Ch11Sol - Chapter 11 Problem Solutions FIND Why are metals...

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Unformatted text preview: Chapter 11 - Problem Solutions FIND: Why are metals opaque, yet other materials may be transparent or translucent? SOLUTION: Metals have no band gap. Incident radiation is absorbed by free electrons. Metals do not allow visible light to pass. FIND: Identify a transparent conductor and the color of conductive polymers. SOLUTION: There are no transparent conductors. Some glasses can have reasonable conductivity and be transparent, but not high conductivity. When polymers with highly conjugated structures are doped with an electron donor of acceptor, the polymer becomes conductive and looks like a metal. FIND: Identify which materials are dielectrics? SOLUTION: Materials with large band gaps and high electrical resistivity are dielectrics - BN, epoxy, and diamond. Sic is a semiconductor but probably has a relatively high impurity level with a lOQ-cm resistivity; CBS is a semiconductor. Ca and Sn are conductors. FIND: Identify the types of polarization in PVdF, NaCl, Na, diamond, PE, and alumina. SOLUTION: There are four types of polarization: 1. electronic, 2. ionic, 3. molecular, and 4. interfacial. Ceramics can exhibit all types, metals only the first, and polymers the first and third. This is the scorecard: terial Polarization Mghanisms PVdF l and 2 NaCl 1-4 Na 1 C 1 PE 1, 3 weak alumina 1-4 FIND: Which dipole has the large moment, C-F or CaN? DATA: The electronegativities are C = 2.55, F = 3.98, and N = 3.04; the bond lengths, which can be obtained (apprordmately) by adding covalent radii, are C-F = 1.49 and C2 N = 1.16A. SOLUTION: The electronegativity difference multiplied by the charge separation is largely responsible for the dipole moment. 1.43 x 1.49 = 2.13 for C-F and 0.49 x 0.57 for CsN. C~F has the larger moment. 277 FIND: Calculate the polarization in tin. GIVEN: The electronic displacement is 2 x lO"°nrn DATA: Tin is BCT; however, it can be reasonably well approximated as FCC with a lattice parameter of6.49A3; atomic number 50; 1.6 x 10‘19 C/e ' SOLUTION: z = (4 atoms/cell)(50e/atom)/a03 = 0.73 3/a3. Thus, P = qu = (0.73 x 103° e/m3)(1.6 x 10"”C/e)(2 x 10-91:.) = 2.34 x 10‘ (:Im2 FIND: The average separatio of charge when copper is placed in an electric field GIVEN: Polarization is 8 x 108 Gina2 DATA: Cu is FCC with a lattice parameter of3.615A; atomic number 29; 1.6 x 10'19C/e 8 x10'8C/m3 d = P/Z 2 m q (2. 455 x 103°e/ mam: x 10-” C/e) =2.04x10"A SOLUTION: z = (4 atoms/cell)(29 e/atom)/a.,’ = 2.455 e/A3 Thus, FIND: Calculate the ionic polarization of MgO GIVEN: The applied electric field causes a displacement of 5 x 10'9 nm between ions. MgO has the NaCl structure. DATA: 4 ions of each type per cell; a, = 22(er + r02.) = 2(0.66 + 1.32) = 3.96A; 1.6 x 10"9 C/charge; 10 electrons in each ion. SOLUTION: Each ion carries a charge of +_2 . there are 4 ion pairs/cell. Thus, Z = (4 dipoles/cell)(20 charges/dipole)/a.,3 = o. 13 charges/A3. Hence, P = qu = (0.13 x 10” charges/max] .6 x 10'19 C/charge)(5 x 10‘“ m) = 1.04 x 10-7 (:Im2 FIND: Percent increase in separation of ions in ZnS. GIVEN: Applied field causes ionic polarization of 7 x 10'8 cm2 DATA: 22118 has the zinc blend structure, 4 ion pairs/unit cell; 1.6 x 10':9 C/charge', aoJE/4= rm+ + r32- = 0.74 +1.84 2:) a0 = 5.96A :s a: = 212 x10'30 m3 SOLUTION: z = (4 dipoles/celer charge 2/dipole)/a.,3 = 7.56 x 102‘ charges/m3 P = qu :> d = Pqu = (7 x 10* C/m2)/[7.56 x 102‘ charges/m3)(l.6 x 10'” orchargen = 5.8 x 10'“ m = 5.8 x 10'8A. Expressing on a fractional basis: 5.8 x 10'3/(rm+ + r52.) = 2.25 x 10'8 = 2.25 x 10‘% COMMENTS: Very small displacement! 278 10. ll. 12. 