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Tetrahedral angles the angles between the tetrahedral bonds of diamond are the same as the angles between the body diagonals of a cube, as in Fig. 10. Use elementary vector analysis to find the value of theangle.

Tetrahedral angles the angles between the tetrahedral bonds of d

Indices of planes consider the planes with indices (100) and (001); the lattice is fee, and the indices refer to the conventional cubic cell. What are the indices of these places when referred to the primitive aces of Fig.11?

Indices of planes consider the planes with indices (100) and

Hcp structure show that the c/a ratio for an ideal hexagonal close-packed structure is (8/3)1/2 = 1.633. If c/a is significantly larger than this value, the crystal structure may be thought of as composed of planes of closely packed atoms, the planes being loosely stacked.

Inter planar separation consider a plane hkl in a crystal lattice.

(a) Prove that the reciprocal lattice vector G = hb1 + kb3 is perpendicular to this plane.

(b) Prove that the distance between two adjacent parallel planes of the lattice is d(hkl) = 2π/|G|.

(c) Show for a simple cubic lattice that d2 = a2 /(h2 + k2 + l2).

Hexagonal space lattice the primitive translation vectors of the hexagonal space lattice may be taken as

a1 = (31/2 a/2)x + (a/2)y       ;           a2 = – (31/2 a/2)x + (a/2)y ;      a3 = cz

(a) Show that the volume of the primitive cell is (31/2 /2) a2c.

(b) Show that the primitive translations of the reciprocal lattice are 

b1 = (2π/31/2 a)x + (2π/a)y       ;                       b2 = – (2π/31/2 a)x + (2π/a) y ; b3 = (2π/c)z,

so that the lattice is its own reciprocal, but with a rotation of axes.

(c) Describe and sketch the first Brillouin zone of the hexagonal space lattice.

Volume of Brillouin zone show that the volume of the first Brillouin zone is (2π)3/Vc, where Vc is the volume of a crystal primitive cell. Recall the vector identity (e x a) x (a x b) = (e ∙ a x b)a.

Width of diffraction maximum we suppose that in a linear crystal there are identical point scattering centers at every lattice point pm = ma, where m is an integer. By analogy with (20), the total scattered radiation amplitude will be proportional of F = Σ exp [– ima ∙ Δk]. The sum over M lattice points is F = 1 – exp [–iM(a ∙ Δk]/ 1 – exp[–i(a ∙ Δk)], by the use of the series 

(a) The scattered intensity is proportional to |F|2. Show that |F|2 ≡ F∙F = sin2 ½ M (a ∙ ΔK)/sin2 ½ (a ∙ Δk)

(b) We know that a diffraction maximum appears when a ∙ Δk = 2πh, where h is an integer, we change Δk slightly and define ε in a ∙ Δk = 2πh + ε such that ε gives the position of the first zero in sin ½ M(a ∙ Δk). Show that ε = 2π/M, so that the width of the diffraction maximum is proportional to 1/M and can be extremely narrow for macroscopic values of M. The same result holds true for a three-dimensional crystal.

Structure factor of diamond the crystal structure of diamond is described Chapter 1. The basis consists of eight atoms if the cell is taken as the conventional cube. 

(a) Find the structure factor S of this basis.

(b) Find the zeros of S and show that the allowed reflections of the diamond structure satisfy v1 + v2 + v3 = 4n, where all indices are even and n is any integer, or else all indices are odd (Fig. 18). Notice that h, k, l may be written for v1, v2, v3 and this is often done).

Structure factor of diamond the crystal structure of diamond is

Form factor of atomic hydrogen for the hydrogen atom is its ground state, the number density is n(r) = (πa03)–1 exp (– 2r/a0), where a0 is the Bohr radius. Show that the form factor is fG = 16/ (4 + G2a20)2.

Diatomic line, consider a line of atoms ABAB. . . AB, with an A––B bond length of 1/2a, the form factors are fA, fB for atoms A, B, respectively. The incident beam of x-rays is perpendicular to the line of atoms. 

(a) Show that the interference condition is nλ = a cos θ, where θ is the angle between the diffracted beam and the line of atoms.

(b) Show the intensity of the diffracted beam is proportional to |fA – fB|2 for n odd, and to |fA + fB) 2 for n even.

(c) Explain what happens if fA = fB.

Quantum solid in a quantum solid the dominant repulsive energy is the zero-point energy of the atoms. Consider a crude one-dimensional model of crystalline He4 with each he atom confined to a line segment of length L. In the ground state the wave function within each segment is taken as a half wavelength of a free particle. Find the zero point kinetic energy per particle.

Cohesive energy of bcc and fcc neon. Using the Lenard-Jones potential, calculate the ratio of the cohesive energies of neon in the bcc and fcc structures (Ans, 0.958). The lattice sums for the bcc structures are

Cohesive energy of bcc and fcc neon. Using the Lenard-Jones

Solid molecular hydrogen for H2 one finds from measurements on the gas that the Lenard-Jones parameters are ε = 50 x 10-16 erg and σ = 2.96 A. Find the cohesive energy in kJ per mole of H2; do the calculation for an icc structure Treat each H2 molecule as a sphere. The observed value of the cohesive energy is 0.751kJ/mol, much less than we calculated; thus, quantum corrections must be very important.

Possibility of ionic crystals R+R Imagine a crystal that exploits for binding the coulomb attraction of the positive and negative ions of the same atom or molecule R. This is believed to occur with certain organic molecules, but it is not found when R is a single atom. Use the data in Tables 5 and 6 to evaluate the stability of such a form of Na in the NaC1 structure relative to normal metallic sodium. Evaluate the energy at the observed inter atomic distance in metallic sodium, and use 0.78eV as the electron affinity of Na.

Linear ionic crystal Consider a line of 2N ions of alternating charge ± q with a repulsive potential energy A/Rn between nearest neighbors. 

(a) Show that at the equilibrium separation (CGS) U (R0) = – 2NqIn2/R0(1 – 1/n).

(b) Let the crystal be compressed so that R0 → R0(l – δ). Show that the work done in compressing a unit length of the crystal has the leading term 1/2Cδ2, where (CGS) C = (n – 1) q2 In2/R0. To obtain the results in SI, replace q2 by q2/4πε0. Note: We should not expect to obtain this result from the expression for U (R0), but we must use the complete expression for U(R).

