# a) The volume of the copper cylinder is given by: 3.14 x (0.5 x 10-6 ) 2 x 1 x 10-6 m 3 = 0.78 x m 3 ;

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Example 1.1: Calculate the number of copper atoms present in a cylinder that has a diameter and a height both equal to 1 µµm. The mass density of copper is 8.93 x 103 kg/m3 and its atomic mass is 63.55 g/mol.

Solution:

a) The volume of the copper cylinder is given by: 3.14 x (0.5 x 10-6)2 x 1 x 10-6 m3 = 0.78 x 10-18 m3;

b) Its mass is: 8.93 x 103 kg/m3 x 0.78 x 10-18 m3 = 7.96 x 10-15 kg = 7.96 x 10-12 g;

c) The number of copper atom present is: 7.96 x 10-12 g x 6.02 x 23 atoms/mol /63.55 g/mol = 6.64 x 1010 atoms.

(This problem shows the relatively large number of copper atoms present in a small volume.)

Example 1.2: a) Give the electronic configurations of the sodium and chlorine atoms (expressed in terms of the s-orbits and p-orbits); and b) repeat this for their ions.

Solution:

a) The electronic configuration of the atoms can be obtained from a standard chemistry handbook. They are:

Na: 1s2, 2s2, 2p6 and 3s1 and Cl: 1s2, 2s2, 2p6, 3s2, 3p5.

(In this configuration, the superscript indicates the number of electrons present in the energy orbits.) Note that both Na and Cl are monovalent. b) In the ion forms: Na+ : 1s2, 2s2 and 2p6 and Cl- : 1s2, 2 s2, 2p6, 3s2, 3p6.

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Example 1.5: Compute the attractive force between a Na+ ion and a Cl- ion at T = 300K.

Solution:

From Table 1.1 (300K), we find that the ion separation = (0.098 + 0.181) nm = 0.279 nm.

Using Eq.(1) in the text, the attractive force is given by: Fa = - 9 x 109 Vm/C x (1.6 x 10-19 C)2/(0.279 x 10-9 m)2 = - 2.96 x 10-9 N.

Example 1.7: Assuming a 1-dimensional solid, the electrostatic attraction between the neighboring ions produces a potential energy given by: EPE(a) = - 1.75 q2/(4ππεεoa), where a is the ion separation. a) Determine ao , the equilibrium separation, if EPE(ao) = - 0.662 eV; and, b) determine the constant A if the potential due to repulsion has the form: Er(a) = A/a4. Solution:

a) Making use of the equation provided: ao = - 1.75 q2/(4ππεεoEPE(ao)) = 1.75 x (1.6 X 10-19C)2/(4 x 3.14 x 8.85 x 10-12 F/m x 0.662 x 1.6 x 10-19 V) = 3.8 x 10-9 m.

b) The totall energy is given by: E’pe(a) = - 1.75 q2/(4ππεεoa) + A/a4. At equilibrium, dE’PE(a)/da = 0 and this leads to: A = 1.75 q2 /(4ππεεoao2) x ao5/4 = 1.75 x (1.6 X 10-19)2/(4 x 3.14 x 8.85 x 10-12 F/m x (3.8 x 10-9)2) x (3.8 x 10-9)5/4 = 5.52 x 10-54 J.m4.

(This problem evaluates the equilibrium bond length and bond energy for a 1-dimensional ionic solid. The factor 1.75 actually accounts for the effect of the more distant ions).

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Example 1.9: Calculate the minimum radius ratio for two different types of ions present in a solid if the coordination number is 8.

Solution:

Consider a simple cube with dimension ao. If the smaller ion (with radius r) is placed at the center and ions of the opposite type (with radius R’) placed in the corners, automatically we have a structure with a coordination number of 8. In addition, for a close-packed structure, we have: The body diagonal of the cube = √√3 x ao = 2 x (r + R’), and ao = 2R’.

Eliminating ao from these equations gives: 1 + r/R’ = √√3 = 1.732, or r/R’ = 0.732.

(Note that if r/R’ is smaller, it will be no longer possible to put in the 8 neighboring ions and yet maintain the close-packed structure.)

