1 00:00:00,030 --> 00:00:02,400 The following content is provided under a creative 2 00:00:02,400 --> 00:00:03,820 commons license. 3 00:00:03,820 --> 00:00:06,850 Your support will help MIT OpenCourseWare continue to 4 00:00:06,850 --> 00:00:10,510 offer high quality educational resources for free. 5 00:00:10,510 --> 00:00:13,390 To make a donation or view additional materials from 6 00:00:13,390 --> 00:00:17,490 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,490 --> 00:00:18,740 ocw.mit.edu. 8 00:00:22,140 --> 00:00:22,830 SAL: Hi. 9 00:00:22,830 --> 00:00:23,340 I'm Sal. 10 00:00:23,340 --> 00:00:25,720 Today we're going to be solving problem one of exam 11 00:00:25,720 --> 00:00:28,790 three of fall 2009. 12 00:00:28,790 --> 00:00:31,680 Now before you attempt the problem, there's a couple of 13 00:00:31,680 --> 00:00:37,040 things that you should know or have background knowledge of 14 00:00:37,040 --> 00:00:39,000 these materials before starting. 15 00:00:39,000 --> 00:00:41,760 First thing is understanding the properties of crystal 16 00:00:41,760 --> 00:00:45,820 defects and for this problem particularly, Frenkel defects, 17 00:00:45,820 --> 00:00:48,800 which we'll talk about. 18 00:00:48,800 --> 00:00:52,420 The formula, which is an Arrhenius relationship, as a 19 00:00:52,420 --> 00:00:56,340 function of temperature and energy enthalpy of the 20 00:00:56,340 --> 00:01:01,880 fraction of vacancies that are formed in a crystal and the 21 00:01:01,880 --> 00:01:03,420 conversion again between one electron 22 00:01:03,420 --> 00:01:05,440 volt and to the joule. 23 00:01:05,440 --> 00:01:08,730 This will help you solve the problem. 24 00:01:08,730 --> 00:01:11,130 So the problem reads as follows. 25 00:01:11,130 --> 00:01:14,520 Silver bromide, which is AgBr, has rock 26 00:01:14,520 --> 00:01:16,440 salt crystal structure. 27 00:01:16,440 --> 00:01:19,800 So it's an FCC Bravais lattice with the ion pair Ag plus and 28 00:01:19,800 --> 00:01:21,320 the Br as the basis-- 29 00:01:21,320 --> 00:01:22,920 Br minus the basis. 30 00:01:22,920 --> 00:01:25,840 The dominant defect in AgBr or silver bromide 31 00:01:25,840 --> 00:01:27,300 is the Frenkel disorder. 32 00:01:27,300 --> 00:01:33,050 So it's telling you what the dominant defective is. 33 00:01:33,050 --> 00:01:39,570 So the question asks, for part A, so I'll put A here. 34 00:01:39,570 --> 00:01:42,860 Part A for the question asks, does the Frenkel disorder in 35 00:01:42,860 --> 00:01:46,520 silver bromide create vacancies of silver plus, 36 00:01:46,520 --> 00:01:49,020 vacancies or Br minus or both? 37 00:01:49,020 --> 00:01:50,220 Explain. 38 00:01:50,220 --> 00:01:54,830 Dionic radii are 0.67 Angstroms for silver plus and 39 00:01:54,830 --> 00:01:58,390 1.96 Angstroms for bromine minus. 40 00:01:58,390 --> 00:01:59,740 So that's data that's given to us. 41 00:01:59,740 --> 00:02:01,820 I'm going to go ahead and write that down. 42 00:02:01,820 --> 00:02:09,180 So I know from reading about point defects and crystals 43 00:02:09,180 --> 00:02:13,380 that a Frenkel defect is pretty much formed when you 44 00:02:13,380 --> 00:02:16,980 have an ion pair of dissimilar sizes. 