Cubic ZnS structure using λ and p from Table 7 and the Madelung constants given in the text, calculate the cohesive energy of KCI in the cubic ZnS structure described in Chapter 1. Compare with the value calculated for KCl in the NaCl structure.

Divalent ionic crystals Barium oxide has the NaC1 structure. Estimate the cohesive energies per molecule of the hypothetical crystals BaO and Ba++ O– – referred to separated neutral atoms. The observed nearest-neighbor inter nuclear distance is R0 = 2.76 A; the first and second ionization potentials of Ba are 5.19 and 9.96eV; and the electron affinities of the first and second electrons added to the neutral oxygen atom are 1.5 and —9.0eV. The first electron affinity of the neutral oxygen atom is the energy released in the reaction O + e → O.The second electron affinity is the energy released in the reaction O + e → O– –. Which valence state do you predict will occur? Assume R0 is the same for both forms, and neglect the repulsive energy.

Young’s modulus and Poisson’s ratio A cubic crystal is subject to tension in the [100] direction. Find expressions in terms of the elastic stiff nesses for Young’s modulus and Poisson’s ratio as defined in Fig. 21.

Longitudinal wave velocity, show that the velocity of a longitudinal wave in the [111] direction of a cubic crystal is given by vs = [1/3(C11 + 2C12 + 4C44)/p]1/2.

Transverse wave velocity show that the velocity of transverse waves in the [111] direction of a cubic crystal is given by vs = [1/3(C11 – C12 + C44/p]1/2.

Effective shear constant show that the shear constant 1/2 (C11 – C12) in a cubic crystal is defined by setting exx = – eyy = ½e and all other strains equal to zero, as in Fig. 22.

Effective shear constant show that the shear

Determinantal approach it is known that an R-dimensional square matrix with all elements equal to unity has roots R and 0, with the R occurring once and the zero occurring R – 1 times. If all elements have the value p, then the roots are Rp arid 0. 

(a) Show that if the diagonal elements are q and all other elements are p. then there is one root equal to (R – 1)p + q and R – 1 roots equal to q – p. 

(b) Show from the elastic equation (57) for a wave in the [111] direction of a cubic crystal that the determinantal equation which gives w2 as a function of K is where q = 1/3K2 (C11 + 2C44) and p = 1/3K2(C12 + C44). This expresses the condition that three linear homogeneous algebraic equations for the three displacement components u, v, w have a solution. Use the result of part (a) to find the three roots of w2; check with the results given for Problems 9 and 10.

General propagation direction 

(a) By substitution in (57) Ibid the determinantal equation which expresses the condition that the displacement R(r) = [u0x + v0y + w0z] exp [i(K∙ r – wt)] be a solution of the elastic wave equations in a cubic crystal. (b) The sum of the roots of a determinantal equation is equal to the sum of the diagonal elements aii Show from part (a) that the sum of the squares of the three elastic wave velocities in any direction in a cubic crystal is equal to (C11 + 2C44)/p. Recall that v2s, = w2/K2.

Stability criteria the criterion that a cubic crystal with one atom in the primitive cell he stable against small homogeneous deformations is that the energy density (43) be positive for all combinations of strain components. What restrictions are thereby imposed on the elastic stiffness constants? (In mathematical language the problem is to find the conditions that a real symmetric quadratic form should be positive definite.

Monatomic linear lattice consider a longitudinal wave us = u cos (wt – sKa) which propagates in a monatomic linear lattice of atoms of mass M, spacing a, and nearest-neighbor interaction C.

(a) Show that the total energy of the wave is where s runs over all atoms.

(b) By substitution of us, in this expression, show that the time-average total energy per atom is where in the last step we have used the dispersion relation (9) for this problem.

Monatomic linear lattice consider a longitudinal wave us = u

Continuum ware equation show that for long wavelengths the equation of motion (2) reduces to the continuum elastic wave equation where v is the velocity of sound

Continuum ware equation show that for long wavelengths

Basis of two unlike atoms for the problem treated by (18) to (26), find the amplitude ratios u/v for the two branches at K max = π/a. Show that at this value of K the two lattices act as if decoupled: one lattice remains at rest while the other lattice moves.

Kohn anomaly we suppose that the inter planar force constant C between planes sands + p is of the form Cp = A sin pk0a/pa, where A and k( are constants and p runs over all integers. Such a form is expected in metals. Use this and Eq. (16a) to find an expression for w2 and also for ∂w2/∂K. Prove that ∂w2/∂K is infinite when K = k0. Thus a plot of w2 versus K or of w versus K has a vertical tangent at k0: there is a kink at k0 in the phonon dispersion relation w(K

Diatomic chain Consider the normal modes of a linear chain in which the force constants between nearest-neighbor atoms are alternately C and 10C. Let the masses he equal, and let the nearest-neighbor separation he a/2. Find w(K) at K = 0 and K = π/a. Sketch in the dispersion relation by eye. This problem simulates a crystal of diatomic molecules such as H2.

Atomic vibrations in a metal consider point ions of mass M and charge e immersed in a uniform sea of conduction electrons. The ions are imagined to be in stable equilibrium when at regular lattice points. If one ion is displaced a small distance r from its equilibrium position, the restoring force is largely due to the electric charge within the sphere of radius r centered at the equilibrium position. Take the number density of ions (or of conduction electrons) as 3/4πR3, which defines R.

(a) Show that the frequency of a single ion set into oscillation is w = (e2/MR3)1/2.

(b) Estimate the value of this frequency for sodium, roughly.

(c) From (a), (b), and some common sense, estimate the order of magnitude of the velocity of sound in the metal.

Soft phonon mode consider a line of ions of equal mass but alternating in charge, with ep = e(–1)p as the charge on the pth ion the inter atomic potential is the sum of two contributions (1) a short-range interaction of force constant C1R = γ that acts between nearest neighbors only, and (2) a coulomb interaction between all ions. 

(a) Show that the contribution of the coulomb interaction to the atomic force constants is CpC = 2(–1)p e2/pa3, where a is the equilibrium nearest-neighbor distance. 

(b) From (16a) show that the dispersion relation may be written as where w20 ≡ 4γ/M and σ = e2/γa3.