13. FIND: Voltage that develops when a hydrostatic pressure is applied to PbZrOs. GIVEN: PbZrOs is piezoelectric SOLUTION: No voltage necessarily develops because the pressure, being” hydrostatic, does not cause charge displacement in the unit cell. Only the scale of the cell is reduced, ala Hooke’s Law, in each direction. COMMENTS: If pressure were applied normal to the appropriate face of the single crystal, voltage would develop. FIND: Describe the properties of PP fiber with imbedded ions. GIVEN: The ions are embedded in molten PP. ' ASSUMPTIONS: PP is a perfect insulator. SOLUTION: The imbedded ions, both negative and positive, provide the fiber with locally charged areas. The fiber is charged for its lifetime. It will attract charged particles fi'om the atmosphere, such as dust. COMMENTS: Fiber with embedded charge is used for dust masks. FIND: Voltage required to produce a polarization of 10‘7Clm2 GIVEN: Material is polycrystalline alumina, l x 104m thick. DATA: J = 8.85 x 10%" m. is about 9 from Table 113-]; the dielectric strength of alumina is > 7 x ro‘Wcm. x=1+P/_a§__P=(rc-1)_o§=(x-1)_9Wd“ V = PD/(rr -1) =W= 0.126V —° (9)(8.85 x 10‘” F/m) SOLUTION: Using equation 11.3-4 as modified in example problem 11.3-2, This translates into a dielectric stress of 0. 126V/0.01 em = 12.6 V/cm, which is insufficiently high to cause breakdown. FIND: How much thicker must a sample of mica be to achieve the same electric polarization as mica exposed to the same electric field? DATA: The dielectric constant of PE and mica are about 2.3 and 7.0. ,_ SOLUTION: In mica the polarization is due to both ionic and electronic mechanisms. In PE only the electron mechanism is active, so the dielectric constant is lower. Taking a ration of the P = (r: - l)_.,§ expression: 279 I4. 15. 16. 17. - 1 g 2.3 - 1 E / d / m: KPH 3:2 PE = I: :2 PE PE PM P” [MWIJLOXEWJ [7-0-1][_0][Em/dm] FIND: Does PAN or PVC have a higher dielectric constant? GIVEN: PAN has a higher dipole strength. DATA: Electronegativities for C, N, and C1 are 2.55, 3.04, and 3.16; bond lengths, 1, for C-Cl and C s N are 1.77 and 1.16A, which can be obtained (approximately) by adding covalent radii; ppm = 1.15 g/cm3 and ppvc = 1.4 g/cm3. SOLUTION: I am going to do this problem backwards. You want to start at the bottom. _=_o+P/§3 P=L"“_O)E.n but “ = Cd/A :> P = (C - Cn)(de)5, but a = V/d => P = (C - COXd/AXV/d) = (CV - COW/A, but Q = cv :> P = (Q - Q.)/A. This expression says that the change in polarization is equal to the extra charge per unit area that can be stored because of the presence of a material between the plates in a capacitor. FIND: Should you build a capacitor fi'om cordierite or PVC? How large will it be? GIVEN: Space is limited. Capacitance need to be 0.05 uF. DATA: 1: is 5.0 and 3.2 for cordierite and PVC, according to Table 113-1. Fields not high; * = 3.35 x 10'l2F/m. SOLUTION: C = _(A/d) = 1c“.,(A/d). the material with higher It is desired. Substituting C=5x10‘3F=5.0x8.85x10'13F/m04/a) _A/d=1.13x10’m for K and Mo: COMMENTS: All we can give for a dimension is Aid. If we knew the voltage to be used, we could put a lower bound on d by implementing the requirement to avoid breakdown. FIND: Show the capacitance of a multiple plate assembly is C = Ic__.,(n-l)A/d. GIVEN: There are n plates separated by dielectric. SKETCH: 280 18. 19. 20. 21. SOLUTION: Count them, there are n plates and n-1 cells. Each cell has characteristic separation d and area A. For one set of plates, C = 1c_.,A/d, so for n-l sets of plates, C = K_,,(n-l)A/d. FIND: Calculate the relative dielectric constant of mica. GIVEN: The sample is placed in a field, E, of 1 lem and the measured polarization, P, i 5.4 x 10-8 (3/1112. ‘ DATA: The relative dielectric constant of mica, given in Table 113-], is K = 5.4 - 8.7. The permittivity ofvacuum, so, is 8.85 x 10-12 F/m. SOLUTION: Using the expression for K developed in Example Problem 11.3-1: K = 1 + (P/aog) = 1 + (5.4 x 10-8 C/m2 )/ [(8.85 x 10-12 F/rn)(103 V/m)] = 7.1 COMMENTS: This is a reasonable value. FIND: Calculate the field strength required in PE. will the PE survive? GIVEN: P = 10'7C/m2 DATA: K = 2.3; _, = 8.85 x 10‘12 Pint; dielectric strength w 200 x 106 V/cm 10" C/ m2 g = PM ' 1L" = (2.3 - 1) (8.85 x 10-” F/m) =8.7x10’V/m SOLUTION: x = I + P/__o§ :> = 87 V/cm << dielectric strength of PE FIND: Why might the dielectric constant of a polycrystalline ceramic decrease with frequency? SOLUTION: x = l P/_.,E,, so as P goes, so goes it. Let us examine P. The factors that contribute to the polarization of a ceramic are electronic, ionic, and space charge. Electronic is generally not frequency dependent; however, 10“5 Hz is very fast switching. Ionic polarization generally shows frequency dependence in this region, since ionic masses may be large. since the material is polycrystalline, space charge efi‘ects can also be significant. FIND: voltage required to develop charge in capacitor. 