45 00:02:16,980 --> 00:02:19,710 So if one of the radii-- 46 00:02:19,710 --> 00:02:23,300 like, say, your anion is a lot bigger than the radius of your 47 00:02:23,300 --> 00:02:26,980 cation, then one should expect that you would have a Frenkel 48 00:02:26,980 --> 00:02:28,670 defect to form. 49 00:02:28,670 --> 00:02:31,680 And what a Frankel defect is-- which you should know from 50 00:02:31,680 --> 00:02:33,020 reading the material-- 51 00:02:33,020 --> 00:02:38,070 is that it's when one of the ions in your crystal leaves 52 00:02:38,070 --> 00:02:39,260 and leaves back a vacancy-- 53 00:02:39,260 --> 00:02:41,470 so just migrates-- 54 00:02:41,470 --> 00:02:44,730 just hops out of its lattice site and leaves back empty 55 00:02:44,730 --> 00:02:47,020 spaces, which is called a vacancy. 56 00:02:47,020 --> 00:02:50,200 So we're given two things. 57 00:02:50,200 --> 00:02:52,660 If I draw a little picture-- 58 00:02:52,660 --> 00:03:03,550 I can call this my silver and I'll call this my bromine. 59 00:03:03,550 --> 00:03:11,670 And I'm given the fact that the radius of Ag plus is equal 60 00:03:11,670 --> 00:03:17,570 to 0.67 Angstroms-- 61 00:03:17,570 --> 00:03:21,200 and an Angstrom is just 10 to the minus 10 meters-- 62 00:03:21,200 --> 00:03:23,280 very small number. 63 00:03:23,280 --> 00:03:31,850 And I'm also given the radius of bromine, the anion, is 1.96 64 00:03:31,850 --> 00:03:38,240 Angstroms. Now I would argue that these two have very 65 00:03:38,240 --> 00:03:39,940 dissimilar radii. 66 00:03:39,940 --> 00:03:44,010 Obviously, the bromine looks to be about three times bigger 67 00:03:44,010 --> 00:03:46,450 than your silver. 68 00:03:46,450 --> 00:03:49,130 So by understanding the definition of a Frenkel 69 00:03:49,130 --> 00:03:55,950 defect, I can claim that I would expect the smaller ion 70 00:03:55,950 --> 00:04:00,680 of the two to be the one that leaves the vacancy behind, 71 00:04:00,680 --> 00:04:02,730 hence forms the Frenkel defect. 72 00:04:02,730 --> 00:04:07,400 So for part A, do I expect it to create silver plus 73 00:04:07,400 --> 00:04:10,520 vacancies or Br minus vacancies or both? 74 00:04:10,520 --> 00:04:13,370 I would expect just to be silver plus, given the size of 75 00:04:13,370 --> 00:04:14,860 your cation. 76 00:04:14,860 --> 00:04:26,130 So if you were to write on this problem, just expect only 77 00:04:26,130 --> 00:04:30,370 Ag plus, which is your silver cation, or the smaller one. 78 00:04:39,830 --> 00:04:50,380 Expect Ag plus smaller ion to create a vacancy and hence, 79 00:04:50,380 --> 00:04:55,770 this leads to the Frenkel defect. 80 00:04:58,660 --> 00:05:03,240 So yes, I should expect it to form a defect now. 81 00:05:03,240 --> 00:05:11,430 If I was given a cation that had a similar radius that I-- 82 00:05:11,430 --> 00:05:14,330 I wouldn't know if I could argue the fact that this will 83 00:05:14,330 --> 00:05:18,350 form a Frenkel defect or not because the conditions are 84 00:05:18,350 --> 00:05:21,570 that you have to have the similar radius between your 85 00:05:21,570 --> 00:05:22,680 cation and your atom-- 86 00:05:22,680 --> 00:05:24,880 and it makes sense because the fact that 87 00:05:24,880 --> 00:05:26,790 your silver is small-- 88 00:05:26,790 --> 00:05:30,070 it has more freedom to hop around. 