(c) Show that w2 is negative (unstable mode) at the zone boundary Ka = π if σ > 0.475 or 4/7ζ (3), where ζ is a Riemann zeta function. Show further that the speed of sound at small Ka is imaginary if σ > (2 In 2)–1 = 0.721. Thus w2 goes to zero and the lattice is unstable for some value of Ka in the interval (0, π) if 0,475 < σ <0.721. Notice that the phonon spectrum is not that of a diatomic lattice because the interaction of any ion with its neighbors is the same as that of any other ion.

Soft phonon mode consider a line of ions of equal mass but

Singularity in density of states 

(a) From the dispersion relation derived for a monatomic linear lattice of N atoms with nearest-neighbor interactions, show the density of modes is D(w) = 2N/π 1 /(w2m – w2)1/2, where wm is the maximum frequency. 

(b) Suppose that an optical phonon branch has the form w(K) = w0 – AK2, near K = 0 in three dimensions. Show that D(w) = (L/2π)3 (2π/A3/2) (w0 – w)1/2 for w < w0 and D(w) = 0 for w > w0. Here the density of modes is discontinuous.

Rms thermal dilation of crystal cell
(a) Estimate for 300 K the root mean square thermal dilation ΔV/V for a primitive cell of sodium. Take the bulk modulus as 7 x 1010 erg cm-3. Note that the Debye temperature 158 K is less than 300 K, so that the thermal energy is of the order of kBT.
(b) Use this result to estimate the root mean square thermal fluctuation Δa/a of the lattice parameter.

Zero point lattice displacement and strain 

(a) In the Debye approximation, show that the mean square displacement of an atom at absolute zero is R) = 3hw2D/8π2 pv3, where v is the velocity of sound. Start from the result (4.29) summed over the independent lattice modes: <R2> = (h/2pV) Σw–1. We have included a factor of ½ to go from mean square amplitude to mean square displacement. 

(b) Show that Σw–1 and (R2) diverge for a one-dimensional lattice, but that the mean square strain is finite. Consider <(∂R/∂x)2> = ½ ΣK2u20 as the mean square strain, and show that it is equal to hw2DL/4MNv3 for a line of N atoms each of mass M, counting longitudinal modes only. The divergence of R2 is not significant for any physical measurement.

Heat capacity of layer lattice

(a) Consider a dielectric crystal made up of layers of atoms, with rigid coupling between layers so that the motion of the atoms is restricted to the plane of the layer, Show that the phonon heat capacity in the Debye approximation in the low temperature limit is proportional to T2.

(b) Suppose instead, as in many layer structures, that adjacent layers arc very weakly bound to each other. What form would you expect the phonon heat capacity to approach at extremely low temperatures?

Gruneisen constant 

(a) Show that the free energy of a phonon mode of frequency w is kBT in [2sinh (hw/2kBT)]. It is necessary to retain the zero-point energy ½hw to obtain this result. 

(b) If Δ is the fractional volume change, then the free energy of the crystal may be written as F(Δ, T) = ½ BΔ2 + kBT Σ In [2sinh (hwK/2kBT)] where B is the hulk modulus. Assume that the volume dependence of wK is δw/w = —γΔ, where γ is known as the Gruneisen constant. If γ is taken as independent of the mode K, show that F is a minimum with respect to Δ when B Δ = γΣ1/2hw coth (hw/2kBT), and show that this may be written in terms of the thermal energy density as Δ = γU(T)/B.

(c) Show that on the Debye model γ = — ∂ In θ/∂ In V. Note: Many approximations are involved in this theory: the result (a) is valid only if w is independent of temperature; γ may he quite different for different modes.

Kinetic energy of electron gas show that the kinetic energy of a three-dimensional gas of N free electrons at 0 K is U0 = 3/5NεF

Pressure and bulk modulus of an electron gas 

(a) Derive a relation connecting the pressure and volume of an electron gas at 0 K. The result may be written as p = 2/3(U0/V). 

(b) Show that the bulk modulus B = – V (∂p/∂V) of an electron gas at 0 K is B = 5p/3 = 10U0/9V. 

(c) Estimate for potassium, using Table 1, the value of the electron gas contribution to B.

Chemical potential in two dimensions, show that the chemical potential of a Fermi gas in two dimensions is given by: μ(T) = kBT in [exp(πth2/mkBT) – 1] for n electrons per unit area. The density of orbital’s of a free electron gas in two dimensions is independent of energy: D(ε) = m/πh2 per unit area of specimen.

Fermi gases in oil astrophysics. 

(a) Given M = 2 x 1033 g for the mass of the Sun, estimate the number of electrons in the Sun. In a white dwarf star this number of electrons may be ionized and contained in a sphere of radius 2 x 109 cm; find the Fermi energy of the electrons in electron volts. 

(b) The energy of an electron in the relativistic limit ε >> mc2 is related to the wave vector as ε ≡ pc = hkc. Show that the Fermi energy in this limit is εF ≈ hc (N/V)1/3, roughly 

(c) If the above number of electrons were contained within a pulsar of radius 10 km, show that the Fermi energy would be ≈ 108eV. This value explains why pulsars are believed to he composed largely of neutrons rather than of protons and electrons, for the energy release in the reaction n → p + e is only 0.8 x 106eV, which is not large enough to enable many electrons to form a Fermi sea. The neutron decay proceeds only until the electron concentration builds up enough to create a Fermi level of 0.8 x l06eV, at which point the neutron, proton, and electron concentrations are in equilibrium.

Liquid He3 the atom He3 has spin ½ and is a fermions, the density of liquid He3 is 0.081 g cm–3 near absolute zero. Calculate the Fermi energy εF and the Fermi temperature TF.

Frequency dependence of the electrical conductivity, use the equation m(dv/dt + v/τ) = – εE for the electron drift velocity v to chow that the conductivity at frequency w is where σ(0) = ne2τ/m.

Frequency dependence of the electrical conductivity, use the

Dynamic magneto conductivity tensor for free electrons a metal with a concentration n of free electrons of charge – e is in a static magnetic field B. The electric current density in the xy plane is related to the electric field by jx = σxxEx + σxyEy;                  jy = σyzET + σyyEy.

Assume that the frequency w >> w0, and where w >> 1/τ, where wc = eB/mc and τ is the collision time. 