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Example 1.18: a) Determine the number of lattice points present in a fcc unit cell, and b) draw a primitive cell within the unit cell.

Solution:

a) There are 4 lattice points in the unit cell. b) The primitive cell is shown in Fig.E1.6.

Example 1.21: For a cubic crystal structure, draw a) the [1 1 1], [1 3 3 ], and [2 3 6] lattice directions, and b) the (1 1 1), (2,3,4), and (3 2 1) lattice planes.

Solution:

a) See Fig. E1.10a.

b) See Fig.E1-10b. For the (1 1 1) plane, the lattice interceptions should be: 1, 1, 1. For the (2 3 4) plane, they should be: 1, 2/3, 1/2. For the (3 2 1) plane, they should be: 1, 3/2, 3.

Example 1.22: a) Determine which of the planes in the fcc crystal structure have the highest density of atoms, and b) evaluate its value for copper. The lattice constant of copper is 0.362 nm.

Solution:

a) The highest density planes for the fcc crystal structure are the {1 1 1} planes. Note that they also have the shortest interplane distance.

b) There is a total of 2 atoms in the (1 1 1) plane in the unit cell. Its area = 1/2 x √√(3/2) x (√√2 x ao)2 = √√3/2 x ao2 = 0.866 ao2.

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Example 1.24: Calculate the atomic density of αα-Fe which has bcc crystal structure. Note that metals have a close-packed structure. Solution:

For the bcc crystal structure, the ratio of the lattice constant to the atomic radius is 4/√√3. Since rFe = 0.124 nm, ao = 4/√√3 x 0.124 nm = 0.286 nm. Since there are 2 atoms per unit cell, the atomic density of αα-Fe is 2/ao3 = 2/(0.286 x 10-9)3 /m3 = 8.55 x 1028 /m3.

Example 1.25: Fig. E1.13 shows the unit cell of a hcp structure. Given that c/ao is 1.63, calculate the APF assuming that all of the atoms have the same radius.

Solution:

The volume of the unit cell = ao x ao/sin(60o) x 1.63 x ao = 1.41 ao3. Since ao = 2r for a close-packed structure, the volume = 1.41 x (2r)3 = 11.3 r3. Since there are 2 atoms per unit cell, the APF = 2 x 4ππ/3 x r3/(11.3 r3) = 0.74.

Example 1.27: Calculate the planar density of atoms in the (1 1 1) plane of germanium.

Solution:

Since the lattice constant of germanium is 0.563 nm, the length of the face diagonal is √√2 x 0.563 nm = 0.796 nm. The area of the triangle in the (1 1 1 ) plane within the unit cell is 0.796/2 x √√3 x 0.796/2 = 0.274 nm2. Since there are 2 atoms in this area, the planar density is 2/(0.274 x 10-18) /m2 = 7.29 x 1018 /m2.

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Example 1.32: Assume that Kd = 3 and ∆∆Ed = 2.4 eV, compute the ratio Nd/Nsite at a) T = 300K, and b) T = 1000K.

Solution:

a) Based on Eq.(4) in the notes, Nd/Nsite = Kd exp(- ∆∆Ed/(kT)) = 3 x exp(- 2.4 x 1.6 x 10-19 J/(1.38 x 10-23 J/K x 300K) = 1.4 x 10-40.

b) Similarly, at 1000K, Nd/Nsite = 3 x exp(- 2.4 x 1.6 x 10-19 J/(1.38 x 10-23 J/K x 1000K) = 2.4 x 10-12.

Example 1.34: In an IC fabrication process, a 100 nm thickness boron-doped surface layer has to be formed on silicon. The required boron density within the surface layer must be no less than 1 x 1024 /m3. If the boron density at the upper silicon surface is kept constant at 1 x 1025 /m3, how long does it take for the surface layer to be formed if a) T = 750oC, and b) 1100oC. Fig.E1.16 provides a plot of the diffusivity of boron as a function of temperature. Assume the original boron density in silicon is negligible.

