WEBVTT 00:00:00.030 --> 00:00:02.400 The following content is provided under a creative 00:00:02.400 --> 00:00:03.820 commons license. 00:00:03.820 --> 00:00:06.850 Your support will help MIT OpenCourseWare continue to 00:00:06.850 --> 00:00:10.510 offer high quality educational resources for free. 00:00:10.510 --> 00:00:13.390 To make a donation or view additional materials from 00:00:13.390 --> 00:00:17.490 hundreds of MIT courses, visit MIT OpenCourseWare at 00:00:17.490 --> 00:00:18.740 ocw.mit.edu. 00:00:22.140 --> 00:00:22.830 SAL: Hi. 00:00:22.830 --> 00:00:23.340 I'm Sal. 00:00:23.340 --> 00:00:25.720 Today we're going to be solving problem one of exam 00:00:25.720 --> 00:00:28.790 three of fall 2009. 00:00:28.790 --> 00:00:31.680 Now before you attempt the problem, there's a couple of 00:00:31.680 --> 00:00:37.040 things that you should know or have background knowledge of 00:00:37.040 --> 00:00:39.000 these materials before starting. 00:00:39.000 --> 00:00:41.760 First thing is understanding the properties of crystal 00:00:41.760 --> 00:00:45.820 defects and for this problem particularly, Frenkel defects, 00:00:45.820 --> 00:00:48.800 which we'll talk about. 00:00:48.800 --> 00:00:52.420 The formula, which is an Arrhenius relationship, as a 00:00:52.420 --> 00:00:56.340 function of temperature and energy enthalpy of the 00:00:56.340 --> 00:01:01.880 fraction of vacancies that are formed in a crystal and the 00:01:01.880 --> 00:01:03.420 conversion again between one electron 00:01:03.420 --> 00:01:05.440 volt and to the joule. 00:01:05.440 --> 00:01:08.730 This will help you solve the problem. 00:01:08.730 --> 00:01:11.130 So the problem reads as follows. 00:01:11.130 --> 00:01:14.520 Silver bromide, which is AgBr, has rock 00:01:14.520 --> 00:01:16.440 salt crystal structure. 00:01:16.440 --> 00:01:19.800 So it's an FCC Bravais lattice with the ion pair Ag plus and 00:01:19.800 --> 00:01:21.320 the Br as the basis-- 00:01:21.320 --> 00:01:22.920 Br minus the basis. 00:01:22.920 --> 00:01:25.840 The dominant defect in AgBr or silver bromide 00:01:25.840 --> 00:01:27.300 is the Frenkel disorder. 00:01:27.300 --> 00:01:33.050 So it's telling you what the dominant defective is. 00:01:33.050 --> 00:01:39.570 So the question asks, for part A, so I'll put A here. 00:01:39.570 --> 00:01:42.860 Part A for the question asks, does the Frenkel disorder in 00:01:42.860 --> 00:01:46.520 silver bromide create vacancies of silver plus, 00:01:46.520 --> 00:01:49.020 vacancies or Br minus or both? 00:01:49.020 --> 00:01:50.220 Explain. 00:01:50.220 --> 00:01:54.830 Dionic radii are 0.67 Angstroms for silver plus and 00:01:54.830 --> 00:01:58.390 1.96 Angstroms for bromine minus. 00:01:58.390 --> 00:01:59.740 So that's data that's given to us. 00:01:59.740 --> 00:02:01.820 I'm going to go ahead and write that down. 00:02:01.820 --> 00:02:09.180 So I know from reading about point defects and crystals 00:02:09.180 --> 00:02:13.380 that a Frenkel defect is pretty much formed when you 00:02:13.380 --> 00:02:16.980 have an ion pair of dissimilar sizes. 00:02:16.980 --> 00:02:19.710 So if one of the radii-- 00:02:19.710 --> 00:02:23.300 like, say, your anion is a lot bigger than the radius of your 00:02:23.300 --> 00:02:26.980 cation, then one should expect that you would have a Frenkel 00:02:26.980 --> 00:02:28.670 defect to form. 00:02:28.670 --> 00:02:31.680 And what a Frankel defect is-- which you should know from 00:02:31.680 --> 00:02:33.020 reading the material-- 00:02:33.020 --> 00:02:38.070 is that it's when one of the ions in your crystal leaves 00:02:38.070 --> 00:02:39.260 and leaves back a vacancy-- 00:02:39.260 --> 00:02:41.470 so just migrates-- 00:02:41.470 --> 00:02:44.730 just hops out of its lattice site and leaves back empty 00:02:44.730 --> 00:02:47.020 spaces, which is called a vacancy. 