材料科学基础课后习题答案10.doc

24Solutions for Chapter 101.FIND: Give specific applications for insulators, conductors, and semiconductors in a CD player.SOLUTION: My father gave me a broken CD player yesterday. I opened it up last night to peer into its workings. I saw a GaAlAs laser/sensor (the part that is not working). I saw many sorts of wires - conductors - generally covered with various materials (polymers) for insulation. I saw a number of resistors, chips, diodes, which contain insulators, conductors, and semiconductors. Several printed circuit boards were in use. 2.Find: Cross sectional area and applied voltage in a polymer rod Given: s = 3.5x10-8 (W-m)-1 , L = 10 m, R = 0.08 W, I = 3.2 A Data: Equation 10.2-1: V=IR Equation 10.2-2: r=(A/L)R Equation 10.2-3: s=1/r Solution: Combining equations 10.2-2 and 10.2-3 gives (1/s)=(A/L)R Solving for A gives A=(L/sR) Substituting value given in the problem statement A= (10m)/(3.5x10-8(W-m)-1)(0.08W) = 3.57x109m2 To find the necessary voltage we use Equation 10.2-1 V=IR=(3.2A)(0.08W)=0.256VIn comparison, Example Problem 10.2-1 shows that a metal rod with the same length and resistance would have a diameter of 3.57x10-7m2 (0.357mm2). The sixteen order of magnitude difference in the areas of the metal and polymer rods is a direct consequence of the corresponding values of their electrical conductivities.Comment: This polymer rod has a diameter of 67.4 kilometers!3.FIND: Which class of materials shows the greatest range of electrical conductivity? GIVEN: Data are provided in Table 10.2-1. SOLUTION: Polymers cover the range of 10-18 to 10-11 (-cm)-1. Metals span the narrow range of 104 - 105 and ceramics the wide range of 10-14 to 105 (-cm)-1. When organic conductors, graphite, or doped polymers are included, then the range of organic materials increases on the high end to about 105 (-cm)-1.4.Find: Show that units of Nqm are equivalent with those of s. Given: 1 A = 1 C/s Data: Units for variables are N=m-3, q=C, m=m2/v-s, s=(W-m)-1. Solution: Nqm has units of (m-3)(C)(m2/V-s) = (C/s)/(V-m) = (A/V)/m = 1/(W-m) = (W-m)-15.Find: Relative conductivity of a metal glass and crystal. Given: Electrons transport charge and Nglass=Ncrystal Data: For a metal, s=Neqeme and (dme/drd)0.Solution: Since N and q are the same for the glass and the crystal, the question reduces to determining which structure has the higher electron mobility. Since the crystalline structure is much more regular than that of the glass, the electron mobility will be much higher in the crystal than in the glass. Therefore, we expect the conductivity of the metallic glass to be lower than that of its crystalline counterpart.6.Find: Proper form of conductivity equation for a material in which charge is transported by electrons and holes. Data: Equation 10.2-5: s=SiNiqimi Solution: For the material described above s = Neqeme+Nhqhmh = q(Neme+Nhmh)7.Find: Mobility of electrons. Given: Ne=Nh=1x1013cm-3=1x1019m-3 r=3.15x10-1W-m me=3mh Data: Equation 10.2-5: s=SiNiqimi and q=1.6x10-19C Solution: If both electrons and holes transport charge then s = Neqeme+Nhqhmh = Nq(me+mh) = Nq(me+(me/3) s = (1/r) = Nq(4/3)me me = 3/(4Nqr) = 3/4(1x1019m-3)(1.6x10-19C)(3.15x10-1W-m) = 1.49 m2/(V-s)8.Find: Band diagrams for conductors, semiconductors and insulatorsSolution: A conductor has a partially filled valence band at all temperatures. Both semiconductors and insulators have filled valence bands and empty conduction bands at O K. The difference between semiconductors and insulators is in the size of their bandgaps. Semiconductors have Eg 2.5 eV.9.FIND: Calculate the resistance of a rod of material 1 m long and 500 m in radius. GIVEN: The resistance of a sample of the same material 1 m long and 50 m in radius is 20 . SOLUTION: We can use equation 10.2-2 to solve the problem. Restating the equation in terms of R:R = (L / A).The length and resistivity of the sample is the same as that of the standard. Hence, the resistance varies with the radius squared. The samples cross-sectional area is 100 times that of the standard. Consequently, the resistance is 100 times less or 20 / 100 = 0.20 .10.Find: Show that f(E) increases exponentially with temperature. Given: Eg 0.2 eV Data: Equation 10.2-11 states: f(E)= 1 exp(E-Ef)/kT + 1 Solution: In a bandgap material Ef is located in the middle of the bandgap. Therefore, if we want to evaluate the occupation probability at Ec we find (E-Ef)=(Ec-Ef)=Eg/2. So the equation reduces to: f(E)= 1 exp(Eg/2kT) + 1 If exp(Eg/2kT)1 then we can neglect the 1 in the denominator and approximate f(E) as: f(E) exp(-Eg/2kT) which shows that f(E) varies exponentially with temperature. Next we must check the validity of our assumption. is exp(Eg/2kT)1 ? If we accept 10 in place of 1 than we require Eg/2kT 10 or equivalently T Eg/(2k ln10) Substituting the known value for k gives T Eg (2519 k/eV)if Eg=0.2 eV, our assumption is valid if T 504 K (231C); if Eg=0.5 eV, our assumption is valid if T 1230 K (957C); and if Eg=1.0 eV, are assumption is valid if T 2519 K (2246C). 