**Example 1.1: Calculate the number of copper atoms present in a cylinder *** that has a diameter and a height both equal to 1 *µµ

**m. The mass density of**

**copper is 8.93 x 10****3****kg/m****3****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**** m****3**** = 0.78 x 10****-18**** m****3****; **

**b) Its mass is: 8.93 x 10****3**** kg/m****3**** x 0.78 x 10****-18**** m****3**** = 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 10****10**** 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: 1s****2****, 2s****2****, 2p****6**** and 3s****1**** and Cl: 1s****2****, 2s****2****, 2p****6****, 3s****2****, 3p****5****. **

**(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****+ ****: 1s****2****, 2s****2**** and 2p****6**** and Cl****-**** : 1s****2****, 2 s****2****, 2p****6****, 3s****2****, ****3p****6****. **

**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 10****9****Vm/C x (1.6 x 10****-19**** C)****2****/(0.279 x 10****-9**** m)****2**** = - 2.96 x 10****-9**** N. **

**The negative sign indicates an attractive force. **

**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/(4*ππεε

**oa), 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/a****4****.**

**Solution:*** a) Making use of the equation provided: ao = - 1.75 q2/(4*ππεε

**oEPE(ao)) = 1.75**

**x (1.6 X 10****-19****C)****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/(4*ππεε

**oa) + A/a****4****.***ππεε*

**At equilibrium, dE’PE(a)/da = 0 and this leads to: A = 1.75 q****2****/(4**

**oao****2****) x**

**ao****5****/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.m****4****.****(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). **

**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.) **

**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 ao****2****= 0.866 ao****2****.*** 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.***αα*

**Since there are 2 atoms per unit cell, the atomic density of**

**-Fe is 2/ao****3****=**

**2/(0.286 x 10****-9****)****3****/m****3****= 8.55 x 10****28****/m****3****.****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(60****o****) x 1.63 x ao = 1.41 ao****3****. Since ao ****= 2r for a close-packed structure, the volume = 1.41 x (2r)****3**** = 11.3 r****3****. Since *** there are 2 atoms per unit cell, the APF = 2 x 4*ππ

**/3 x r****3****/(11.3 r****3****) = 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 nm****2****. Since**

**there are 2 atoms in this area, the planar density is 2/(0.274 x 10****-18****) /m****2****=**

**7.29 x 10****18****/m****2****.*** 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 10****24**** /m****3****. If the ****boron density at the upper silicon surface is kept constant at 1 x 10****25**** /m****3****, ****how long does it take for the surface layer to be formed if a) T = 750****o****C, ****and b) 1100****o****C. 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 (750****o****C) = 1 x 10****-21**** m****2****/s and Di (1100****o****C) = 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 = 750****o****C. This gives 0.1 = 1 - erf(1 x 10****-7**

* /(2 x *√√

**(1 x 10****-21****x t))), or t = 1.86 x 10****6****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).****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 2*θθ

*θθ*

**and the Bragg angle**

**assuming that the crystals have a cubic**

**lattice structure and the order of diffraction n = 1.****Solution: **

* The diffraction angle is 2*θθ

**= 180****o****- tan****-1****(0.01 m/0.03 m) = 161.6****o****. Thus, the***θθ*

**Bragg angle**

**= 80.8****o****.****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 10****5****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. **

**Example 2.8: A hypothetical solid has 1 x 10****26**** 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 10****26**** electrons. Since each energy state can accomodate ****only 2 electrons, there must be at least 3 x 10****26**** 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 10****26**** 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 10****7**** S/m. Assume that ****the electron mobility is 1.26 x 10****-3**** m****2****/V.s and the atomic density of copper ****is 2 x 10****29**** /m****3**** (metals have electrons only). **

**Solution: **

* The (free) electron density in copper is *σσ

*µµ*

**/(q**

**n) = 1.7 x 10****7****S/m/(1.6 x 10****-19****C**

**x 0.00126 m****2****/V.s) = 0.84 x 10****29****/m****3****. The total number of electrons present**

**in copper is 2 x 10****29****x 29 = 5.8 x 10****30****. The fraction of (free) electrons =**

**0.84 x 10****29****/m****3****/(5.8 x 10****30****/m****3****) = 0.014.****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**** m****2****/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***σσ*