(a) Solve the drift velocity equation (51) to find the components of the magneto conductivity tensor:

σxx = σyy + iw2p/4πw;               σxy = – σyx = wcw2p/4πw2, where w2p = 4πne2/m. 

(b) Note from a Maxwell equation that the (dielectric function tensor of the medium is related to the conductivity tensor as ε = 1 + i(4π/w)σ. Consider an electromagnetic wave with wave vector k = kz. Show that the dispersions relation for this wave in the medium is c2k2 = w2 – w2p ± wc w2p/w 

At a given frequency there are two modes of propagation with different wave vectors and different velocities. The two modes correspond to circularly polarized waves. Because a linearly polarized wave can he decomposed into two circularly polarized waves, it follows that the plane of polarization of a linearly polarized wave will he rotated by the magnetic field.

Cohesive energy of free electron Fermi gas, we define the dimensionless length r1, as r0/aH, where r0 is the radius of a sphere that contains one electron, and aH is the bohr radius h2/e2m. 

(a) Show that the average kinetic energy per electron in a free electron Fermi gas at 0 K is 2.21/r2s, where the energy is expressed in rydbergs, with 1 Ry = me4/2h2

(b) Show that the coulomb energy of a point positive charge e interacting with the uniform electron distribution of one electron in the volume of radius r0 is – 3e2/2r0, or – 3/rs in rydbergs. 

(c) Show that the coulomb self-energy of the electron distribution in the sphere is 3e2/5r0, or 6/5rs in rydbergs. 

(d) The sum of (b) and (c) gives – 1.80/rfor the total coulomb energy per electron. Show that the equilibrium value of r is 2.45. Will such a metal be stable with respect to separated El atoms?

Static magneto conductivity tensor for the drift velocity theory of (51), show that the static current density can be written in matrix form as in the high magnetic field limit of wcτ >> 1, show that σyx = nec/B = σxy in this limit σxx = 0, to order 1/wcτ. The quantity σyx is called the Hall conductivity

Static magneto conductivity tensor for the drift velocity

Maximum surface resistance considers a square sheet of side L, thickness d, and electrical resistivity p. The resistance measured between opposite edges of the sheet is called the surface resistance: Rsq = pL/Ld = p/d, which is independent of the area L2 of the sheet. (Rsq is called the resistance per square and is expressed in ohms per square, because p/d has the dimensions of ohms.) If we express p by (44), then Rsq = m/nde2τ. Suppose now that the minimum value of the collision time is determined by scattering from the surfaces of the sheet, so that τ ≈ d/vF, where vF is the Fermi velocity. Thus the maximum surface resistivity is Rsq ≈ mvp/nd2e2. Show for a monatomic metal sheet one atom in thickness that Rsq ≈ h/e2 = 4.1kΩ

Square lattice, free electron energies 

(a) Show for a simple square lattice (two dimensions) that the kinetic energy of a free electron at a corner of the first zone is higher than that of an electron at midpoint of a side face of the zone by a factor of 2. 

(b) What is the corresponding factor for a simple cubic lattice (three dimensions)? 

(c) What bearing might the result of (b) have on the conductivity of divalent metals?

Free electron energies in reduced zone. Consider the free electron energy bands of an fcc crystal lattice in the approximation of an empty lattice, hut in the reduced zone scheme in which all k’ s are transformed to lie in (he first Brillouin zone. Plot roughly in the [111] direction the energies of all bands up to six times the lowest band energy at the zone boundary at k = (2π/a) (½, ½, ½). Let this he the unit of energy. This problem shows why band edges need not necessarily be at the zone center. Several of the degeneracies (band crossings) will be removed when account is taken of the crystal potential.

Kronig-Penney model

(a) For the delta-function potential and with P << 1, find at k = 0 the energy of the Lowest energy band.

(b) Fur the same problem find the band gap at k = π/a.

Potential energy in the diamond structure

(a) Show that for the diamond structure the Fourier component UG of the crystal potential seen by an electron is equal to zero for G = 2A, where A is a basis vector in the reciprocal lattice referred to the conventional cubic cell.

(b) Show that in the usual first-order approximation to the solutions of the wave equation in a periodic lattice the energy gap vanishes at the zone boundary plane normal to the end of time vector A.

Complex wave vectors in the energy gap find an expression for the imaginary part of the wave vector in the energy gap at the boundary of the first Brillouin zone, in the approximation that led to Eq. (46). Give the result for the Im (k) at the center of the energy gap. The result for small Im (k) is (h2/2m)[Im(k)]2 ≈ 2mU2/h2G2. The form as plotted in Fig. 12 is of importance in the theory of Zener tunneling from one band to another in the presence of a strong electric field.

Complex wave vectors in the energy gap find an expression for

Square lattice consider a square lattice in two dimensions with the crystal potential U(x, y) = – 4U cos (2πx/a) cos (2πy/a). Apply the central equation to find approximately the energy gap at the corner point (π/a, π/a) of the Brillouin zone. It will suffice to solve a 2 x 2 determinantal equation.

Impurity orbits indium antimonidc has Eg = 0.23eV; dielectric constant ε = 18; electron effective mass me = 0.015m. Calculate 

(a) The donor ionization energy; 

(b) The radius of the ground state orbit.

(c) At what minimum donor concentration will appreciable overlap effects between the orbits of adjacent impurity atoms occur? This overlap tends to produce an impurity band–a band of energy levels which permit conductivity presumably by a hopping mechanism in which electrons move from one impurity site to a neighboring ionized impurity site.

Ionization of donors in a particular semi conductor there are 1013 donors/cm3 with an ionization energy Ed of 1meV and an effective mass 0.01m.

(a) Estimate the concentration of conduction electrons at 4k.

(b) What is the value of the Hall coefficient? Assume no acceptor atoms are present and that Eg >> kBT?

Hall effect with two carrier types assuming concentration n, p; relaxation times τe, τh; and masses me, mh, show that the Hall coefficient in the drift velocity approximations is (CGS) RH = 1/e∙ p – nb2/(p + nb)2, where b = μch is the mobility ratio. In the derivation neglect terms of order B2 in SI we drop the c.