00:02:47.020 --> 00:02:50.200 So we're given two things. 00:02:50.200 --> 00:02:52.660 If I draw a little picture-- 00:02:52.660 --> 00:03:03.550 I can call this my silver and I'll call this my bromine. 00:03:03.550 --> 00:03:11.670 And I'm given the fact that the radius of Ag plus is equal 00:03:11.670 --> 00:03:17.570 to 0.67 Angstroms-- 00:03:17.570 --> 00:03:21.200 and an Angstrom is just 10 to the minus 10 meters-- 00:03:21.200 --> 00:03:23.280 very small number. 00:03:23.280 --> 00:03:31.850 And I'm also given the radius of bromine, the anion, is 1.96 00:03:31.850 --> 00:03:38.240 Angstroms. Now I would argue that these two have very 00:03:38.240 --> 00:03:39.940 dissimilar radii. 00:03:39.940 --> 00:03:44.010 Obviously, the bromine looks to be about three times bigger 00:03:44.010 --> 00:03:46.450 than your silver. 00:03:46.450 --> 00:03:49.130 So by understanding the definition of a Frenkel 00:03:49.130 --> 00:03:55.950 defect, I can claim that I would expect the smaller ion 00:03:55.950 --> 00:04:00.680 of the two to be the one that leaves the vacancy behind, 00:04:00.680 --> 00:04:02.730 hence forms the Frenkel defect. 00:04:02.730 --> 00:04:07.400 So for part A, do I expect it to create silver plus 00:04:07.400 --> 00:04:10.520 vacancies or Br minus vacancies or both? 00:04:10.520 --> 00:04:13.370 I would expect just to be silver plus, given the size of 00:04:13.370 --> 00:04:14.860 your cation. 00:04:14.860 --> 00:04:26.130 So if you were to write on this problem, just expect only 00:04:26.130 --> 00:04:30.370 Ag plus, which is your silver cation, or the smaller one. 00:04:39.830 --> 00:04:50.380 Expect Ag plus smaller ion to create a vacancy and hence, 00:04:50.380 --> 00:04:55.770 this leads to the Frenkel defect. 00:04:58.660 --> 00:05:03.240 So yes, I should expect it to form a defect now. 00:05:03.240 --> 00:05:11.430 If I was given a cation that had a similar radius that I-- 00:05:11.430 --> 00:05:14.330 I wouldn't know if I could argue the fact that this will 00:05:14.330 --> 00:05:18.350 form a Frenkel defect or not because the conditions are 00:05:18.350 --> 00:05:21.570 that you have to have the similar radius between your 00:05:21.570 --> 00:05:22.680 cation and your atom-- 00:05:22.680 --> 00:05:24.880 and it makes sense because the fact that 00:05:24.880 --> 00:05:26.790 your silver is small-- 00:05:26.790 --> 00:05:30.070 it has more freedom to hop around. 00:05:30.070 --> 00:05:33.420 So it requires less energy for this smaller atom to then 00:05:33.420 --> 00:05:36.680 start hopping around your lattice site without penalty 00:05:36.680 --> 00:05:40.240 or without major penalty compared to your bromine ion. 00:05:40.240 --> 00:05:43.000 So therefore, this is the one that should be 00:05:43.000 --> 00:05:45.180 expected to form that. 00:05:45.180 --> 00:05:45.640 OK. 00:05:45.640 --> 00:05:46.580 So that's part A. 00:05:46.580 --> 00:05:49.140 And part B reads-- 00:05:49.140 --> 00:05:52.000 calculate the temperature at which the fraction of Frenkel 00:05:52.000 --> 00:05:54.850 defects in a crystal with silver bromide exceeds one 00:05:54.850 --> 00:05:56.690 part per billion. 00:05:56.690 --> 00:05:59.600 The enthalpy of Frankel defects and formation, delta h 00:05:59.600 --> 00:06:04.960 sub f, has a value of 1.16 electron volts per defect. 00:06:04.960 --> 00:06:11.220 And the entropic pre-factor A has a value of 3.091. 00:06:11.220 --> 00:06:12.440 So it's giving you data. 00:06:12.440 --> 00:06:16.370 Now the way I would solve this problem is that the first 00:06:16.370 --> 00:06:18.780 thing I would do is that I would write down 00:06:18.780 --> 00:06:19.880 what my data is. 00:06:19.880 --> 00:06:28.400 So I'll call this data and the first thing is that our 00:06:28.400 --> 00:06:31.270 fraction of vacancies that are formed in silver bromide is 00:06:31.270 --> 00:06:32.670 one part per billion-- 00:06:32.670 --> 00:06:34.540 so 10 to the -9. 