281 22. GIVEN: C = 2.5 x 10"”C; plates 20 x 20 nm; (1 = 0.01 mm; dielectric: vacuum or PTFE. DATA: mm... = 1; mm = 2; ,0 = 8.85 x ions/m SKETCH: SOLUTION: Q = C x V and C = x“.,(A/d). Thus, Q = K_.,V(A/d) :> V = dQ/K_.,A (10"m) (2.5 x 10""C) _m = a. 71V (8.85 x 10-” F/m) (4 x 10“ m2) Vvacuum = Since the dielectric constant of PTFE is double that of vacuum and 1c appears in the denominator, the voltage will be just half: Vm = 0.36V. FIND; What is an appropriate material and wall thickness for spark plug wire? GIVEN: The voltage carried is 15,000 V. SKETCH: SOLUTION: A spark plug wire needs the following properties: high dielectric strength (to avoid breakdown), low dielectric loss (to avoid losses and heat buildup), flexible, resistant to gas and oil, heat resistant, durable and long wearing. Silicone has a dielectric strength of 220 x 10‘ V/cm, low dielectric loss, and all the other required properties. It is the material of choice. The minimum thickness can be assessed by taking the ratio of V to dielectric strength: 1sooow(220 x 10‘ V/cm) = 6.8 x 10'5 cm COMMENTS: The actual wall thickness is 2—5mm, which adds a degree of safety and durability. 282 23. 24. 25. 26. 27. 28. FIND: Why might the dielectric constant of a polar polymer be temperature dependent? SOLUTION: Increasing the temperature may allow the polar groups more“ mobility, increasing the dielectric constant with temperature. Increasing the temperature is equivalent to decreasing the frequency. Both effects give the dipoles more time to respond. COMMENTS: We find that most polar polymers show time and frequency dependent dielectric properties. FIND: Why is the dielectric constant of PE frequency independent, whereas that of PVC is not. SOLUTION: The solution to this problem is contained in that of number 21. PE shows only electronic polarization. The electron’s response is so fast the PE shows no frequency response at all but very high frequencies. PVC contains dipoles that do not respond as quickly as do electrons. FIND: What is the approximate value of x for Al? 7 SOLUTION: Metals cannot store separated charge, so C a: 0 and x e 0. FIND: Is dielectric strength an intrinsic property? relative permittivity? polarization? SOLUTION: Dielectric strength, like mechanical strength depends on the weakest link. Hence, it is an extrinsic property? Permittivity and polarization are intrinsic properties. FIND: Bakelite® or nylon, which generates more heat in a fluctuating electric field? GIVEN: Work, W 0: s x tan 8. W is displayed as heat. DATA: According to Table 114-1, tan 6 for bakelite and nylon are 0.08 and 0.01. SOLUTION: The heat generation in Bakelite® will be 8 times higher than that in nylon. COMMENTS: A a distributor cap in a car does not experience a high the field, so the loss is not large. In addition, the part is generally subjected to a temperature dictated by its proximity to the engine. More important is the resistance to gasoline and oil, its ability to maintain shape at temperature, its durability, and, of course, cost. FIND: Maximum voltage that can be used in it PS capacitor so that power loss does not exceed 0.1W. GIVEN: Capacitor size is 25 x 25 x 0.01 mm; frequency is 106 Hz. DATA: For PS i: = 2.5 and tan 8 = 0.0003. SOLUTION: We use equation 11.4-1 to solve this problem: W a: rt__.,f‘é2tan6. Substituting for E, = V/d and solving for V, we get V = [Wd2/1t_o.,ftan8] m. We note that in this equation, W is per unit volume, so we need to normalize to sample volume, A X d, 283 29. 