**gives a conductivity***ΩΩ*

**= L/(AcsR) = 1 m/(3.14 x (0.5 x 10****-3****m****)****2****x 0.0432**

**) =**

**29.5 x 10****6****S/m.*** b) The electron density n is *σσ

*µµ*

**/(q**

**n) = 29.5 x 10****6****S/m/(1.6 x 10****-19****C x 3 x 10****-3**

**m****2****/V.s) = 6.14 x 10****28****/m****3****.*** c) The drift velocity vdrift is *µµ

**nE’ = 3 x 10****-3****m****2****/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 *σσ

**I for****copper). ****Solution: **

* Since *σσ

*σσ*

**=***ββκκ*

**I /(1 +***ρρ*

**). For impure copper,***σσ*

**= 1/**

**= 1/(5.8 x 10****-11****S/m) x (1 +***ΩΩ*

**0.1 x 2.47) = 2.19 x 10****10**

**.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**** m****2****, the Hall voltage is 2.5 mV, and the magnetic ****induction is 0.1 T. **

**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*

**‘/***µµ*

**= 1 +**

**n**

**2*** Bz2 = 1.01. Bz = *√√

*µµ*

**(0.01)/**

**n = 0.1/0.00126 m****2****/V.s = 79.8 T***µµ*

**(see Table 2.5 for the value of**

**n).****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 ****o****C, calculate the furnace **

**temperature. ****Solution: **

* The temperature of the furnace is Th = *∆∆

**VAA/(KsA - KsB) = 10 mV/0.040 mV/****o****C**

**= 250****o****C (see Table 2.7 for the value of KsA - KsB).****Example 2.23: An intrinsic semiconductor has a conductivity of 250 S/m ****at 20****o****C and 1100 S/m at 100****o****C. What is the value of the energy gap? ****Solution: **

* Since ln(*σσ

*σσ*

**1/**

**2) = Eg/(2kT2) - Eg/(2kT1). Solving for the energy gap gives:***σσ*

**Eg = 2kln(***σσ*

**1/**

**2)/(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 50****o****C. ****Solution: **

* For GaAs, the intrinsic conductivity *σσ

*µµ*

**(intrinsic) = niq(***µµ*

**n +**

**p) = 1.8 x 10****12**

**/m****3****x 1.6 x 10****-19****C x (0.85 m****2****/V.s + 0.04 m****2****/V.s) = 2.56 X 10****-7****S/m (see Table**

**2.8 for the parameters).*** Since *σσ

*σσ*

**=***σσ*

**o’ exp(- Eg/(2kT)),**

**(50****o****C) = 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.****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 10****22****/m****3****. At what temperature will the intrinisic carrier density equal to ND? ****We are given that NC = 2.46 x 10****25**** /m****3****, Nv = 1 x 10****25**** /m****3**** 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 10****25****/m****3****x 1 x 10****25****/m****3****)/(1 x 10****22****/m****3****)) = 867K.****Example 2.30A: List the electron and hole densities of silicon at 300K if it ****is a) intrinsic; b) doped with 1 x 10****20**** /m****3**** of donors (ND); c) doped with 1 x ****10****20**** /m****3**** of acceptors (NA); and d) doped with 1 x 10****20**** /m****3**** of donors (ND) ****and 0.5 x 10****20**** /m****3**** of acceptors (NA). Note that ni = 1.45 x 10****16**** /m****3****. ****Solution: ****For silicon: ND NA n p ni **** 0 0 ni ni ni **** 10****20**** 0 10****20**** ni****2****/10****20**** ni **** 0 10****20**** ni****2****/10****20**** 10****20**** ni **** 1.0 x 10****20**** 0.5 x 10****20**** 0.5 x 10****20**** ni****2****/0.5 x 10****20**** ni **

**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 *εε

**r = 6. a) Find the capacitance.**

**b) If there are 1 x 10****15****dipoles/m****3****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 = *εε

**iAcs/d = 6 x 8.86 x 10****-12****F/m x 2 x 10****-3****m****2****/(1 x 10****-2**

**m) = 1.06 x 10****-11****F.*** b) The relative permittivity *εε

*αα*

**r = 6 = 1 +***εε*

**‘Ndipole/**

**o . Therefore, the*** polarizability *αα

**‘ = (6 - 1) x 8.85 x 10****-12****F/m/(1 x 10****15****/m****3****) = 4.43 x 10****-26****F.m****2****.**

**c) The stored charge QAcs = CVa = 1.06 x 10****-11****F x 100 V = 1.06 x 10****-9****C.***εε*

**d) The stored charge without the insulator QoAcs = CoV = CVa/**

**r = 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 10****5**** V/m, and T = 300K. For ionic **

* molecules, *αα

**‘ = po****2****/(3kT).**

**Solution:*** For a water molecule, the polarizability *αα

**‘ = po****2****/(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.m****2****.****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 Ba****2+**** ions ****are stationary. **