Solution:

From Fig.E1.16, the diffusivity Di (750oC) = 1 x 10-21 m2/s and Di (1100oC) = 1 x 10-17 m2/s. From Eq.(7): co = 0, ci(x)/cis = 1 - erf(x/(2√√(Dit))).

a) ci(x)/cis = 0.1 at x =100 nm and T = 750oC. This gives 0.1 = 1 - erf(1 x 10 -7

/(2 x √√(1 x 10-21 x t))), or t = 1.86 x 106 s (516 hrs).

b) Similarly, at 1100oC, 0.1 = 1 - erf(1 x 10-7/(2 x √√(1 x 10-17 x t))), or t = 186 s (0.052 hrs).

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Example 1.36: In a diffraction experiment involving a polycrystalline sample (see Fig.E1.17), the (1 1 1) diffraction ring measured on a

photographic plate is 0.01 m from the center. The sample is 0.03 m away from the photographic plate on which the diffraction pattern is projected and the lattice constant of the crystals is 0.42 nm. Calculate the diffraction angle 2θθ and the Bragg angle θθ assuming that the crystals have a cubic lattice structure and the order of diffraction n = 1.

Solution:

The diffraction angle is 2θθ = 180o - tan-1(0.01 m/0.03 m) = 161.6o. Thus, the Bragg angle θθ = 80.8o.

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Example 2.2: Compare the thermal energy Eth of a free electron at 300K with the value of the excitation energy (1.12 eV) for silicon.

Solution:

At 300K, Eth = kT = 1.38 x 10-23 J/K x 300 K = 4.14 x 10-21 J. Eexcite = 1.12 eV = 1.12eV x 1.6 x 10-19 J/eV = 1.79 x 10-19 J. This is about 50 times larger than Eth and one can therefore expect only a small fraction of the electrons in silicon to be free electrons.

Example 2.3: Distinguish between the thermal velocity and the drift

velocity, i.e., velocity due to an applied electric field, for electrons inside a solid.

Solution:

Thermal velocity is temperature dependent has no specific direction, while drift velocity for an electron is opposite to the direction of the applied field. Quantitatively, thermal velocity is √√(2kT/me*) ≈≈ 1 x 105 m/s, while the drift velocity is proportional to the electric field (and is zero when no current is flowing).

Example 2.6: Verify that f(E) = 0.047 when a) E - EF = + 3kT, and b) f(E) = 0.952 when E - Ef = - 3kT.

Solution:

a) Based on the Fermi function, f(E) = 1/[exp((E - EF )/(kT) + 1)] = 1/[exp(3) + 1] = 0.047.

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Example 2.8: A hypothetical solid has 1 x 1026 atoms and each atom has 6 electrons. a) Suggest what is the minimum number of energy states

needed to accomodate all the electrons within a single energy band, and b) if the lowest energy state in the band is 0 eV and the energy states are separated by 0.5 x 10-26 eV, compute the energy of the most energetic electron.

Solution:

a) There are 6 x 1026 electrons. Since each energy state can accomodate only 2 electrons, there must be at least 3 x 1026 states.

b) If these energy states are separated by 0.5 x 10-26 eV, the energy of the most energetic electron is 3 x 1026 x 0.5 x 10-26 eV = 1.5 eV.

Example 2.11: A copper atom has 29 electrons. Compute the fraction of (free) electrons in copper if its conductivity is 1.7 x 107 S/m. Assume that the electron mobility is 1.26 x 10-3 m2/V.s and the atomic density of copper is 2 x 1029 /m3 (metals have electrons only).

Solution:

The (free) electron density in copper is σσ/(qµµn) = 1.7 x 107 S/m/(1.6 x 10-19 C x 0.00126 m2/V.s) = 0.84 x 1029 /m3. The total number of electrons present in copper is 2 x 1029 x 29 = 5.8 x 1030. The fraction of (free) electrons = 0.84 x 1029 /m3/(5.8 x 1030 /m3) = 0.014.

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Example 2.12: A wire made of an aluminum alloy is 1 mm in diameter and 1 m in length. a) If there is a current of 10 mA flowing and the voltage across the wire is 0.432 mV, determine its conductivity. b) If the electron mobility is 3 x 10-3 m2/V.s, what is the electron density? c) What is the drift velocity for the electrons?