11.Find: Explanation for different temperature dependence of r in metals and semiconductors Solution: In metals, charge is transported by electrons so that s=Neqeme. Since metals have a partially filled valence band at all temperatures, the Neqe product is not a function of T. Since me decreases with increasing T (due to more frequent collisions between electrons and vibrating ion cores) the conductivity of a metal decreases with increasing T. Thus, the resistivity of a metal (rm=1/sm) increases with increasing T. In semiconductors, me+mh deceases with increasing T but N increases exponentially with increasing T. This exponential increase in N overshadows the roughly linear decrease in mobility such that the conductivity of a semiconductor increases with increasing T. The result is that in contrast to metals (and other conductors), the resistivity of a semiconductor decreases with increasing T.12. FIND: Go to the library and ascertain a value for the electrical conductivity of pyrolytic graphite. Calculate the resistivity of a carbon fiber. GIVEN: The carbon fiber is 10 micrometers in diameter and its use length is 1 m.ASSUMPTIONS: Assume the conductivity of a carbon fiber roughly equals that of pyrolytic graphite. DATA: Diamond has a conductivity of 10-18 (-cm)-1. Graphites conductivity depends somewhat on structure, but graphite is a good conductor. Its conductivity is on the order of 105 (-cm)-1. SOLUTION: The resistance of a carbon fiber is approximated using equation 10.2-2:R = (L / A) = 10-7 -m 1 m / (5 x 10-6 m)2 = 1300 . COMMENTS: The current that flows under a potential of 12 V is determined using Ohms law (equation 10.2-1):I = V / R = 12 V / 1300 = 9 milliamps.Carbon fibers are sometimes used in composites to prevent charge build-up. 13. Find: Influence of T on R for the various classes of materials. Given: R(metal)=R(semi)=R(insulator) at 20C. which has highest R at 50C? Data: R = r(L/A) = (1/s)(L/A) Solution: For an insulator, s is a weak function of T (it is very low at all T). Therefore, resistance of the insulating sample will not change. For a conductor, s decreases with increasing T (see Problem 8) so that the resistance of the metal will be higher at 50C. For a semiconductor, s increases with increasing T, so the resistance of the semiconductor will be lower at 50C. Therefore, the metal will have the highest R at 50C.14. FIND: How can ReO3, a ceramic, have a conductivity as high as 5 x 105 (-cm)-1? SOLUTION: The basic equation governing electrical conductivity of materials is given in equation 10.2-4: = N q .To have a high conductivity, there must be a number of charge carriers and they must be mobile. Charge carriers may be ions or electrons, but generally ionic mobility is low. The high conductivity of ReO3 suggests electronic carriers are present. Detailed studies confirm this hypothesis. COMMENTS: Some ceramics structures have channels that act as paths for ions. Such fast ion conductors have conductivities in the range of 10-3 to 103 (-cm)-1.16. FIND: Calculate the resistivity of Cu at 200C and compare to that at 20C. DATA: See Table 10.2-2. Note the reference temperature for Cu is 20C. SOLUTION: We use equation 10.2-16 to complete the problem: (T) = o1 + e T).Proceeding for Cu,(T) = 1.67 x 10-6 -cm 1 + (6.8 x 10-3 / C) T(100C) = 1.67 x 10-6 -cm 1 + (6.8 x 10-3 / C) 80C = 9.1 x 10-7 -cm(20C) = 1.67 x 10-6 -cm 1 + (6.8 x 10-3 / C) 0C = 1.67 x 10-6 -cm. COMMENTS: Resistivity decreases with increasing temperature, which means that losses that cause heating decrease with temperature. A bar carrying a high current will heat up. As it heats up, the resistance decreases, so heat is generated at a decreasing rate. Hence, the bar may reach steady state before it melts.17. Find: Form of the conductivity equation for a solid with composition m2x. Solution: s = SiNiqimi = Neqeme+Nhqhmh+Ncqcmc+Naqama where the subscripts c and a correspond to cations and anions. If we assume N=Na=(1/2)Nc and Zc and Za are the valences of the ions then s = qeNeme+Nhmh+N(Zama+2Zcmc)18. Find: Influence of grain size on electron/hole conduction. Solution: A reduction in grain size increases the number of grain boundaries in the crystal. This increase in the defect density reduces me and mh but has no significant effect on either N or q. Therefore, a reduction in the grain size decreases the conductivity of most materials in which charge is transported by electrons or holes. 