Cyclotron resonance for a spheroidal energy surface considers the energy surface where mt is the transverse mass parameter and m1 is the longitudinal mass parameter. A surface on which ε(k) is constant will be a spheroid. Use the equation of motion (6), with v = h–1 Δkε, to show that wc = eB/ (m1mt)1/2 c when the static magnetic field B lies in the xy plane. This result agrees with (34) when θ = π/2, the result is in CGS; to obtain SI, omit the c.

Cyclotron resonance for a spheroidal energy surface considers

Magneto resistance with two carrier types, Problem 6.9 shows that in the drift velocity approximation the motion of charge carriers in electric and magnetic fields does not lead to transverse magneto resistance. The result is different with two carrier types. Consider a conductor with a concentration n of electrons of effective mass me and relaxation time τe; and a concentration p of holes of effective mass mh and relaxation time τA. Treat the limit of very strong magnetic fields wcτ>> 1.

(a) Show in this limit that σxy = (n – p)ec/B.

(b) Show that the Hall field is given by, with Q ≡ wcτ, which vanishes if n = p. 

(c) Show that the effective conductivity on the x direction is if n = p, σ ∞ B–2. If n ≠ p, σ saturates in strong fields; that is, it approaches a limit independent of B as B → ∞

Magneto resistance with two carrier types

Brillouin zones of rectangular lattice make a plot of the first two Brillouin zones of a primitive rectangular two-dimensional lattice with axes a, b = 3a.

Brillouin zone, rectangular lattice a two-dimensional metal has one atom of valency one in a simple rectangular primitive cell a = 2 A; b = 4 A. 

(a) Draw the first Brillouin zone give its dimensions, in cm-1

(b) Calculate the radius of the free electron Fermi sphere, in cm–1

(c) Draw this sphere to scale on a drawing of the first Brillouin zone, make another sketch to show the first few periods of the free electron band in the periodic zone scheme, for both the first and second energy bands. Assume there is a small energy gap at the zone boundary.

Hexagonal emit-packed structure Consider first Brillouin zone of a crystal with a simple hexagonal lattice in three dimensions with lattice constants a and c. Let Gc denote the shortest reciprocal lattice vector parallel to the c axis of the crystal lattice. 

(a) Show that for a hexagonal-close-packed crystal structure the Fourier component U(Gc) of the crystal potential U(r) is zero. 

(b) Is U (2Gc) also zero? 

(c) Why is it possible in principle to obtain an insulator made up of divalent atoms at tile lattice points of a simple hexagonal lattice? 

(d) Why is it nut possible to obtain an insulator made up of monovalent atoms in a hexagonal-close-packed structure?

Brillouin zones of two-dimensional divalent metal. A two-dimensional metal in the form of a square lattice has two conduction electrons per atom. In the almost free electron approximation, sketch carefully the electron and hole energy surfaces. For the electrons choose a zone scheme such that the Fermi surface is shown as closed.

Open orbits an open orbit in monovalent tetragonal metal connects opposite faces of the boundary of a Brillouin zone. The faces are separated by G = 2 x 10cm1. A magnetic field B = 103 gauss = 10-1 tesla is normal to the plane of the open orbit. 

(a) What is the order of magnitude of the period of the motion in k space? Take c ≈ 108 cm/sec. 

(b) Describe in real space the motion of an electron on this orbit in the presence of flue magnetic field.

Cohesive energy for a square well potential 

(a) Find an expression for the binding energy of an electron in one dimension in a single square well of depth U0 and width a. (This Is the standard first problem m elementary quantum mechanic) Assume that the solution is symmetric about the midpoint of the well.

(b) Find a numerical result for the binding energy in terms of U0 for the special case |U0| =2h2 /ma2 and compare with the appropriate limit of Fig. 20. In this limit of widely separated wells the band width goes to zero, so the energy for k = 0 is the same as the energy for other k in the lowest energy band. Other bands are formed from lime excited states of the well, in this limit

De Haas-van Alphen period of potassium

(a) Calculate the period Δ(1/B) expected for potassium on the free electron model.

(b) What is the area in real space of the external orbit, for B 10kG = 1 T? The same period applies to oscillations in the electrical resistivity, known as the Shubnikow-de Haas effect.

Band edge structure on k • p perturbation theory consider a non-degenerate orbital ψnk at k = 0 in time band n of a cubic crystal. Use second-order perturbation theory to find the result where the sum is over all other orbital’s ψjk at k = 0. The effective mass at this point is this problem somewhat difficult. The mass at the conduction band edge in a narrow gap semi conductor is often dominated by the effect of the valence band edge, whence where the sum is over the valence bands; Eg is the energy gap. For given matrix elements, small gaps lead to small masses.

Wannier function the Wannier functions of a hand are defined In terms of the Bloch functions of the same baud by where rn is a lattice point.

(a) Prove that Wannier functions about different lattice points n, m are orthogonal:

f dV w*(r – r) w (r - rm) = 0, n ≠ m. (43). This orthogonality property makes the functions often of greater use than atomic orbital’s centered on different lattice sites, because the Tatter are not generally orthogonal. 

(b) The Wannier functions are peaked around the lattice sites. Show that for ψk = N-1/2 elkx u0(x) the Wannier function is for N atoms on a line of lattice constant a.

Wannier function the Wannier functions of a hand are defined In terms of the Bloch

Open orbit and magneto resistance. We considered the transverse magneto resistance of free electrons in Problem 6.9 and of electrons and holes in Problem 8.5. In some crystals the magneto resistance saturates except in special crystal orientations. An open orbit carries current only in a single direction in the plane normal to the magnetic field; such carriers are not deflected by the field. In the arrangement of Fig. 6.14, let the open orbits be parallel to k in real space these orbits carry current parallel to the y as. Let σyy = sσ0 be the conductivity of the open orbits; this defines the constant s. The magneto conductivity tensor in the high field limit w0τ >> 1 is with Q = wcτ. 

(a) Show that the Hall field is Ey = – Ex/sQ. 

(b) Show that the effective resistivity in the x direction is p (Q20) (s/s + 1), so that the resistivity does not saturate, hut increases as B2.

Open orbit and magneto resistance. We considered the transverse

Magnetic field penetration in a plate the penetration equation may be written as λ2Δ2B = B, where λ is the penetration depth. 