Solution:

a) The resistance R is Va/I = 0.432 x 10-3 V/(10 x 10-3 A) = 0.0432 ΩΩ. This gives a conductivity σσ = L/(AcsR) = 1 m/(3.14 x (0.5 x 10-3 m)2 x 0.0432 ΩΩ) = 29.5 x 106 S/m.

b) The electron density n is σσ/(qµµn) = 29.5 x 106 S/m/(1.6 x 10-19 C x 3 x 10-3 m2/V.s) = 6.14 x 1028 /m3.

c) The drift velocity vdrift is µµnE’ = 3 x 10-3 m2/V.s x 0.432 x 10-3 V/1 m = 1.3 x10-6 m/s.

Example 2.16: From Table 2.6, estimate the resistivity of copper if there is 0.1 wt % of chromium present (see Table II-6 for the value of σσI for

copper). Solution:

Since σσ = σσI /(1 + ββκκ). For impure copper, ρρ = 1/σσ = 1/(5.8 x 10-11 S/m) x (1 + 0.1 x 2.47) = 2.19 x 1010ΩΩ.m.

Example 2.17: Determine the Hall coeffient RH if the current passing through a solid is 1 mA, the width of the solid is 0.005 m, the cross-sectional area is 1 x 10-6 m2, the Hall voltage is 2.5 mV, and the magnetic induction is 0.1 T.

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Example 2.18: If the fractional increase in resistivity of copper due to magnetoresistance effect is 1%, determine the magnetic induction. Solution:

Since ρρ‘/ρρ = 1 + µµn 2

Bz2 = 1.01. Bz = √√(0.01)/µµn = 0.1/0.00126 m2/V.s = 79.8 T (see Table 2.5 for the value of µµn).

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Example 2.19: A chromel-alumel thermocouple is used to monitor the temperature of a hot furnace. If the voltage output from the thermocouple is 10 mV and the cold junction is at 0 oC, calculate the furnace

temperature. Solution:

The temperature of the furnace is Th = ∆∆VAA/(KsA - KsB) = 10 mV/0.040 mV/oC = 250oC (see Table 2.7 for the value of KsA - KsB).

Example 2.23: An intrinsic semiconductor has a conductivity of 250 S/m at 20oC and 1100 S/m at 100oC. What is the value of the energy gap? Solution:

Since ln(σσ1/ σσ2) = Eg/(2kT2) - Eg/(2kT1). Solving for the energy gap gives: Eg = 2kln(σσ1/σσ2)/(1/T1 - 1/T2) = 2 x 86.2 x 10-6 eV/K x ln(1100 S/m/250

S/m)/(1/293 K - 1/373 K) = 0.349 eV.

Example 2.24: Calculate the intrinsic conductivity of GaAs at 50oC. Solution:

For GaAs, the intrinsic conductivity σσ(intrinsic) = niq(µµn + µµp) = 1.8 x 1012 /m3 x 1.6 x 10-19 C x (0.85 m2/V.s + 0.04 m2/V.s) = 2.56 X 10-7 S/m (see Table 2.8 for the parameters).

Since σσ = σσo’ exp(- Eg/(2kT)), σσ(50oC) = 2.56 x 10-7 S/m x exp(1.42 x (1/300K - 1/323K)/(2 x 8.62 x 10-5 eV/K)) = 1.8 x 10-6 S/m.

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The Fermi level is EF = Ec - EF(intrinsic) + 0.1 eV = Ec - (1.11/2) + .1 eV = Ec - 0.455 eV. The probability of finding an electron in the conduction band is f(Ec) = 1/(exp(Ec - EF)/(kT) + 1) = 1/(exp(0.455 eV/(86.2 x 10-6 eV/K x 298K)) + 1) = 2.2 x 10-8.