89 00:05:30,070 --> 00:05:33,420 So it requires less energy for this smaller atom to then 90 00:05:33,420 --> 00:05:36,680 start hopping around your lattice site without penalty 91 00:05:36,680 --> 00:05:40,240 or without major penalty compared to your bromine ion. 92 00:05:40,240 --> 00:05:43,000 So therefore, this is the one that should be 93 00:05:43,000 --> 00:05:45,180 expected to form that. 94 00:05:45,180 --> 00:05:45,640 OK. 95 00:05:45,640 --> 00:05:46,580 So that's part A. 96 00:05:46,580 --> 00:05:49,140 And part B reads-- 97 00:05:49,140 --> 00:05:52,000 calculate the temperature at which the fraction of Frenkel 98 00:05:52,000 --> 00:05:54,850 defects in a crystal with silver bromide exceeds one 99 00:05:54,850 --> 00:05:56,690 part per billion. 100 00:05:56,690 --> 00:05:59,600 The enthalpy of Frankel defects and formation, delta h 101 00:05:59,600 --> 00:06:04,960 sub f, has a value of 1.16 electron volts per defect. 102 00:06:04,960 --> 00:06:11,220 And the entropic pre-factor A has a value of 3.091. 103 00:06:11,220 --> 00:06:12,440 So it's giving you data. 104 00:06:12,440 --> 00:06:16,370 Now the way I would solve this problem is that the first 105 00:06:16,370 --> 00:06:18,780 thing I would do is that I would write down 106 00:06:18,780 --> 00:06:19,880 what my data is. 107 00:06:19,880 --> 00:06:28,400 So I'll call this data and the first thing is that our 108 00:06:28,400 --> 00:06:31,270 fraction of vacancies that are formed in silver bromide is 109 00:06:31,270 --> 00:06:32,670 one part per billion-- 110 00:06:32,670 --> 00:06:34,540 so 10 to the -9. 111 00:06:34,540 --> 00:06:38,690 That's a fraction so no units. 112 00:06:38,690 --> 00:06:42,500 The second thing that we're given in the problem is that 113 00:06:42,500 --> 00:06:44,320 our energy of formation-- 114 00:06:44,320 --> 00:06:46,390 our enthalpy energy-- 115 00:06:46,390 --> 00:06:54,960 is 1.16 electron volts per defect. 116 00:06:54,960 --> 00:06:58,770 This is the energy penalty for every time one of your silver 117 00:06:58,770 --> 00:07:01,270 cations jumps out of sight to create that. 118 00:07:01,270 --> 00:07:02,270 Nothing is free. 119 00:07:02,270 --> 00:07:05,240 Everything requires energy. 120 00:07:05,240 --> 00:07:09,090 We're also given the fact that A, the entropic 121 00:07:09,090 --> 00:07:13,620 pre-factor, is 3.091-- 122 00:07:13,620 --> 00:07:16,510 with no units-- 123 00:07:16,510 --> 00:07:19,680 and given our equation that I showed you 124 00:07:19,680 --> 00:07:20,250 on the bullet point. 125 00:07:20,250 --> 00:07:22,170 That you should know-- 126 00:07:22,170 --> 00:07:24,160 you can go ahead and see what's missing. 127 00:07:24,160 --> 00:07:27,570 So if I write down my equation-- 128 00:07:27,570 --> 00:07:28,390 this is all-- 129 00:07:28,390 --> 00:07:33,080 I'll box this off as my data because I'm going to refer to 130 00:07:33,080 --> 00:07:35,280 this to solve the problem. 131 00:07:35,280 --> 00:07:37,880 And the problem talks about temperature. 132 00:07:37,880 --> 00:07:45,450 So I know that f sub v is going to equal to the 133 00:07:45,450 --> 00:07:51,003 pre-factor times the exponent of negative delta h, of 134 00:07:51,003 --> 00:07:55,990 formation over kb t. 135 00:07:55,990 --> 00:07:59,895 Now you notice that kb wasn't given to you and kb is both in 136 00:07:59,895 --> 00:08:01,040 its constant. 137 00:08:01,040 --> 00:08:04,230 Now a lot of student forget that they have a table of 138 00:08:04,230 --> 00:08:06,820 contents in front of them and that value is in there. 139 00:08:06,820 --> 00:08:10,510 So if you don't know it by heart, then you want to make 140 00:08:10,510 --> 00:08:13,870 sure that you reference to that table because a lot of 141 00:08:13,870 --> 00:08:16,570 information will be given to you, because you're expected 142 00:08:16,570 --> 00:08:18,513 to look at the table of contents or 143 00:08:18,513 --> 00:08:19,830 your periodic table. 