00:06:34.540 --> 00:06:38.690 That's a fraction so no units. 00:06:38.690 --> 00:06:42.500 The second thing that we're given in the problem is that 00:06:42.500 --> 00:06:44.320 our energy of formation-- 00:06:44.320 --> 00:06:46.390 our enthalpy energy-- 00:06:46.390 --> 00:06:54.960 is 1.16 electron volts per defect. 00:06:54.960 --> 00:06:58.770 This is the energy penalty for every time one of your silver 00:06:58.770 --> 00:07:01.270 cations jumps out of sight to create that. 00:07:01.270 --> 00:07:02.270 Nothing is free. 00:07:02.270 --> 00:07:05.240 Everything requires energy. 00:07:05.240 --> 00:07:09.090 We're also given the fact that A, the entropic 00:07:09.090 --> 00:07:13.620 pre-factor, is 3.091-- 00:07:13.620 --> 00:07:16.510 with no units-- 00:07:16.510 --> 00:07:19.680 and given our equation that I showed you 00:07:19.680 --> 00:07:20.250 on the bullet point. 00:07:20.250 --> 00:07:22.170 That you should know-- 00:07:22.170 --> 00:07:24.160 you can go ahead and see what's missing. 00:07:24.160 --> 00:07:27.570 So if I write down my equation-- 00:07:27.570 --> 00:07:28.390 this is all-- 00:07:28.390 --> 00:07:33.080 I'll box this off as my data because I'm going to refer to 00:07:33.080 --> 00:07:35.280 this to solve the problem. 00:07:35.280 --> 00:07:37.880 And the problem talks about temperature. 00:07:37.880 --> 00:07:45.450 So I know that f sub v is going to equal to the 00:07:45.450 --> 00:07:51.003 pre-factor times the exponent of negative delta h, of 00:07:51.003 --> 00:07:55.990 formation over kb t. 00:07:55.990 --> 00:07:59.895 Now you notice that kb wasn't given to you and kb is both in 00:07:59.895 --> 00:08:01.040 its constant. 00:08:01.040 --> 00:08:04.230 Now a lot of student forget that they have a table of 00:08:04.230 --> 00:08:06.820 contents in front of them and that value is in there. 00:08:06.820 --> 00:08:10.510 So if you don't know it by heart, then you want to make 00:08:10.510 --> 00:08:13.870 sure that you reference to that table because a lot of 00:08:13.870 --> 00:08:16.570 information will be given to you, because you're expected 00:08:16.570 --> 00:08:18.513 to look at the table of contents or 00:08:18.513 --> 00:08:19.830 your periodic table. 00:08:19.830 --> 00:08:23.490 But if I include this as my data-- 00:08:23.490 --> 00:08:25.340 I'm going to go ahead and write it over here-- 00:08:25.340 --> 00:08:27.760 that kb-- 00:08:27.760 --> 00:08:37.350 half the value of 1.38 times 10 to the -23 and this has 00:08:37.350 --> 00:08:42.020 units of joules for degree Kelvin. 00:08:42.020 --> 00:08:43.700 So this is something that you should pay particular 00:08:43.700 --> 00:08:48.940 attention to because now kb is in joules for Kelvin, but our 00:08:48.940 --> 00:08:51.200 energy was given in electron volts. 00:08:51.200 --> 00:08:53.270 Now this is the number one thing that 00:08:53.270 --> 00:08:54.770 will take off points. 00:08:54.770 --> 00:08:57.490 You'll get points taken off if you don't notice this-- that 00:08:57.490 --> 00:09:01.040 you need to go ahead and do the conversion from electron 00:09:01.040 --> 00:09:03.540 volts to joules or joules to electron volts. 00:09:03.540 --> 00:09:06.500 Either which way, you're going to get the answer. 00:09:06.500 --> 00:09:09.800 But the problem asks, calculate the temperature. 00:09:09.800 --> 00:09:11.210 The very first sentence. 