30. 31. V = [Wdz/Adzgf *,0: tan 51‘” m [Wd/Adztf _,x tan a j” 122 J = 277V COMMENTS: We should check that this is not in excess of the dielectric strength of the PS: 2.4 x 101°V/m x 10‘5m = 240,000 v. No problem. z 0. 1(10") (2.5 x 10-2 f 3. 14( 10‘ ) (8. 85 x 10-”) 2.5 (00003) FIND: Is PE 3 good choice for material to be used for a container in which to reheat food in a microwave oven? If not, suggest a better material. DATA: According to Table 11.4-1, tan 6 for PE is very small. Heat generation in the PE will be minimal. SOLUTION: Oils can reach very high temperature in a microwave oven. PE melts below 140°C, and it has been reported to melt in a microwave oven when in contact with oils. The ideal material generates little heat, is stable to about 180°C, does not absorb smells, does not degrade, does not release chemicals into the food, and so on. COMMENTS: To impart the function of browning to the top, you need a material that absorbs heavily in the microwave region and re—radiates to cause browning. Few ceramics absorb heavily, so a browning top may be a composite material. FIND: Calculate e’ for PC. GIVEN: Use the data in Tables “3—1 and 11.4-1. DATA: 1: = 3.0 and tan a = 9 x10'4. so = 8.85 x10'12 F/m SOLUTION: We solve the problem using equation 1 1.4-3: tans = e'leo :> s' = 9x 10-4 x 8.85 x10'12 F/m = 8.0 x 10-15 F/m. COMMENTS: Recall that tan 6 is frequency dependent. The K and tan 8 values used are at a specific frequency. FIND: Determine the refractive index of PP GIVEN: The critical angle in it is 42.86” SKETCH: SOLUTION: Using the concept developed in sample problem 11.5-1 based on equation 11.52: in, = sin" (10.) => n = (sin a)" = (sin 42.80")'1 = 1.470 284 32. 7 33. 34. 35. 36. FIND: Why is firsed silica clear and window glass green when viewed on edge? Assumptions: One common impurity in glass is iron. Iron makes oxide silica glasses appear green, such as in Coke® bottles. FIND: Determine the angle of light passing through glass and PE in layered composite material. GIVEN: Materials are Soda—lime-silicate glass and polyethylene; angles of incidence are (a) 10° and (b) 70°. The thickness of the PE film is 0.2 mm and that of the glass is 1 mm. DATA: For PE and this glass, 11 = 1.54 and 1.51 SKETCH: SOLUTION: All angles are defined relative to the surface normal. The angle of incidence is (a) 20° and (b) 70". we first need to determine the angle of propagation through the PE film on top, then use this, the angle of reflection, as the angle of incidence for the glass. The thicknesses are irrelevant information for this problem. Proceeding for the PE layer at 20° sin i/sin r= 112/1113 sin I = 1.54 sin 20“ :> r = 31.7” For the glass layer 31 .7“ is i: sin i/sin r = 113/112 :> sin r = (151/154) sin 31.7" :> r = 31.1" Part (b) is done exactly the same way; however, 70" is greater than the critical angle, so the light never enters the PE (1' = 90°) or the glass. FIND: Calculate the reflective index normal to the axis of a PET fiber I GIVEN: Birefringence is 0.90 and nun-1.. = 1.660 Assumptions: We assume that the fiber is isotropic in its cross-section , SOLUTION: Using the basic definition of birefiingence: An = n.-n,, rearranging and substituting, nr = n. - An = 1.660 — 0.090 a 1.570 FIND: Show how a headlight reflects from dry and wet pavement. GIVEN: The road is rough on a scale of the wavelength of light. SOLUTION: The light incident on the rough or dry road is in large part absorbed, with the remainder being reflected diflirsely. 0n the other hand, the fight from the wet road will 285 37. 38. 39. 40. 