19. Find: Examples of materials to illustrate influence of rd on s. Solution: A. Materials in which an increase in rd results in a decrease in s include all metals and ceramic conductors in which charge is transported via electrons.B. An increase in rd tends to increase s in ionic conductors and, if the defects happen to be effective dopants they can increase the conductivity of semiconductors. In addition, some polymers can have their conductivity increased significantly by impurity additions.20. FIND: Plot the electrical conductivity of an epoxy loaded with ionic impurities as a function of extent of cure. SKETCH: SOLUTION: The impurities are mobile when the curing begins, as the resin is fluid-like. At the gel point, the ions lose a substantial part of their mobility. Eventually the epoxy becomes a glass, where all long-range motion is lost. COMMENTS: This technique is used to monitor the cure of thermoset resins in thick composites, such as submarine hulls. The sensors are small and thin electronic capacitors. They are used once each.21. Find: Discuss applications of superconductors. Solution: i) Low voltage power transmission- Superconducting wires will decrease the energy loss during long distance transmission and permit the use of low voltages to minimize health risks. ii) High speed computers- Superconducting components will permit faster computational operations by providing higher charge carrier mobilities. iii) Fusion- Power generation from a fusion reaction requires the use of extremely high temperature plasmas. These plasmas must be contained within an intense magnetic field. The role of superconductors will be to generate and maintain this field within economic constraints. iv) Battery-powered cars- Superconducting components will reduce the weight and increase the range of these cars to the point where they may be commercially viable. v) Low-friction trains- The Meissner effect is being used to minimize friction (and associated power loss) in high speed commuter trains.22. Find: List applications that favor each of the various types of conductors. Solution: A) Conducting polymers offer better specific conductivity values. Applications include weight critical components in the aerospace industry and in battery powered cars. B) Metals are favored in many routine applications including appliance cords. They are also the materials of choice when high conductivity values are the major design factor. C) Ceramic conductors offer advantages in high temperature applications and in aggressive environments.23. Find: Compare influence of fine and coarse precipitates on s. Solution: Defects influence s through their effect on the mobility of charge carriers (electrons). This is primarily a function of the number of obstacles and the average distance between obstacles. Since a fine precipitate distribution offers both mere obstacles and a shorter distance between obstacles, such a distribution will result in a larger reduction in the conductivity of the alloy. Comment: Note the similarity with the effect of precipitate size on the strength of the alloy.24. Find: Why is Ti a conductor while Ge is a semiconductor? Data: From Appendix B, the electron configurations for Ti and Ge are Ti: (1s22s22p63s23p63d24s2) Ge: (1s22s22p63s23p63d104s24p2) Solution: Although Ti and Ge both have 4 valence electrons, their electron configurations are very different. Ge is a Group IV element with the characteristic xs2xp2 configuration that results in sp3 hybridization and a corresponding filled valence band. In contrast, in Ti the 3d and 4s bands overlap so that Ti has a partially filled valence band at all temperatures. The result is that Ti is a metal and Ge is a semiconductor (since it has a small bandgap). 25. Find: Why is an electron in level Ed not a viable charge carrier? Solution: When an electron accelerates in response to an external field its kinetic energy increases. In order for an electron to accept this extra energy there must be an empty energy level located just above the filled level. This is in fact the case when the electron is located in the conduction band but is not true when the electron is in the donor level.26. Find: Characteristics of 1-2-3 superconductors. Solution: A. A 1-2-3 superconductor is an oxide with three types of cations an a 1:2:3 ratio. B. An example is YBa2Cu3O7-