00:09:11.210 --> 00:09:13.040 So what does that mean? 00:09:13.040 --> 00:09:15.880 Well, I'm going to go ahead and look at my equation and 00:09:15.880 --> 00:09:18.620 I'm going to look at the data that I have and the constants 00:09:18.620 --> 00:09:20.800 and see if I'm missing anything else because 00:09:20.800 --> 00:09:24.010 obviously you can't solve an equation that has two unknowns 00:09:24.010 --> 00:09:25.290 and just one equation. 00:09:25.290 --> 00:09:28.100 Solve the equation, you've got to only have one unknown for 00:09:28.100 --> 00:09:29.280 the equation. 00:09:29.280 --> 00:09:31.320 So f sub v-- 00:09:31.320 --> 00:09:31.830 no. 00:09:31.830 --> 00:09:32.530 Check-- 00:09:32.530 --> 00:09:32.860 no. 00:09:32.860 --> 00:09:34.310 That's given. 00:09:34.310 --> 00:09:35.620 Is A given? 00:09:35.620 --> 00:09:36.430 It's right here. 00:09:36.430 --> 00:09:38.520 That's given as well. 00:09:38.520 --> 00:09:40.640 What about delta A sub F? 00:09:40.640 --> 00:09:43.100 Well, that's the energy-- 00:09:43.100 --> 00:09:47.870 the energy penalty to create a defect and kb, which we got 00:09:47.870 --> 00:09:49.740 from our table of contents, and t. 00:09:49.740 --> 00:09:54.340 So this concludes that the only thing we're not given is 00:09:54.340 --> 00:09:56.730 temperature because that's what we're asked to solve. 00:09:56.730 --> 00:10:00.580 So I need to do some math on here to go ahead and solve for 00:10:00.580 --> 00:10:01.220 temperature. 00:10:01.220 --> 00:10:03.670 And the first thing I want to do is take the natural log of 00:10:03.670 --> 00:10:04.550 both sides. 00:10:04.550 --> 00:10:07.270 So by taking the natural log of both sides-- 00:10:07.270 --> 00:10:10.710 so natural log of f sub v-- 00:10:10.710 --> 00:10:20.210 this equals to natural log of A plus the natural log of the 00:10:20.210 --> 00:10:27.260 exponent part and we know from math that the natural log of 00:10:27.260 --> 00:10:30.510 an exponent cancels each other out because they're inverses. 00:10:30.510 --> 00:10:37.740 So the natural log of exponent, of negative delta h 00:10:37.740 --> 00:10:40.860 sub f, over kb t-- 00:10:43.360 --> 00:10:49.870 this cancels that and we just get the natural log of A plus 00:10:49.870 --> 00:10:54.700 negative because you have a negative up here-- 00:10:54.700 --> 00:10:59.900 delta h of formation divided by kb t. 00:10:59.900 --> 00:11:03.600 So the only thing we don't know here is temperature. 00:11:03.600 --> 00:11:06.320 So if I can rearrange this equation and solve for 00:11:06.320 --> 00:11:08.480 temperature, I can get an answer. 00:11:08.480 --> 00:11:10.552 So if I do that-- 00:11:10.552 --> 00:11:11.940 what's the best way of doing that? 00:11:11.940 --> 00:11:12.150 Well. 00:11:12.150 --> 00:11:12.900 I can add-- 00:11:12.900 --> 00:11:15.770 I can move this to the other side and move this to the 00:11:15.770 --> 00:11:23.780 other side and that gives me delta h sub f divided by kb t 00:11:23.780 --> 00:11:28.800 equals natural log of A minus natural log of your vacancy 00:11:28.800 --> 00:11:36.990 fraction and I can then multiply both sides by t so it 00:11:36.990 --> 00:11:40.580 cancels this one and it arrives over here and then 00:11:40.580 --> 00:11:43.150 divide both sides by ln of this. 00:11:43.150 --> 00:11:44.360 So I'll go ahead and do that. 00:11:44.360 --> 00:11:46.680 So I'll multiply this by t. 00:11:46.680 --> 00:11:48.120 Multiply that by t. 00:11:48.120 --> 00:11:58.990 So that cancel that and I end up having delta hf over kb t-- 00:11:58.990 --> 00:12:00.790 the t got canceled-- 00:12:00.790 --> 00:12:06.740 this equals to the natural log of A minus the natural log of 00:12:06.740 --> 00:12:09.490 your vacancy fraction times t. 00:12:09.490 --> 00:12:15.940 So now if I divide both sides by this factor, I solve for 00:12:15.940 --> 00:12:16.490 temperature. 00:12:16.490 --> 00:12:29.320 So this gives me an isolation that t ends up being delta hf 00:12:29.320 --> 00:12:39.245 over kb times the natural log of A minus the natural log of 00:12:39.245 --> 00:12:41.160 the vacancy fraction. 