41. be reflected specularly, hardly illuminating the road and blinding oncoming cars. FIND: What is required for a transparent object to disappear when immersed in water? SOLUTION: To disappear, the object must stop reflecting light off its surfaces. Hence, the refractive index difference between the object and water must be zero. The refractive index of the material in the object must be 1.33. COMMENTS: There are no limitations on shape, but a hollow object cannot disappear. FIND: Explain why optical waveguides are fabricated with a refractive index gradient. GIVEN: Light sources are not perfectly collimated. SKETCH: See Fig. 11.5-6 SOLUTION: Light that travels along the center of the fiber in a straight line takes a shorter route than light that efi'ectively bounces 011‘ surfaces. A refractive index gradient is used to speed up the light that takes the longer paths, thereby insuring that all light sent at any instant of time arrives together at the destination. FIND: How does 7t. depend on n? SOLUTION: referring to equation 11.5-2, we note that n = Jim/llama“ ::> ll. material = lam/n. Since n > 1, the wavelength of the light decreases as it passes from air into a material. FIND: Critical angle of light traveling into and out of water. DATA: From Table 115-2, the tell-active index of air and water are 1.0 and 1.33. Assumption: The water is quiescen SKETCH: ' SOLUTION: The concept of critical angle applies only when light is traveling from a medium of high to low index. (You can try to work out the mathematics for the reverse and you will need to take the inverse sine of a number larger than 1). Therefore, we need calculate the solution only from a fish’s point of view. Mum = sin i/sin r. Ar the critical angle sin r = 1‘ Sin ic = l.0/1.33 z) ic = 48.75”. COMMENTS: Next time you are swimming underwater, try the experiment. FIND: Can a fish see you on shore or in a boat? DATA: n for water is 1.33 and for air is 1.0. 286 42. 43. SKETCH: '\ SOLUTION: We use equation 11.5-3 to work this problem: vwater / vair = nai, I nwater = sin i / sin r 1.0/ 1.33 = sini/sinr. Continuing as in Example Problem 115-], sin r = 1.33 sin i. Hence, r = sin"1 (1.33 sin i). This expression requires that 1 2 1.33 sin i. Hence, 7 sin ism = 1/133 :> i = 48.8". So long as the angle from which the fish looks at the point on the surface is less than 48.8" , the fish can theoretically see out of the water. When i = 0°, r = 0°; when i = 48.8°, r : 90°. Hence, the limitation is on the fish's view of the surface, not your position on land or in a boat. COMMENTS: I don't know enough about fish vision to know just what fish do see. FIN D: Determine the reflectivity from ice and water at normal incidence DATA: From Table 115-5, the refractive index of ice and water are 1.31 and 1.33 2 SOLUTION: Fresnel Formula is 12:93?) For ice,R=[0.31/2.31]2= 1.8%, For n water, R = [033/233]2 = 2.0% FIND: Show the principle of a refractometer GIVEN: It uses the principle of critical angle SKETCH: 287 45. SOLUTION: Light is collimated and directed to propagate through a sample film. The angle of propagation is increased until the detector, which sits at 90°, senses a strong intensity. COMMENTS: Monochromatic light at different wave lengths can be used to determine dispersion as well. FIND: Design a waveguide assembly that can bear high tensile loads. GIVEN: Seven optical fiber are required. DATA: Pound for pound, Kevlar® and Spectra® fibers are stronger than steel. SKETCH: SOLUTION: this is just one of many possible solutions. The large fibers are the optical fibers, which are coated with a protective polymer. The small fibers are the polymeric reinforcement fibers. The entire assembly is protected with another polymer coating. FIND: Why are immersion lenses used in optical microscopy. DATA: The refractive index of oil is greater than that of air. SKETCH: - ,. . SOLUTION: Light travels from the sample on the stage upwards through the cover slip through air and into the glass compound objective lens. Reflection will occur at each surface. The oil fills the space between the cover slip and the objective lens. If the indices match - that of the oil equals that of the glass - then no retraction occurs and more light enters the objective lens. (Normally as you increase magnification, your light intensity decreases with the area of viewing.) COMMENTS: With better dry objectives - better coatings, higher aperture - oil immersion techniques, which can be messy, are being used less and less. 288 46. 