144 00:08:19,830 --> 00:08:23,490 But if I include this as my data-- 145 00:08:23,490 --> 00:08:25,340 I'm going to go ahead and write it over here-- 146 00:08:25,340 --> 00:08:27,760 that kb-- 147 00:08:27,760 --> 00:08:37,350 half the value of 1.38 times 10 to the -23 and this has 148 00:08:37,350 --> 00:08:42,020 units of joules for degree Kelvin. 149 00:08:42,020 --> 00:08:43,700 So this is something that you should pay particular 150 00:08:43,700 --> 00:08:48,940 attention to because now kb is in joules for Kelvin, but our 151 00:08:48,940 --> 00:08:51,200 energy was given in electron volts. 152 00:08:51,200 --> 00:08:53,270 Now this is the number one thing that 153 00:08:53,270 --> 00:08:54,770 will take off points. 154 00:08:54,770 --> 00:08:57,490 You'll get points taken off if you don't notice this-- that 155 00:08:57,490 --> 00:09:01,040 you need to go ahead and do the conversion from electron 156 00:09:01,040 --> 00:09:03,540 volts to joules or joules to electron volts. 157 00:09:03,540 --> 00:09:06,500 Either which way, you're going to get the answer. 158 00:09:06,500 --> 00:09:09,800 But the problem asks, calculate the temperature. 159 00:09:09,800 --> 00:09:11,210 The very first sentence. 160 00:09:11,210 --> 00:09:13,040 So what does that mean? 161 00:09:13,040 --> 00:09:15,880 Well, I'm going to go ahead and look at my equation and 162 00:09:15,880 --> 00:09:18,620 I'm going to look at the data that I have and the constants 163 00:09:18,620 --> 00:09:20,800 and see if I'm missing anything else because 164 00:09:20,800 --> 00:09:24,010 obviously you can't solve an equation that has two unknowns 165 00:09:24,010 --> 00:09:25,290 and just one equation. 166 00:09:25,290 --> 00:09:28,100 Solve the equation, you've got to only have one unknown for 167 00:09:28,100 --> 00:09:29,280 the equation. 168 00:09:29,280 --> 00:09:31,320 So f sub v-- 169 00:09:31,320 --> 00:09:31,830 no. 170 00:09:31,830 --> 00:09:32,530 Check-- 171 00:09:32,530 --> 00:09:32,860 no. 172 00:09:32,860 --> 00:09:34,310 That's given. 173 00:09:34,310 --> 00:09:35,620 Is A given? 174 00:09:35,620 --> 00:09:36,430 It's right here. 175 00:09:36,430 --> 00:09:38,520 That's given as well. 176 00:09:38,520 --> 00:09:40,640 What about delta A sub F? 177 00:09:40,640 --> 00:09:43,100 Well, that's the energy-- 178 00:09:43,100 --> 00:09:47,870 the energy penalty to create a defect and kb, which we got 179 00:09:47,870 --> 00:09:49,740 from our table of contents, and t. 180 00:09:49,740 --> 00:09:54,340 So this concludes that the only thing we're not given is 181 00:09:54,340 --> 00:09:56,730 temperature because that's what we're asked to solve. 182 00:09:56,730 --> 00:10:00,580 So I need to do some math on here to go ahead and solve for 183 00:10:00,580 --> 00:10:01,220 temperature. 184 00:10:01,220 --> 00:10:03,670 And the first thing I want to do is take the natural log of 185 00:10:03,670 --> 00:10:04,550 both sides. 186 00:10:04,550 --> 00:10:07,270 So by taking the natural log of both sides-- 187 00:10:07,270 --> 00:10:10,710 so natural log of f sub v-- 188 00:10:10,710 --> 00:10:20,210 this equals to natural log of A plus the natural log of the 189 00:10:20,210 --> 00:10:27,260 exponent part and we know from math that the natural log of 190 00:10:27,260 --> 00:10:30,510 an exponent cancels each other out because they're inverses. 