Example 2.30: A n-type semiconductor has a donor density ND = 1 x 1022 /m3. At what temperature will the intrinisic carrier density equal to ND? We are given that NC = 2.46 x 1025 /m3, Nv = 1 x 1025 /m3 and Eg = 1.1 eV (these parameters are assumed to be temperature independent). Solution:

The intrinisic carrier density ni = √√(NcNv) exp(- Eg/(2kT)). This leads to T = Eg/(2k ln(√√(NcNv)/ND)) = 1.1 eV x 1.6 x 10-19 J/eV/(2 x 1.38 x 10-23 J/K x ln(√√(2.46 x 1025 /m3 x 1 x 1025 /m3)/(1 x 1022 /m3)) = 867K.

Example 2.30A: List the electron and hole densities of silicon at 300K if it is a) intrinsic; b) doped with 1 x 1020 /m3 of donors (ND); c) doped with 1 x 1020 /m3 of acceptors (NA); and d) doped with 1 x 1020 /m3 of donors (ND) and 0.5 x 1020 /m3 of acceptors (NA). Note that ni = 1.45 x 1016 /m3. Solution: For silicon: ND NA n p ni 0 0 ni ni ni 1020 0 1020 ni2/1020 ni 0 1020 ni2/1020 1020 ni 1.0 x 1020 0.5 x 1020 0.5 x 1020 ni2/0.5 x 1020 ni

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Example 2.36: A parallel plate capacitor has the following properties: Acs = 2 x 10-3 m2, d = 0.01 m, Va = 100 V and εεr = 6. a) Find the capacitance. b) If there are 1 x 1015 dipoles/m3 in the insulator, what is the

polarizability? c) Determine the amount of surface charge on the capacitor plates. d) Repeat part c) in the absence of the insulator. e) What is the amount of polarization charge?

Solution:

a) The capacitance C = εεiAcs/d = 6 x 8.86 x 10-12 F/m x 2 x 10-3 m2/(1 x 10-2 m) = 1.06 x 10-11 F.

b) The relative permittivity εεr = 6 = 1 + αα‘Ndipole/εεo . Therefore, the

polarizability αα‘ = (6 - 1) x 8.85 x 10-12 F/m/(1 x 1015 /m3) = 4.43 x 10-26 F.m2. c) The stored charge QAcs = CVa = 1.06 x 10-11 F x 100 V = 1.06 x 10-9 C. d) The stored charge without the insulator QoAcs = CoV = CVa/εεr = 1.06 x 10-9 C/6 = 0.176 x 10-9 C.

e) The polarization charge ∆∆QAcs = (Q - Qo)Acs = (1.06 - 0.176) x 10-9 C = 0.88 x 10-9 C.

Example 2.39: Compute the polarizability of a water molecule if its dipole moment po = 1 x 10-29 C.m, E’ = 1 x 105 V/m, and T = 300K. For ionic

molecules, αα‘ = po2/(3kT). Solution:

For a water molecule, the polarizability αα‘ = po2/(3kT) = (1 x 10-29 C.m)2/(3 x 1.38 x 10-23 J/K x 300K) = 8.05 x 10-39 F.m2.

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Example 2.40: Defining the dipole moment as the product of the dipole charge times the separation, compute the dipole moment for the

tetragonal BaTiO3 unit cell shown in Fig.2.45. Assume that the Ba2+ ions are stationary.

Solution:

For the given unit cell structure, there is the equivalence of 3 O2- ions, 1 Ti4+ ion and 1 Ba2+ ion. In the figure, the Ti4+ ion is shifted upward and the O2- ions are shifted downward (note that the equatorial O2- ions are shifted by a smaller amount when compared with the corner O2- ions).

The dipole moment related to the shift in the Ti4+ ion is pTi4+ = 4 x 1.6 x 10-19 C x 0.006 x 10-9 m = 3.84 x 10-30 C.m. The dipole moment related to the shift in the O2- ion is pO2- = 2 x 1.6 x 10-19 C x (2 x 0.006 x 10-9 m + 0.009 x 10-9 m) = 6.72 x 10-30 C.m.

Since the Ba2+ ions are considered to be stationary, the total dipole moment is pTi4+ + pO2- = (3.84 + 6.72) x 10-30 C.m = 1.06 x 10-29 C.m.

Example 2.42: The equivalent circuit for a lossy capacitor is shown in Fig.E2.1, calculate the loss factor. Assume Va = Vo sin(ωωt).