47. 48. FIND: Estimate the amount of light that would reach the film in a camera that uses a 5 lens assembly. GIVEN: Lenses are not treated with anti reflective coating. DATA: From Table 115-5, Miami,“ = 1.51 Assumptions: The glass is a soda-lime-silicate of n — 1.51. Normal incidence. SKETCH: R=[”' ]=(o.51/2.51f =95.87% n+ 1 SOLUTION: Light is reflected from each lens. Assuming normal incidence, is transmitted from each of 5 lenses (0.9587)’ = 0.81 = 81% is transmitted to the film. COMMENTS: This effect is significant and cameras often have more lenses than this. In addition, we have ignored the effects of the reflected light. Consider, for example, light inflected off an interior lens and then reflected back by a prior lens onto the film. The image will not be sharp. FIND: What glass makes the best beveled glass window? GIVEN: The purpose of the bevel is to produce mini rainbows inside the house. SKETCH: See Fig. “5-5 SOLUTION: Unfortunately refractive index does not determine which glass provides the best rainbow. The cut of the glass and the optical dispersion are the critical issues. for any given geometry of cut, select the glass with the highest dispersion. COMMENTS: Lead glass is used for crystal. Because of its high refractive index, it has high reflectivity. It sparkles. FIND: Why can people who normally wear glasses see better under water? SOLUTION: Light traveling through your eye is traveling though an aqueous solution. When it emerges from your eye into the air, it is refracted. Ifit goes into water from your eye, refraction is greatly reduced. Errors in the eye's lens are magnified when the interface facilitates significant light refraction, since the light rays passing through water subtend a 289 49. 50. 51. 52. smaller angle than do those passing through air. COMMENTS: This is the principle of oil immersion microscopy. FIND: What determines the color separation of the sunlight passing through beveled glass? SKETCH: Refer to Fig. 11.5-5. SOLUTION: The color separation is determined first a few factors: 1. The refractive index of the glass 2. The perfection of the surfaces of the glass 3. The distance the light travels through the glass 4. The various incident angles 5. The distance between the lens and the point of analysis. COMMENTS: It is beautiful when the light is resolved by a beveled glass window. The best windows are "crystal", or high lead silica glasses. FIND: Why does light effect only electronic polarization? SOLUTION: The response time of a process is largely effected by the masses that must move. The larger the mass, the longer the response time. Only electrons with their small mass can respond to frequencies associated with light in the visible region. FIND: What is the maximum wavelength to stimulate photoconduction in tin? GIVEN: The bandgap in tin is 0.09 eV DATA: 1.6 x 10'191/eV; c = 3 x 10“ n/s; h 2 6.62 x 10'3‘J-s SOLUTION: Equation 11.7-2 states that E = helix. = E; :> a = hc/EG = (6.62 x 106‘ J-s)(3 x 103 m/s)/(l .6 x 10'19 J/eV)(0.08 eV) = 15.5 urn =15512.5 nm COMMENTS: This is longer and hence less energetic than the visible region. Hence, energy from the visible region is absorbed, leading to the generalization that electrical conductors are opaque. FIND: What materials are transparent? SOLUTION: High band gap materials are transparent and we can calculate the bandgap necessary to avoid absorption of visible light. From Fig. 11.1—1, the most energetic visible light has a wavelength of 0.4 pm. We substitute this into E = hell and solve for E: EG = (6.62 x 10‘“ J-s)(3 x 10" m/s)/(1.6 x 10"” J/eV)(4 x 10'1 m) = 3.16v All materials with bandgaps greater than 3.1 eV have the chance of being transparent to visible light. ' 290 53. 54. 55. 56. 