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ME280 Homework #12 PROBLEM SOLUTIONS 12.5 (a) Calculate the drift velocity of electrons in silicon at room temperature and when the magnitude of the electric field is 500 V/m. (b) Under these circumstances, how long does it take an electron to traverse a 25-mm (1-in.) length of crystal? Solution (a) The drift velocity of electrons in Si may be determined using Equation 12.7. Since the room temperature mobility of electrons is 0.14 m2/V-s (Table 12.3), and the electric field is 500 V/m (as stipulated in the problem statement), vd = e E = (0.14 m2/V- s)(500 V/m) = 70 m/s (b) The time, t, required to traverse a given length, l (= 25 mm), is just t = l 25 103 m = = 3.6 10-4 s vd 70 m /s 12.31 Briefly explain why the ferroelectric behavior of BaTiO3 ceases above its ferroelectric Curie temperature. Solution The ferroelectric behavior of BaTiO3 ceases above its ferroelectric Curie temperature because the unit cell transforms from tetragonal geometry to cubic; thus, the Ti4+ is situated at the center of the cubic unit cell, there is no charge separation, and no net dipole moment. 12.D4 One of the procedures in the production of integrated circuits is the formation of a thin insulating layer of SiO2 on the surface of chips (see Figure 12.26). This is accomplished by oxidizing the surface of the silicon by subjecting it to an oxidizing atmosphere (i.e., gaseous oxygen or water vapor) at an elevated temperature. The rate of growth of the oxide film is parabolic—that is, the thickness of the oxide layer (x) is a function of time (t) according to the following equation: x 2 Bt (12.37) Here the parameter B is dependent on both temperature and the oxidizing atmosphere. (a) For an atmosphere of O2 at a pressure of 1 atm, the temperature dependence of B (in units of μm2/h) is as follows: 1.24 eV B 800 exp kT (12.38a) where k is Boltzmann’s constant (8.62 × 10–5 eV/atom) and T is in K. Calculate the time required to grow an oxide layer (in an atmosphere of O2) that is 100 nm thick at both 700°C and 1000°C. (b) In an atmosphere of H2O (1 atm pressure), the expression for B (again in units of μm 2/h) is 0.70 eV B 215 exp kT (12.38b) Now calculate the time required to grow an oxide layer that is 100 nm thick (in an atmosphere of H2O) at both 700°C and 1000°C, and compare these times with those computed in part (a). Solution (a) In this portion of the problem we are asked to determine the time required to grow a layer of SiO 2 that is 100 nm (i.e., 0.100 m) thick on the surface of a silicon chip at 1000C, in an atmosphere of O2 (oxygen pressure = 1 atm). Thus, using Equation 12.37, it is necessary to solve for the time t. However, before this is possible, we must calculate the value of B from Equation 12.38a as follows: 1.24 eV 1.24 eV B 800 exp = (800) exp -5 kT (8.62 10 eV/atom - K)(1000 + 273 K) = 0.00990 m2/h Now, solving for t from Equation 12.37 using the above value for B and that x = 0.100 m, we have x2 (0.100 m) 2 = B 0.00990 m2 / h t = = 1.01 h Repeating the computation for B at 700C: 1.24 eV B = (800) exp -5 (8.62 10 eV/atom - K)(700 + 273 K) = 3.04 10-4 m2/h And solving for the oxidation time as above t = (0.100 m) 2 3.04 10-4 m2 / h = 32.9 h (b) This part of the problem asks for us to compute the heating times to form an oxide layer 100 nm thick at the same two temperatures (1000C and 700C) when the atmosphere is water vapor (1 atm pressure). At 1000C, the value of B is determined using Equation 12.38b, as follows: 0.70 eV 0.70 eV B 215 exp = (215) exp kT (8.62 10-5 eV/atom - K)(1000 + 273 K) = 0.365 m2/h And computation of the time t from the rearranged form of Equation 12.37, leads to t = And at 700C, the value of B is x2 (0.100 m) 2 = B 0.365 m2 / h = 0.0274 h = 98.6 s 0.70 eV B = (215) exp = 0.0510 m2 / h -5 (8.62 10 eV/atom - K)(700 + 273 K) Whereas the time required to grow the 100 nm oxide layer is t = x2 (0.100 m) 2 = B 0.0510 m2 / h = 0.196 h = 706 s From the above computations, it is very apparent (1) that the 100 nm oxide layer forms more rapidly at 1000C (than at 700C) in both O2 and H2O gaseous atmospheres, and (2) that the oxide layer formation is more rapid in water vapor than in oxygen. 