Solution:

The energy stored per cycle in the capacitor is Estored = ∫∫cycle Vai dt = ∫∫cycle Vo2ωωC sin(ωωt) cos(ωωt) dt = - Vo2C (cos(2ωωt) - 1)/4. Its peak value = Vo2C/2. Energy loss due to the resistor is given by E’’ = Vo2/Rac x ∫∫cycle sin(ωωt) sin(ωωt) dt = Vo2ππ/(ωωRac). The loss factor tan(δδ) = Vo2 ππ/((ωωRac ) x (2 ππVo2C/2)) = 1/(ωωCRac).

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Example 4.1: What are the energies of light with wavelengths equal to: a) 400 nm, and b) 800 nm? What are their colors?

Solution:

The photon energy is: Ephoton = hc/λλL = 6.63 x 10-34 J.s x 3 x 108 m/s/(1.6 x 10-19 C x λλL ) = 1.24 eV/λλL.

a) When λλL = 400 nm or 0.4 µµm, Ephoton = 3.1 eV (blue). b) When λλL = 800 nm or 0.8 µµm, Ephoton = 1.55 eV (red).

Example 4.2: What is the number of photons per unit time emerging from a solid if the power output (of the photons) is 10 W and the (average)

wavelength is 0.65 µµm. Solution:

Similar to Example 4.1, the photon energy at the wavelength of 0.65 µµm is 1.9 eV. The number of photons emerging per unit time in an output of 10 W = 10 W/(1.9 eV x 1.6 x 10-19 V/eV) = 0.33 x 1020 (photons)/s.

Example 4.4: When light passes from a medium with a high index of refraction to one with a lower value, total reflection occurs at the critical angle θθcr. Compute the critical angle for light transmission from glass to air. Assume ηηs(glass) = 1.46.

Solution:

Based on Snell’s law: sin(θθglass) = sin(θθair)/ηηs(glass). Since sin(θθair) = 1 and

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Example 4.5: Compare R* , the reflectance in silica glass (ηηs = 1.46) to that in pure PbO (ηηs = 2.60). Explain briefly why ornamental glassware has a small percentage of PbO present.

Solution:

Based on the reflectance formula: R* (glass) = (1.46 - 1)2/(1.46 + 1)2 = 0.035. Similarly, R* (PbO) = (2.60 - 1)2/(2.60 + 1)2 = 0.198. PbO is therefore more reflective, which gives the ornamental value.

Example 4.6: Green light at a wavelength of 0.5 µµm is allowed to shine on a piece of silicon. Estimate the distance inside the silicon when the light intensity is dropped by a factor of 100 (see Fig.E4.2).

Solution:

From Fig.E4.2, αα* is 9 x 103 /cm when the wavelength is 0.5 µµm. The distance when the light intensity is dropped by 100 is x = ln(φφLo/φφ)/αα* = ln(100)/(9 x 103 /cm) = 5.1 x 10-4 cm.

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Example 4.8: A silicon sample has a thickness of 100 µµm and a

conductivity of 10 S/m. If we assume that light is absorbed primarily within 2 µµm from the surface and gives a photoconductivity of 1000 S/m,

estimate the fractional change in the sample conductivity if the measurement is made along the semiconductor surface. Ignore any change in conductivity beyond the surface layer.

Solution:

The lateral conductivity of the sample will be the parallel combination of the conductivity of the surface layer and that of the rest of the silicon sample. The fractional change is: (σσ1t1 + σσ2t2)/(σσ1 x (t1 + t2)) = (2 µµm x 1000 S/m + 98 µµm x 10 S/m)/(100 µµm x 10 S/m) = 2.98.

Example 4.11: If we want to form GaInP crystal with an energy gap of 1.5 eV, what percent of GaP has to be used? For simplicity, we assume that the energy gap of GaInP depends linearly on the composition of GaP and InP and that the energy gaps of GaP and InP are 2.26 eV and 1.35 eV, respectively.