(a) Show that B(x) inside a super conducting plate perpendicular to the x axis and of thickness δ is given by where Ba, is the field outside the plate and parallel to it; here x = 0 is at the center of the plate. 

(b) The effective magnetization M(x) in the plate is defined by B(x) – Ba, = 4πM(x). Show that, in CGS, 4πM(x) = – Ba, (1/8λ2) (δ2 – 4x2), for δ << λ A. In SI we replace the 4π by μ0.

Magnetic field penetration in a plate the penetration equation

Critical field of thin films

(a) Using the result of Problem lb, show that the free energy density at T = 0 K within a superconducting film of thickness δ in an external magnetic field Ba is given by, for δ << A,

(CGS) FS (x, Ba) = US (0) + (δ2 – 4x2) B2a/64πλ2. In SI the factor π is replaced by ¼μ0. We neglect a kinetic energy contribution to the problem. 

(b) Show that the magnetic contribution to FS when averaged over the thickness of the film is B2(δ /λ)2/96π.

(c) S how that the critical field of the thin film is proportional to (λ/δ) Hc, where Hc, is the bulk critical field, if we consider only the magnetic contribution to US.

Two-fluid model of a superconductor on the two-fluid model of a super conductor we assume that at temperatures 0 < T < T0, the current density may be written as the sum of the contributions of normal and superconducting electrons: j = jN, + jS, where jN, = σ0E and js is given by the London equation. Here σ0 is an ordinary normal conductivity, decreased by the reduction in the number of normal electrons at temperature T as compared to the normal state. Neglect inertial effects on both jN, and jS

(a) Show from the Maxwell equations that the dispersion relation connecting wave vector k and frequency w for electromagnetic waves in the superconductor is (CGS) k2c2 = 4πσ02wt – c2λL–2 + w2; or (SI) k2c2 = (σ0/ε0)wt c2λL–2 + wwhere A; is given by (148) with n replaced by ns. Recall that curl B = – Δ2B.

(b) If T is the relaxation time of the normal electrons and nN is their concentration, show by use of the expression σ0 = nNe2τ/m that at frequencies w << 1/τ the dispersion relation does not involve the normal electrons in an important way, so that the motion of the electrons is described by the London equation alone. The super current short-circuits the normal electrons. The London equation itself only holds true if hw is small in comparison with the energy gap. Note: The frequencies of interest are such that w << wp, where wp is the plasma frequency.

Structure of a vortex 

(a) Find a solution to the London equation that has cylindrical symmetry and applies outside a line core. In cylindrical polar coordinates, we want a solution of B – λ2Δ2B = 0 this problem is somewhat difficult that is singular at the origin and for which the total flux is the flux quantum: The equation is in fact valid only outside the mornla1 core of radius ζ. 

(b) Show that at the solution has the limits

Structure of a vortex  (a) Find a solution to the London

London penetration depth 

(a) Take the time derivative of the London equation (10) to show that ∂j/∂t = (c2/4πλ2L) E. 

(b) If mdv/dt = qE, as for free carriers of charge q and mass m, show that λ2L = mc2 / 4πnq2.

Diffraction effect of Josephson junction, consider a junction of rectangular cross-section with a magnetic field B applied in the plane of the junction, normal to an edge of width w. Let the thickness of the junction be T. Assume for convenience that the phase difference of the two superconductors is π/2 when B = 0. Show that the dc current in the presence of the magnetic field is

Meissner effect in sphere considers a sphere of a type 1 super conductor with critical field Hc

(a) Show that in the Meissner regime the effective magnetization M within the sphere is given by –8πM/3 = Ba, the uniform applied magnetic field. 

(b) Show that the magnetic field at the surface of the sphere in the equatorial plane is 3Bn/2. (It follows that the applied field at which the Meissner affect starts to break down is 2Hc/3.) Reminder: The demagnetization field of a uniformly magnetized sphere is – 4πM/3.

Diamagnetic susceptibility of atomic hydrogen the wave function of the hydrogen atom in its ground state (Is) is ψ = (πa30)-1/2 exp (– r/a0), where a0, = h2/me2 = 0.529 x 10-8 cm. The charge density is p(x, y, z) = – e|ψ|2, according to the statistical interpretation of the wave function. Show that for this state (r2) = 3a02, and calculate the molar diamagnetic susceptibility of atomic hydrogen (–2.36 x 10-6 cm3/mole).

Hund rules apply the Hund rules to find the ground state (the basic level in the notation of Table 1) of 

(a) Eu++, in the configuration 4f7 5sp6

(b) Yb3+

(c) Tb3+

The results for (b) and (c) are in Table I, but you should give the separate steps in applying the rules.

Triplet excited states some organic molecules have a triplet (S = 1) excited state at an energy kBΔ above a singlet (S = 0) ground state. 

(a) Find an expression for the magnetic moment (μ) in a field B. 

(b) Show that the susceptibility for T >> Δ is approximately independent of A. 

(c) With the help of a diagram of energy levels versus field and a rough sketch of entropy versus field, explain how this system might be cooled by isentropic magnetization (not demagnetization).

Heat capacity from internal degrees of freedom

(a) Consider a two-level system with an energy splitting kBΔ between upper and lower states; the splitting may arise from a magnetic field or in other ways. Show that the heat capacity per system is the function is plotted in Fig. 11. Peaks of this type in the heat capacity are often known as Schottky anomalies. The maximum heat capacity is quite high, but for T << Δ and for T >> Δ the heat capacity is low. 

(b) Show that for T >> Δ we have C ≡ k(Δ/2T)2 + . . . . The hyperfine interaction between nuclear and electronic magnetic moments in paramagnetic salts (and in systems hating electron spin order) causes splittings with Δ = 1 to 100mK. These splittings are often detected experimentally by the presence of a ten11 in 1/P in the heat capacity in the region T >> Δ. Nuclear electric quadrupole interactions wit11 crystal fields also cause splittings, as in Fig. 12.

Heat capacity from internal degrees of freedom

Pauli spin susceptibility the spin susceptibility of a conduction electron gas at alsolute zero may be approached by another method. Let N+ = ½N (1 + ζ); N = ½ N (1 - ζ) be the concentrations of spin-up and spin-down electrons. 