57. COMMENTS: As you know, other efi'ects can prevent a materials from being transparent. FIND: Might the materials in problem 441ase? SOLUTION: To lase materials needs a AE that corresponds to an energy of 0.4-0.7 um, which assessed using E = hem. We calculated in problem 44 what bandgap is required to absorb radiation at 0.4 pm = 3.1 eV. For it = 0.7 urn, the bandgap is 1.8eV. We need to have 3 energy levels, with an energy rise of 1.8 - 3.1 EV from ground to a metastable excited state. From the table, only tin cannot lase in the visible and it may not be possible with CDS. FIND: Determine the concentration of Fe ions in a Coke® bottle. GIVEN: The extinction coefficient is 573 mm'1 Assumptions: Assume a Coke® bottle is about 3 mm thick and about 20% of incident light is transmitted. SOLUTION: The Beer-Lambert equation, 11.6-8 is 111., = exp(-_cx). Substituting values, 0.20 = exp(-573 x c x 3). Taking the natural log of each side and solving for c: c = ln(0.20)/(573 x 3) = 0.94% FIND: Why is beer normally sold in brown or green bottles or cans? GIVEN: Beer degrades in UV light SOLUTION: Clearly, the brown or green glass absorbs UV radiation, thereby preventing it from reaching the beer. COMMENTS: When you buy beer in clear bottles, make sure that it has not been on the shelf long. FIND: Might polymers be photosensitive glasses? SOLUTION: The silver halide crystals in photosensitive glasses disassociate. The fragments need to remain relatively close to one another so that they can recombine in a reasonably short period of time. Since the diffusion coefficient of ions in polymers may be high, the fragments may move far apart. Glasses with low diffusion coefficients need to be selected. COMMENTS: Some photochromic polymer glasses have been made, but the colors were not pleasing. FIND: Sunburn through a car window? SOLUTION: Sunburn is chiefly caused by UV radiation that gets through the Earth’s atmosphere. Fig. 11.6-1 shows that dielectric materials, such as oxide glasse and ceramics, are generally opaque to UV light. Hence, a sunburn will take many long hours behind ordinary glass. COMMENTS: The bare arm in the open window will tan much more rapidly. 291 58. 59. 60. 61. FIND: What are the required characteristics of a high power laser window? SOLUTION: No absorption at the lasing frequency, since high power will give rise to high heating. No scattering to attenuate or degrade the beam’s collimation. Stable to the environmean conditions. COMMENTS: Materials that satisfy these requirements are hard to come by. FIND: Plot optical transmission as a function of grain size for silica crystals. SKETCH: 1 Transmitttvity Grain Slze SOLUTION: The sample will pass the light at small grain size and large grain size. COMMENTS: Many gemstones, which are often single crystals, and Lucalox®, which is small grained alumina, are transparent. FIND: Explain how sol-gel based glassy silica can be both transparent and porous. SOLUTION: Typically in these zerogels, as they are called, the pores are very small, on the order of 100A. The pores are so small that they do not scatter visible light. COMMENTS: These glasses have potential for use as window glass, since the thermal conductivity is very low. FIND: How does doubling the concentration of a chromophore afi‘ect the absorption of light at the absorption fiequency? GIVEN: At a concentration of 0.001 g/cm3, the intensity decreases 10 %. ASSUMPTIONS: All the intensity decrease is due to the chromophore. SOLUTION: Equation 11.6-8 shows the relationship between intensity and concentration: I / Io = exp (-ecx). Hence, when c = 0.00! g/cm3, I/ I0 = 0.90. Hence, ex = 105.4 cm3/g, a constant in this problem. Thus, substituting the new values, m0 = exp (—ecx) = exp (405.4 x 0.002) = 31 %. The intensity decreases 19 % when the concentration is doubled. COMMENTS: The problem could have been completed faster by using the ratio technique used throughout the text. 292 62. 63. 65. 66. 67. 