Solution: The required energy gap of GaInP is Eg = 1.5 eV = (y x 2.26 + (1 - y) x 1.35) eV, where y is the percent of InP. This leads to: y = 0.17. The composition should be 17% GaP and 83% InP.

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Example 4.12: InSb is an IR detector and it has an energy gap of 0.17 eV. What is the probability that an electron in the valence band is thermally excited across the energy gap at 300K.

Solution:

Since Eg > kT (T = 300K), we can assume that the probability of carrier excitation to be: exp(- Eg/(kT)) = exp(- 0.17 eV/0.026 eV) = 1.4 x 10-3. This suggests that there will be 14 electrons excited from the valence band to the conduction band per 10,000 electrons (in the valence band).

Example 4.18: Suggest why Se thin films are unsuitable for use in electrophotography of documents with a red background.

Solution:

Although Se has an absorption threshold near 1.7 eV, the

electrophotography sensitivity or photoconductivity threshold may not coincide with the absorption threshold. In fact, the photoconductivity threshold in Se is in the green (2.4 eV) and is not sensitive to red or yellow. Reflected red or yellow light from the background therefore appears as black, the same as the prints in the document.

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Example 4.19: Compute the photoconductivity in a p-type photoconductor if the hole lifetime is 50 µµs and the length of the

photoconductor is 1000 µµm. Assume the average velocity of the holes is 1 x 105 m/s.

Solution:

The transit time of the photoconductor is ττt = L/v = 1 x 10-3 m/105 m/s = 10 ns. The photoconductivity gain is ττp/ττt = 50 x 10-6 s/10-8 s = 5000.

Example 4.24: If a TV set has 500 x 500 pixels and the sweep frequency is 5 Mhz, how long does it take to scan the whole screen once.

Solution:

There are 500 x 500 pixels or 250,000 pixels. To achieve a sweep

frequency of 5 Mhz, the maximum time delay per pixel is less than 0.1 µµs (twice the sampling rate). For 1/4 million pixels, the total time delay is: 0.25 x 106 x 0.1 x 10-6 s = 0.025 s.

Example 4.27: What is the minimum number of layers of liquid crystals required to give a 90o twist if the maximum angle between the layers is 12o.

Solution:

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Example 5.17: Compare the resistivity of magnetic solids in the crystalline state, as oxides, and in the amorphous state.

Soltuion:

Crystalline metals have a resistivity of about 10-7ΩΩ.m. Metallic amorphous Fe-related solids have a value of about 10-6ΩΩ.m, whereas a ferrite has a value of about 100 ΩΩ.m.

Example 5.19: If the B-H plot of a ferromagnetic solid is given by: B = Ho - H, determine (BH)max.

Solution:

To obtain the peak value, take ∂∂(BH)/∂∂H = 0. Since BH = HoH - H2, Ho = 2H, or H = Ho/2. Therefore, (BH)max = Ho/2(Ho - Ho/2) = Ho/4.

Example 5.21: Silicon is known to increase the resistivity of Fe. What will be the reduction in the eddy current loss if the silicon content of a Fe transformer core is increased from 3% to 10%? ββ for silicon in Fe is 117. Solution:

The fractional increase in resistivity of the core is: ((1 + 0.1ββ) - (1 + 0.03ββ))/ (1 + 0.03ββ) = ((1 + 0.1 x 117ββ) - (1 + 0.03 x 117))/ (1 + 0.03 x 117) = 1.81. If

nothing else changes, the eddy current loss will be reduced by 1.81 times.

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Example 5.22: Which of the following materals are more suitable for a transformer core: i) Sheets of SiFe; ii) MnZn ferrite; or iii) dry air? Solution:

Both ii) and iii) have low permeability and are not suitable. Their principal advantage, if any, would be low loss and possibly a high breakdown

voltage.

Example 5.24: If a magnetic tape contains granular particles of diameter 0.1 µµm, determine the speed of the rotor if a full cycle of the recording requires at least 10 grains and the audio (signal) range is 2 kHz.

Solution:

The shortest recording wavelength will be 0.1 µµm x 10 = 1 µµm. Speed = wavelength x frequency = 1 x 10-6µµm x 2 x 103 /s = 2 x 10-3 m/s.

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