**Solution: **

**For the given unit cell structure, there is the equivalence of 3 O****2-**** ions, 1 ****Ti****4+**** ion and 1 Ba****2+**** ion. In the figure, the Ti****4+**** ion is shifted upward and the ****O****2-**** ions are shifted downward (note that the equatorial O****2-**** ions are shifted ****by a smaller amount when compared with the corner O****2-**** ions). **

**The dipole moment related to the shift in the Ti****4+**** 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 O****2-**** 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 Ba****2+**** 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(*ωω

**t).****Solution: **

* The energy stored per cycle in the capacitor is Estored = *∫∫

*∫∫*

**cycle Vai dt =**

**cycle***ωω*

**Vo****2***ωω*

**C sin(***ωω*

**t) cos(***ωω*

**t) dt = - Vo****2****C (cos(2**

**t) - 1)/4. Its peak value = Vo****2****C/2.***∫∫*

**Energy loss due to the resistor is given by E’’ = Vo****2****/Rac x***ωω*

**cycle sin(**

**t)***ωω*

**sin(***ππ*

**t) dt = Vo****2***ωω*

**/(***δδ*

**Rac). The loss factor tan(***ππ*

**) = Vo****2***ωω*

**/((***ππ*

**Rac ) x (2**

**Vo****2****C/2))***ωω*

**= 1/(**

**CRac).****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/*λλ

**L = 6.63 x 10****-34****J.s x 3 x 10****8****m/s/(1.6 x***λλ*

**10****-19****C x***λλ*

**L ) = 1.24 eV/**

**L.*** a) When *λλ

*µµ*

**L = 400 nm or 0.4**

**m, Ephoton = 3.1 eV (blue).***λλ*

**b) When***µµ*

**L = 800 nm or 0.8**

**m, 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 *µµ

**m.**

**Solution:*** Similar to Example 4.1, the photon energy at the wavelength of 0.65 *µµ

**m 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 10****20****(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 *θθ

**cr. Compute the critical angle for light transmission from glass to***ηη*

**air. Assume**

**s(glass) = 1.46.****Solution: **

* Based on Snell’s law: sin(*θθ

*θθ*

**glass) = sin(***ηη*

**air)/***θθ*

**s(glass). Since sin(**

**air) = 1 and*** Example 4.5: Compare R* , the reflectance in silica glass (*ηη

**s = 1.46) to***ηη*

**that in pure PbO (**

**s = 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 *µµ

**m 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

*µµ*

*******is 9 x 10****3****/cm when the wavelength is 0.5**

**m. The***φφ*

**distance when the light intensity is dropped by 100 is x = ln(***φφ*

**Lo/***αα*

**)/**

*******=**

**ln(100)/(9 x 10****3****/cm) = 5.1 x 10****-4****cm.*** Example 4.8: A silicon sample has a thickness of 100 *µµ

**m and a****conductivity of 10 S/m. If we assume that light is absorbed primarily within *** 2 *µµ

**m 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: (*σσ

*σσ*

**1t1 +***σσ*

**2t2)/(***µµ*

**1 x (t1 + t2)) = (2**

**m x 1000***µµ*

**S/m + 98***µµ*

**m x 10 S/m)/(100**

**m 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. **

**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. **

**Example 4.19: Compute the photoconductivity in a p-type *** photoconductor if the hole lifetime is 50 *µµ

**s and the length of the*** photoconductor is 1000 *µµ

**m. Assume the average velocity of the holes is 1**

**x 10****5****m/s.****Solution: **

* The transit time of the photoconductor is *ττ

**t = L/v = 1 x 10****-3****m/10****5****m/s = 10***ττ*

**ns. The photoconductivity gain is***ττ*

**p/**

**t = 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 *µµ

**s**

**(twice the sampling rate). For 1/4 million pixels, the total time delay is:**

**0.25 x 10****6****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 90****o**** twist if the maximum angle between the layers is ****12****o****. **

**Solution: **

**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 10-7*ΩΩ

**.m. Metallic amorphous***ΩΩ*

**Fe-related solids have a value of about 10****-6**

**.m, whereas a ferrite has a***ΩΩ*

**value of about 100**

**.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 - H****2****, 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 + 0.1*ββ

*ββ*

**) - (1 + 0.03**

**))/***ββ*

**(1 + 0.03***ββ*

**) = ((1 + 0.1 x 117**

**) - (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. **** **

**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 *µµ

**m, 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 *µµ

*µµ*

**m x 10 = 1**

**m. Speed =***µµ*

**wavelength x frequency = 1 x 10****-6**

**m x 2 x 10****3****/s = 2 x 10****-3****m/s.**