68. FIND: Why is snow white? SOLUTION: Snow is a collection of small ice crystals. The total surface area of ice crystals in snow is enormous. Light is scattered from the surfaces of the crystals, giving snow a white appearance. Thus, even though ice is transparent, new fallen snow is white. COMMENTS: If you shine a blue light on snow, it will appear bluish. FlN D: Why are there no K; for elements with low atomic number? SOLUTION: A K; implies an outer as well as an inner electron shell. The elements of . interest have so few electrons that only one shell is occupied. FIND: Why do some materials glow white in UV light? SOLUTION: Many materials contain “impurities” that fluoresce in the UV, that is, they absorb UV radiation and re-ernit in the visible, causing them to glow. COMMENTS: To paper and other materials often are added optical brighteners that give this effect and make the material look “whiter”. Fluorescence microscopy uses this effect too. FIND: Why are household mirrors silvered on the back side and most laser mirrors silvered on the front surface? SOLUTION: Lasers are usually used for their special properties - monochromatic, coherent, well-collimated. If the laser passes through a material, these properties can only degrade. In addition, any losses in the glass will cause it to heat up. Even low losses in high power lasers can be important. FIND: How does laser or LED light interact with a prism? SOLUTION: Both lasers and LED’s are monochromatic sources. Thus, the light cannot be firrther resolved into wavelengths. It simply retracts as it passes through the prism. FIND: How long will it take for the intensity of a phosphor to degrade to Lie? GIVEN: The intensity after 10 sis 1J2. SOLUTION: Using the equation: 1/1., = exp(-t/1), where ‘t is a relaxation or decay time. We use the given information to calculate 1:: 0.5 = exp(-lO/1:) :> t = 14.4 3 Hence, when t = 14.4 sec, I = lJe. FIND: What is required to make a colored LED? DATA: h = 6.62 x 10‘34 1—3; c = 3 x ro‘m/s; 1.6 x 10'” MN SOLUTION: We need to make a p—n junction with an E5 that corresponds to an emission at wavelengths of 5.3 (green), 6.8 (red), and 4.6 (blue) all x 10'7m. The conversion from AEG is done using: HG = rte/7t = (6.62 x 10'3‘J-sx3 x lO’m/s)/(l .6 x 10“”J/evxr in 10'7m) 293 69. 70. 71. 72. a 2.3 (green), 1.8 (red), and 2.7 (blue) eV FIND: What is a suitable detector for laser light? SOLUTION: An LED, since electrons will be excited, producing a voltage that can be amplified. FIND: What property of a laser are utilized in bar code readers? SOLUTION: Presumably a He-Ne laser beam is split. Half the beam reflects OR the bar code on the package and it is somehow recombined with the reference half. The coherence of the laser allows a pattern to establish between the two beams that can be sensed. FIND: Show how doping with different ions produces different color emission. GIVEN: ZnS doped with Cl' is violet and doped with Cu” is red. :1 SKETCH: II] gap-f- Cl ' levels SOLUTION: The dopant introduces new levels. The new gap in the Cl' doped ZnS is larger than that of the Cu2+ doped ZnS, since the latter emits at a longer wavelength. Note that Cl' is doner-type and Cu2+ is acceptor-type. FIND: What is the operating principle of a UV sensor for estimating temperature? SOLUTION: You have probably seen pictures of houses and other objects that show, via a color map, the sources of heat loss. Those sources are generally shown as the red areas, the ones that are "hot". When an object is above OK, it emits radiation characteristic of its temperature. An instrument that can measure the wavelength of the emitted radiation can be used to determine an object's temperature. This is the principle of black body radiation, which you can study in heat transfer classes. COMMENTS: The same principle is used to image soldiers using night vision glasses, to find people in smoke-filled rooms, or to non-intrusiver measure the temperature of a moving hot fiber. 294 ...
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This note was uploaded on 09/22/2011 for the course MSE 2001 taught by Professor Tannebaum during the Fall '08 term at Georgia Tech.

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Ch11Sol - Chapter 11 Problem Solutions FIND Why are metals...

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