAND STRUCTURE OF CRYSTALS C6) Study the band gaps ∆1 and ∆2 as a function of A. What happens to the bands as A increases? Draw graphs of the bands for some different values of A. What happens to the velocities and effective masses? What happens to the width of the bands? P7) Predict the energies for the three lowest bands in the limit A → ∞ sketch the probability distribution of the electrons in this case. C8) Compare with a numerical calculation. The dependence of the band structure on A in our model problem can be interpreted in the following way. For A→ ∞ the barriers between the atoms become very large, the electrons become localised, and hence the electrons are bound to the atoms. The probability for an electron to tunnel from one atom to another atom is negligible, the effective mass is infinitely large. As the barrier height decreases the localised electron wave–functions begin to overlap, electrons begin to tunnel from atom to atom , the localised wave–functions begin to de localise and the effective mass decreases, electrons move more easily. Finally as A → 0 we regain the free-electron picture. This interpretation corresponds to what happens as we put together atoms into a crystal. The correspondence is not exact as for putting real atoms together we both change the height and the width of the potential, whereas in our model we have only changed the height and kept the width constant. Both case however give qualitatively similar results. 7 Appendix: Time independent perturbation calcu- lation In quantum mechanics one often encounters a problem (with Hamilton operator H) that we cannot solve exactly. In many cases however we have a solution to a simpler (H0), but related problem H0ψ0p = E 0 pψ 0 p (E.32) (We assume for simplicity that the eigenvalue spectrum for the Hamilton operator H0 is discrete, which also applies to H) if the perturbation is small.) We try to split the the Hamilton operator H of the full problem into two parts H = H0 +H′ (E.33) where H′ in some sense is small and can be treated as a weak perturbation of the un- perturbed Hamilton operator H0. Let H′ be proportional to a parameter λ, a coupling constant that represents the strength of the perturbation. Further we assume, that eigen- values E and eigen functions ψ of our problem Hψ = Eψ (E.34) can be written as a power series in λ. Up to second order in λ one finds for the eigenvalue Ej, that corresponds to the unperturbed eigenvalues E 0 j Ej = E 0 j + 〈j | H′ | j〉+ ∑ n 6=j | 〈j | H′ | n〉 |2 E0j − E0n . (E.35) F Lab: Magnetism 1 Purpose The purpose of this lab session is to measure the paramagnetic susceptibility χ of some salts and Aluminium with Gouy’s method. Keywords: Magnetism, para magnetism, dia magnetism, ferro magnetism, Gouy’s method. 2 Literature C. Kittel, Introduction to Solid State Physics, chapter 14 and 15, J.R. Hook and H.E. Hall, Solid State Physics: chapter 7 and 8. 3 Magnetic properties The earliest known application of magnetism is the compass, known to the chinese. The origin of the word magnetism is however from Greece: ’lithos Magnetes’ means stone from Magnesia, a province in Greece. Since the 19th century we are familiar with more important applications of magnetism like: electrical engines, transformers, generators and the hard disc of computer just to mention a few. The magnetic properties of a material are described by the magnetic susceptibility χ, CK page 417 eq (14.1) or HH eq (7.1). With it we can make the following classification of magnetic materials. • For a diamagnetic material we have χ < 0 and for a paramagnetic material we have χ > 0. • For a ferromagnetic material there is no proportionality between M and B and χ >> 0 is not a well defined property. • Other kinds of magnetism are antiferromagnetism and ferrimagnetism. The different kinds of magnetism are described in more detail in chapter 14 in CK and chapters 7 and 8 in HH. 91 92 F. LAB: MAGNETISM The purpose of this laboratory exercise is to measure the magnetic susceptibility of some paramagnetic and diamagnetic substances. The substances are aluminium and the following salts CuCl2 · 2 H2O, CuSO4 · 5 H2O, FeSO4 · 7 H2O, NiSO4 · 6 H2O, MnCl2 · 2 H2O and Ce(NO3)3 · 6 H2O. 4 Diamagnetism The physical mechanism of diamagnetism (CK page 417–420) can be described with Lenz law. When the magnetic flux is changed in an electric circuit, an electric current is induced in such a way as to counteract the applied change in flux. The induced magnetic flux is directed opposite to the applied flux and hence the susceptibility is negative, see CK page 418 or HH eq 7.39. Diamagnetism is present in general but it is a very weak effect and can be observed in cases where the atoms do not have a strong permanent magnetic moment. 5 Paramagnetism The physical mechanism of paramagnetism (CK page 420–429, HH page 198–211) is that the material contains permanent magnetic moments (in molecules, atoms, ions or electrons). In the salts we use here the magnetic moments arrise from the metal ions in the salts, the metal ions have partially filled shells and hence a net magnetic moment. These microscopic magnetic dipoles are usually randomly oriented. Hence there total magnetic moment adds up to zero. The sample will respond to an external magnetic field in such a way that the magnetic dipoles will tend to align along the field. The phonons on the other hand will try to disalign them. So there is a competition between these two effects, the thermal disorientation and the organising external field. A calculation of the paramagnetic susceptibility of atoms or ions in a solid gives χ ∝ 1/T , Curies law, see page 422 (eq 22) in CK or eq 7.18 and 22 in HH. χ = np2µ2B 3kBT µ0 (F.1) Where n is the density of the ion in question. The expression for χ also contains the quantity p = g √ J(J + 1), the effective number of Bohr magnetons. If we know g and J a theoretical value for p can be calculated. By measuring the susceptibility χ we can also estimate an experimental value for p. A comparison of the two p values gives information on angular momentum coupling. Landé’s g–factor can be calculated by (CK page 420 (eq 13) or HH eq 7.10 and eq(13)) g = 1 + J(J + 1) + S(S + 1)− L(L+ 1) 2J(J + 1) (F.2) where L is the total orbital angular momentum and S is the total spin. The quantum numbers can be determined by the use of Hund’s rules, CK page 424 or HH page 201. The problem below illustrates how the connection of L, S and J works and how to apply Hund’s rules. 6. GOUY’S METHOD 93 Problem 1 1.1 How many electrons can at maximum be placed in a d–shell? 1.2 Determine with Hund’s rules the quantum numbers J, L and S for the following electron configurations: 3d9, 3d8, 3d6, 3d5 and 4f ( = 4f1 ) ? 1.3 What theoretical p–values do you get for the electron configurations above ? 1.4 In some cases we have ”quenching” of the orbital angular momentum (CK page 426). This means that L = 0. What values for p do you get if ”quenching” occurs? For non–ferromagnetic metals (like Aluminium) Curie’s law does not apply for the susceptibility (see CK page 433–436 or HH 7.2.4). An experimental study of the temper- ature dependence one finds that χ is nearly independent of temperature. One also finds that the experimental values are a factor of 100 less than an estimate from Curie’s law. This is because for the electron gas in a metal one has to use the Fermi–Dirac distribution. Besides the paramagnetic susceptibility there is also a diamagnetic contribution (Landau–magnetism) due to the free electron orbits. The susceptibility of the electron gas is given by eq 43 page 436 in CK or eq 7.23 and 7.24 in HH. χ = nµ2B kBTF µ0 (F.3) 6 Gouy’s method A determination of the susceptibility is done by measuring the force exerted by an inho- mogeneous magnetic field on our sample. The force is extremely small and we have to use a very sensitive method. In the experiment we make use of Gouy’s method. In it the sample is freely suspended in an inhomogeneous magnetic field according to figure F.1. 2                 N S B z 1B Figure F.1: Gouy’s method to determine the susceptibility