191 00:10:30,510 --> 00:10:37,740 So the natural log of exponent, of negative delta h 192 00:10:37,740 --> 00:10:40,860 sub f, over kb t-- 193 00:10:43,360 --> 00:10:49,870 this cancels that and we just get the natural log of A plus 194 00:10:49,870 --> 00:10:54,700 negative because you have a negative up here-- 195 00:10:54,700 --> 00:10:59,900 delta h of formation divided by kb t. 196 00:10:59,900 --> 00:11:03,600 So the only thing we don't know here is temperature. 197 00:11:03,600 --> 00:11:06,320 So if I can rearrange this equation and solve for 198 00:11:06,320 --> 00:11:08,480 temperature, I can get an answer. 199 00:11:08,480 --> 00:11:10,552 So if I do that-- 200 00:11:10,552 --> 00:11:11,940 what's the best way of doing that? 201 00:11:11,940 --> 00:11:12,150 Well. 202 00:11:12,150 --> 00:11:12,900 I can add-- 203 00:11:12,900 --> 00:11:15,770 I can move this to the other side and move this to the 204 00:11:15,770 --> 00:11:23,780 other side and that gives me delta h sub f divided by kb t 205 00:11:23,780 --> 00:11:28,800 equals natural log of A minus natural log of your vacancy 206 00:11:28,800 --> 00:11:36,990 fraction and I can then multiply both sides by t so it 207 00:11:36,990 --> 00:11:40,580 cancels this one and it arrives over here and then 208 00:11:40,580 --> 00:11:43,150 divide both sides by ln of this. 209 00:11:43,150 --> 00:11:44,360 So I'll go ahead and do that. 210 00:11:44,360 --> 00:11:46,680 So I'll multiply this by t. 211 00:11:46,680 --> 00:11:48,120 Multiply that by t. 212 00:11:48,120 --> 00:11:58,990 So that cancel that and I end up having delta hf over kb t-- 213 00:11:58,990 --> 00:12:00,790 the t got canceled-- 214 00:12:00,790 --> 00:12:06,740 this equals to the natural log of A minus the natural log of 215 00:12:06,740 --> 00:12:09,490 your vacancy fraction times t. 216 00:12:09,490 --> 00:12:15,940 So now if I divide both sides by this factor, I solve for 217 00:12:15,940 --> 00:12:16,490 temperature. 218 00:12:16,490 --> 00:12:29,320 So this gives me an isolation that t ends up being delta hf 219 00:12:29,320 --> 00:12:39,245 over kb times the natural log of A minus the natural log of 220 00:12:39,245 --> 00:12:41,160 the vacancy fraction. 221 00:12:41,160 --> 00:12:45,460 Now all we have to do is plug in the numbers, but again, if 222 00:12:45,460 --> 00:12:48,260 you look at the units of Boltzmann's constant, which 223 00:12:48,260 --> 00:12:51,750 are joules per Kelvin and the value that we're given up here 224 00:12:51,750 --> 00:12:55,640 is electron volts for defect, we want to go ahead and 225 00:12:55,640 --> 00:12:59,200 convert the electron volt to the joule because it'll just 226 00:12:59,200 --> 00:13:00,910 be easier to do the math that way. 227 00:13:00,910 --> 00:13:06,110 So I can go ahead and write it out and I do that by 228 00:13:06,110 --> 00:13:11,470 multiplying the top by that conversion factor and if I do 229 00:13:11,470 --> 00:13:16,170 the math, t ends up being-- 230 00:13:16,170 --> 00:13:19,580 we have 1.16 of this-- 231 00:13:19,580 --> 00:13:29,090 has units of electron volts for defect and then I want to 232 00:13:29,090 --> 00:13:32,400 get joules because that's what Boltzmann's constant has. 233 00:13:32,400 --> 00:13:35,730 So in order to get joules, if I multiply this by the 234 00:13:35,730 --> 00:13:41,980 conversion factor which is 1.6 times 10 to the -19-- 235 00:13:41,980 --> 00:13:45,770 this has units of joules per electron volt. 236 00:13:45,770 --> 00:13:48,540 Now the electron volts cancel and I get-- 237 00:13:48,540 --> 00:13:51,810 the numerator has units of joules per defect. 