16.3 An electrochemical cell is composed of pure copper and pure cadmium electrodes immersed in solutions of their respective divalent ions. For a 6.5 × 10 –2 M concentration of Cd2+, the cadmium electrode is oxidized yielding a cell potential of 0.775 V. Calculate the concentration of Cu 2+ ions if the temperature is 25°C. Solution The electrochemical reaction that occurs within this cell is just Cd + Cu2+ Cd2+ + Cu while V = 0.775 V and [Cd2+] = 6.5 10-2 M. Thus, Equation 16.20 is written in the form V = (VCu VCd ) 0.0592 [Cd 2 ] log 2 [Cu2 ] This equation may be rewritten as V (VCu VCd ) 0.0296 log [Cd 2 ] [Cu2 ] Solving this expression for [Cu2+] gives V (VCu VPb) [Cu2+ ] = [Cd 2+ ] exp (2.303) 0.0296 The standard potentials from Table 16.1 are VCu = +0.340 V and VCd = – 0.403 V. Therefore, 0.775 V {0.340 V (0.403 V)} [Cu2+ ] = (6.5 10-2 M ) exp (2.303) 0.0296 = 0.784 M 16.7 A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weight loss of 485 g due to corrosion. To what rate of corrosion, in both mpy and mm/yr, does this correspond? Solution This problem asks for us to calculate the CPR in both mpy and mm/yr for a thick steel sheet of area 100 in.2 which experiences a weight loss of 485 g after one year. Employment of Equation 16.23 leads to CPR(mm/yr) = = KW A t (87.6)(485 g)(10 3 mg / g) (7.9 g /cm 3 )(100 in.2 ) (2.54 cm /in.)2 (24 h /day)(365 day / yr)(1 yr) = 0.952 mm/yr Also CPR(mpy) = (534)(485 g)(10 3 mg / g) (7.9 g /cm 3)(100 in.2 ) (24 h /day)(365 day / yr)(1 yr) = 37.4 mpy 16.12 For each form of corrosion, other than uniform, do the following: (a) Describe why, where, and the conditions under which the corrosion occurs. (b) Cite three measures that may be taken to prevent or control it. For each of the forms of corrosion, the conditions under which it occurs, and measures that may be taken to prevent or control it are outlined in Section 16.7. 16.13 Briefly explain why, for a small anode-to-cathode area ratio, the corrosion rate will be higher than for a large ratio. Solution For a small anode-to-cathode area ratio, the corrosion rate will be higher than for a large ratio. The reason for this is that for some given current flow associated with the corrosion reaction, for a small area ratio the current density at the anode will be greater than for a large ratio. The corrosion rate is proportional to the current density (i) according to Equation 16.24. 16.15 For each of the metals listed in the table, compute the Pilling–Bedworth ratio. Also, on the basis of this value, specify whether or not you would expect the oxide scale that forms on the surface to be protective, and then justify your decision. Density data for both the metal and its oxide are also tabulated. Metal Metal Density (g/cm3) Metal Oxide Oxide Density (g/cm3) Mg 1.74 MgO 3.58 V 6.11 V2O5 3.36 Zn 7.13 ZnO 5.61 Solution The general form of the equation used to calculate the Pilling-Bedworth ratios is Equation 16.32 (or Equation 16.33). For magnesium, oxidation occurs by the reaction Mg + and therefore, from Equation 16.32 1 O MgO 2 2 P B ratio = = AMgO Mg AMg MgO (40.31 g / mol)(1.74 g /cm 3) (24.31 g / mol)(3.58 g /cm 3) = 0.81 Thus, this would probably be a nonprotective oxide film since the P-B ratio is less than unity; to be protective, this ratio should be between one and two. The oxidation reaction for V is just 2V + 5 O V2O5 2 2 and the P-B ratio is (Equation 16.33) P B ratio = AV O V 2 5 (2) AV V O 2 5 = (181.88 g / mol)(6.11 g /cm 3) (2)(50.94 g / mol)(3.36 g /cm 3) = 3.25 Hence, the film would be nonprotective since the ratio does not lie between one and two. Now for Zn, the reaction for its oxidation is analogous to that for Mg above. Therefore, P B ratio = = AZnO Zn AZn ZnO (81.39 g / mol)(7.13 g /cm 3) (65.39 g / mol)(5.61 g /cm 3) = 1.58 Thus, the ZnO film would probably be protective since the ratio is between one and two.