(a) Show that in a magnetic field B the total energy of the spin-up band in a free electron gas is E1 = E0 (1 + ζ)5/3 – 1/2NμB (1 + ζ) where E0, = 1/10NεF, in terms of the Fermi energy εF in zero magnetic field. Find a similar expression for E

(b) Minimize E total = E+ + E with respect to ζ and solve for the equilibrium value of ζ in the approximation ζ << 1. Go on to show that the magnetization is M = 3Nμ2B/2εp, in agreement with Eq. (45).

Conduction electron ferromagnetism we approximate the effect of exchange interactions among the conduction electrons if we assume that electrons with parallel spins interact with each other with energy – V, and V is positive, while electrons with anti parallel spins do not interact with each other.

(a) Show with the help of Problem 5 that the total energy of the spin-up hand is E+ = E0 (1 + ζ) 5/3 – 1/8VN(1 + ζ)2 – ½NμB (1 + ζ); (find a similar expression for E-. 

(b) Minimize the total energy and solve for ζ in the limit ζ << 1. Show that the magnetization is so that the exchange interaction enhances the susceptibility. 

(c) Show that with B = 0 the total energy is unstable at ζ = 0 when V > 4εF/3N. If this is satisfied, a ferromagnetic state (ζ ≠ 0) will have a lower energy than the paramagnetic static. Because of the assumption ζ <<1, this is a sufficient condition for ferromagnetism, but it may not be a necessary condition. It is known as the Stoner condition.

Two-level system the result of Problerrl4 is often seen in another form. 

(a) If the two energy levels are at Δ and – Δ, show that the energy and heat capacity are U = - Δ tanh (Δ/kBT); C = kB (Δ/kBT)2 sech2 (Δ/kBT). 

(b) If the system has a random composition such that all values of Δ are equally likely up to some limit Δ0, show that the heat capacity is linearly proportional to the temperature, provided kBT << Δ0. This result was applied to the heat capacity of dilute magnetic alloys by W. Marshall, Phys. Rev. 118, 1519 (1960). It is also used in the theory of glasses.

Para-magnetism of S = 1 system. 

(a) Find the magnetization as a function of magnetic field and temperature for a system of spins with S = 1, moment μ and concentration n. 

(b) Show that in the limit μB << kT the result is M ≡ (2nμ2/3kT)B

Configurational heat capacity derive an expression in terms of P(T) for the heat capacity associated with order/disorder effects in an AB alloy. [The entropy (8) is called the Configurational entropy or entropy of mixing.]

Magnon dispersion relation derive the magnon dispersion relation (24) for a spin S on a simple cubic lattice, z = 6 is replaced by where the central atom is at p and the six nearest neighbors arc connected to it by six vectors δ. Look for solutions of the equations for dSxp/dt and dSyp/dt of the form exp (ik ∙ p – iwt).

Heat capacity of magnons use the approximate magnon dispersion relation w = Ak2 find the leading term in the heat capacity of a three-dimensional Ferro magnet at low temperatures kBT << J. The result is 0.113 kB (kBT/hA)3/2, unit volume. The zeta function that enters the result may he estimated numerically; it is tabulated in Jahnke-Emde.

Neel temperature taking the effective field on the two-sub-lattice model of an anti Ferromagnetic as BA = Ba – μMB – εMA; BB = Ba – μMA – μMB, show that θ/TN = μ + ε/μ – ε.

Coercive force of a small particle 

(a) Consider a small spherical single-domain particle of a uniaxial Ferro-magnet. Show that the reverse field along the axis required to reverse the magnetization is Ba, = 2K/MS, in CGS units. The coercive force of single-domain particles is observed to be of this magnitude. Take UK, = K sin2 θ as the anisotropy energy density and UM = – BaM cos θ as the interaction energy density with the external field; here θ is the angle between B, and M. 

(b) Show that the magnetic energy of a saturated sphere of diameter d is ≈ M2sd3. An arrangement with appreciably less magnetic energy has a single wall in an equatorial plane. The domain wall energy will be πσad2/4, where σw is the wall energy per unit area. Estimate for cobalt the critical radius below which the particles are stable as single domains, taking the value of JS2/n as for iron.

Saturation magnetization near Tc shows that in the mean field approximation the saturation magnetization just below the Curie temperature has the dominant temperature dependence (Tc – T)1/2. Assume the spin is ½.  The result is the same as those for a second-order transition in a Ferro electric crystal the experimental data for Ferro magnets (Table 1) suggest that the exponent is closer to 0.33.

Giant magneto resistance in a ferromagnetic metal, the conductivity σp for electrons whose magnetic moments are oriented parallel to the magnetization is typically larger than σa for those anti parallel to the magnetization. Consider a ferromagnetic conductor consisting of two separate regions of identical dimensions in series whose magnetizations can be independently controlled. Electrons of a given spin flow first through one region and then through the other. It is observed that the resistance when both magnetizations point upwards, R↑↓ is lower than the resistance when they point opposite, R↑↓. This resistance change can be large for σpa >> 1, and the phenomenon is called giant magneto resistance (GMR). A small external magnetic field can switch the resistance from R↑↓ to R↑↑ by reorienting the magnetization of the second layer. This effect is increasingly used in magnetic storage applications such as the magnetic bit readout in hard drives. The giant magneto resistance ratio is defined as:

(a) If there is no spin-flip scattering for the conduction electrons, show that

(b) If σa → 0, explain physically why the resistance in the ↑↓ magnetization configuration is infinite.

Giant magneto resistance in a ferromagnetic metal, the

Neel wall the direction of magnetization change in a domain wall goes from that of the Bloch wall to that of a Neel wall (Fig. 36) in thin films of material of negligible crystalline anisotropy energy, such as Permalloy. The intercept of the Bloch wall with the surface of the film creates a surface region of high demagnetization energy. The Neel wall avoids this intercept contribution, but at the expense of a demagnetization contribution throughout the volume of the wall. The Neel wall becomes energetically favorable when the film becomes sufficiently thin. Consider, however, the energetic of the Neel wall in bulk material or negligible crystalline anisotropy energy. There is now a demagnetization contribution to the wall energy density. By a qualitative argument similar to (56), show that σ=(π2JS2/Na2) + (2πM2sNa) find N for which σw is a minimum. Estimate the order of magnitude of σw for typical values of J, Ms. and a.