238 00:13:51,810 --> 00:13:55,000 That's good because that's going to cancel with the units 239 00:13:55,000 --> 00:13:56,130 in Boltzmann's constant. 240 00:13:56,130 --> 00:14:02,860 And then I put in 1.38 times 10 to the -23-- 241 00:14:02,860 --> 00:14:05,500 this is joules per Kelvin-- 242 00:14:08,570 --> 00:14:19,840 and this factor is multiplied by the natural log of 3.091 243 00:14:19,840 --> 00:14:26,920 minus the natural log of 10 to the -9. 244 00:14:26,920 --> 00:14:28,900 So unitless-- 245 00:14:28,900 --> 00:14:33,700 so if you do the math out, you end up getting a value that at 246 00:14:33,700 --> 00:14:36,490 the end of the day, your temperature-- 247 00:14:36,490 --> 00:14:37,790 this t right here-- 248 00:14:37,790 --> 00:14:39,070 will be equal to-- 249 00:14:42,540 --> 00:14:43,810 let's go ahead and write it over here. 250 00:14:43,810 --> 00:14:45,670 So now that we go ahead and do the math-- 251 00:14:45,670 --> 00:14:47,020 that way it's nice and clean. 252 00:14:47,020 --> 00:14:49,940 We know that the t, the temperature that we got from 253 00:14:49,940 --> 00:14:53,950 plugging all that math-- 254 00:14:53,950 --> 00:15:00,340 ends up being just 615 Kelvin. 255 00:15:00,340 --> 00:15:01,630 And where did the Kelvin come from? 256 00:15:01,630 --> 00:15:05,570 Well, it came from your joules per Kelvin, from your 257 00:15:05,570 --> 00:15:08,290 Boltzmann's constant, because that's what the dimensional 258 00:15:08,290 --> 00:15:09,880 analysis does. 259 00:15:09,880 --> 00:15:11,650 So this is the answer. 260 00:15:11,650 --> 00:15:12,330 This is good. 261 00:15:12,330 --> 00:15:17,870 The problem doesn't say show the answer in degrees Celsius, 262 00:15:17,870 --> 00:15:21,040 but since everybody knows degrees Celsius-- 263 00:15:21,040 --> 00:15:23,260 everybody in science relies on degrees Celsius-- 264 00:15:23,260 --> 00:15:27,300 you can just convert this over to degrees Celsius, which 265 00:15:27,300 --> 00:15:31,830 happens to be 342 degrees Celsius. 266 00:15:31,830 --> 00:15:36,740 So this right here is your answer to part B. 267 00:15:36,740 --> 00:15:41,020 And again, I want to express again that the crucial part in 268 00:15:41,020 --> 00:15:45,750 getting this is knowing the fact that you're given an 269 00:15:45,750 --> 00:15:51,000 energy in electron volts, but the constant that you use has 270 00:15:51,000 --> 00:15:52,500 units of joules. 271 00:15:52,500 --> 00:15:59,080 So you need to convert to get the right unit out or else you 272 00:15:59,080 --> 00:16:00,930 will get not Kelvin-- 273 00:16:00,930 --> 00:16:03,180 you would get something different here, which wouldn't 274 00:16:03,180 --> 00:16:06,850 make any sense given the question that was asked. 275 00:16:06,850 --> 00:16:09,600 So with that, again I advise that-- always read 276 00:16:09,600 --> 00:16:10,760 the problem in detail. 277 00:16:10,760 --> 00:16:14,760 Look at the units that you're working with and make the 278 00:16:14,760 --> 00:16:17,050 appropriate conversions because everything should be 279 00:16:17,050 --> 00:16:19,280 on your table of contents.