Neel wall the direction of magnetization change in a domain

Equivalent electrical circuit considers an empty coil of inductance L0 in a series with a resistance R0; show if the coil is completely filled with a spin system characterized by the susceptibility components X (w) and X"(w) that the inductance at frequency w becomes L = [1 + 4πX'(w)]L0, in series with an effective resistance R = 4πwX" (w) + L0 + R0. In this problem X = X' + iX" is defined for a linearly polarized rf field.

Rotating coordinate system we define the vector F(t) = Fs(t)x + Fy (t) y + Fs(t)z. Let the coordinate system of the unit vectors x, y, z rotate with an instantaneous angular velocity Ω, so that dx/dt = Ωyz – Ωzy, etc. 

(a) Show that dF/dt = (dF/dt)R + Ω x F, where (dF/dt)n is the time derivative of F as viewed in the rotating frame R.

(h) Show that (7) may be written (dM/dt)R = γM x (Ba + Ω/γ). This is the equation of motion of M in a rotating coordinate system. The transformation to a rotating system is extraordinarily useful; it is exploited widely in the literature. 

(c) Let Ω = – γB0z; thus in the rotating frame there is no static magnetic field. Still in the rotating frame we now apply a dc pulse B1x for a time t. If the magnetization is initially along 9, find an expression for the pulse length t such that the magnetization will be directed along – z at the end of the pulse. (Neglect relaxation effects.) 

(d) Describe this pulse as viewed from the laboratory frame of reference.

Hyperfine effects on ESR in metals we suppose that the electron spin of a conduction electron in a metal sees an effective magnetic field from the hyperfine interaction of the electron spin with the nuclear spin. Let the z component of the field seen, by the conduction electron be written where Izf equally likely to be ± ½. 

(a) Show that (B2t) = (a/2N)3N. 

(b) Show that (B*i = 3(a/2N) 4N2, for N >> 1.

Hyperfine effects on ESR in metals we suppose that the

FMR in the anisotropy field consider a spherical specimen of a uniaxial ferromagnetic crystal with an anisotropy energy density of the form UK = K sin2 θ, where θ is the angle between the magnetization and the z axis. We assume that K is positive. Show that the ferromagnetic resonance frequency in an external magnetic field B0z is w0 = γ(B0 + BA), where BA = 2K/Ms.

Exchange frequency resonance consider a Ferrimagnet with two sub-lattices A and B of magnetizations MA, and MB, where MB, is opposite to MA when the spin system is at rest. The gyro magnetic ratios are γA, γB, and the molecular fields are BA = – λMB; BB = – λMA. Show that there is a resonance at w20 = λ2A|MB| - γB|MA|2 this is called the exchange frequency resonance.

Surface Plasmon’s consider a semi-infinite plasma on the positive side of the plane z = 0. A solution of Laplace's equation Δ2φ = 0 in the plasma is φi (x, z) = A cos kx e – kz whence Esi = kA cos kx e –kz Esi = kA sin kx e –kz

(a) Show that in the vacuum φ0 (x, z) = A cos kx ekx for z < 0 satisfies the boundary condition that the tangential component of E be continuous at the boundary; that is, find Ex0.

(b) Note that D, = e(w)Ei; Do = E,. Show that the boundary condition that the normal component of D be continuous at the boundary requires that ε(w) = – 1, whence from (10) we have the Stern-Ferrell result: w2s = ½ w2for the frequency w , of a surface plasma oscillation

Interface Plasmon’s we consider the plane interface z = 0 between metal 1 at z > 0 and metal 2 at z < 0. Metal 1 has bulk Plasmon frequency wp1; metal 2 has wp2. The dielectric constants in both metals are those or free-electron gases. Show that surface Plasmon’s associated with the interface have the frequency

Interface Plasmon’s we consider the plane interface z = 0

Alfven waves consider a solid with an equal concentration n of electrons of mass m, and holes of mass mh. This situation may arise in a semimetal or in a compensated semiconductor. Place the solid in a uniform magnetic field B = Bz. Introduce the coordinate ζ = x + iy appropriate for circularly polarized motion, with ζ having time dependence e-iwt. Let w = eB/mℓC and wh = eB/mhc. (

a) In CGS units, show that ζe = eE+/mℓw(w + wℓ); ζh = – eE+/rnhw(w – wh) are the displacements of the electrons and holes in the electric field E e–iwt = (Ex + iEy) e- iwt.

(b) Show that the dielectric polarization P+ = ne(ζh – ζe) in the regime w << wℓ, wh may be written as P+ = nc2(mh + me)E+/B2, and the dielectric function ε(w) = ε1 + 4πP+/E+ = ε1 + 4πc2p/B2, where ε1, is the dielectric constant of the host lattice and p = n(m + mh) is the mass density of the carriers. If ε1 may be neglected, the dispersion relation w2ε (w) = c2Kbecomes, fur electromagnetic waves propagating in the z direction, w2 = (B2/4πp)K2. Such waves are known m Alfven waves; they propagate with the constant velocity B/ (4πp)1/2. If B = 10kG; n = 1018 cm-3; m = 10-27g, the velocity is ~108 cm s-1. Alfven waves have been observed in semi-metals and in electron-hole drops in germanium.

Helicon wanes 

(a) Employ the method of Problem 3 to treat a specimen with only one carrier type, say holes in concentration p, and in the limit w << wh = eB/mhc. Show that ε(w) = 4πpe2/mhwwh, where D(w) = ε(w)E+ (w). The term εin ε has been neglected. 

(b) Show further that the dispersion relation becomes w = (BC/4πpe)K2, the helicon dispersion relation; in CGS. For K = 1 cm–1 and B = 1000G, estimate the helicon frequency in sodium metal. (The frequency is negative; with circular-polarized modes the sign of the frequency refers to the sense of the rotation.)

Plasmon mode of a sphere the frequency of the uniform Plasmon mode of a sphere is determined by the depolarization field E = – 4πP/3 of a sphere, where the polarization P = –ner, with r as the average displacement of the electrons of concentration n. Show from F = ma that the resonance frequency of the electron gas is w20 = 4πne2/3m. Because all electrons participate in the oscillation, such an excitation is called a collective excitation or collective mode of the electron gas.

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