00:12:41.160 --> 00:12:45.460 Now all we have to do is plug in the numbers, but again, if 00:12:45.460 --> 00:12:48.260 you look at the units of Boltzmann's constant, which 00:12:48.260 --> 00:12:51.750 are joules per Kelvin and the value that we're given up here 00:12:51.750 --> 00:12:55.640 is electron volts for defect, we want to go ahead and 00:12:55.640 --> 00:12:59.200 convert the electron volt to the joule because it'll just 00:12:59.200 --> 00:13:00.910 be easier to do the math that way. 00:13:00.910 --> 00:13:06.110 So I can go ahead and write it out and I do that by 00:13:06.110 --> 00:13:11.470 multiplying the top by that conversion factor and if I do 00:13:11.470 --> 00:13:16.170 the math, t ends up being-- 00:13:16.170 --> 00:13:19.580 we have 1.16 of this-- 00:13:19.580 --> 00:13:29.090 has units of electron volts for defect and then I want to 00:13:29.090 --> 00:13:32.400 get joules because that's what Boltzmann's constant has. 00:13:32.400 --> 00:13:35.730 So in order to get joules, if I multiply this by the 00:13:35.730 --> 00:13:41.980 conversion factor which is 1.6 times 10 to the -19-- 00:13:41.980 --> 00:13:45.770 this has units of joules per electron volt. 00:13:45.770 --> 00:13:48.540 Now the electron volts cancel and I get-- 00:13:48.540 --> 00:13:51.810 the numerator has units of joules per defect. 00:13:51.810 --> 00:13:55.000 That's good because that's going to cancel with the units 00:13:55.000 --> 00:13:56.130 in Boltzmann's constant. 00:13:56.130 --> 00:14:02.860 And then I put in 1.38 times 10 to the -23-- 00:14:02.860 --> 00:14:05.500 this is joules per Kelvin-- 00:14:08.570 --> 00:14:19.840 and this factor is multiplied by the natural log of 3.091 00:14:19.840 --> 00:14:26.920 minus the natural log of 10 to the -9. 00:14:26.920 --> 00:14:28.900 So unitless-- 00:14:28.900 --> 00:14:33.700 so if you do the math out, you end up getting a value that at 00:14:33.700 --> 00:14:36.490 the end of the day, your temperature-- 00:14:36.490 --> 00:14:37.790 this t right here-- 00:14:37.790 --> 00:14:39.070 will be equal to-- 00:14:42.540 --> 00:14:43.810 let's go ahead and write it over here. 00:14:43.810 --> 00:14:45.670 So now that we go ahead and do the math-- 00:14:45.670 --> 00:14:47.020 that way it's nice and clean. 00:14:47.020 --> 00:14:49.940 We know that the t, the temperature that we got from 00:14:49.940 --> 00:14:53.950 plugging all that math-- 00:14:53.950 --> 00:15:00.340 ends up being just 615 Kelvin. 00:15:00.340 --> 00:15:01.630 And where did the Kelvin come from? 00:15:01.630 --> 00:15:05.570 Well, it came from your joules per Kelvin, from your 00:15:05.570 --> 00:15:08.290 Boltzmann's constant, because that's what the dimensional 00:15:08.290 --> 00:15:09.880 analysis does. 00:15:09.880 --> 00:15:11.650 So this is the answer. 00:15:11.650 --> 00:15:12.330 This is good. 00:15:12.330 --> 00:15:17.870 The problem doesn't say show the answer in degrees Celsius, 00:15:17.870 --> 00:15:21.040 but since everybody knows degrees Celsius-- 00:15:21.040 --> 00:15:23.260 everybody in science relies on degrees Celsius-- 00:15:23.260 --> 00:15:27.300 you can just convert this over to degrees Celsius, which 00:15:27.300 --> 00:15:31.830 happens to be 342 degrees Celsius. 00:15:31.830 --> 00:15:36.740 So this right here is your answer to part B. 00:15:36.740 --> 00:15:41.020 And again, I want to express again that the crucial part in 00:15:41.020 --> 00:15:45.750 getting this is knowing the fact that you're given an 00:15:45.750 --> 00:15:51.000 energy in electron volts, but the constant that you use has 00:15:51.000 --> 00:15:52.500 units of joules. 00:15:52.500 --> 00:15:59.080 So you need to convert to get the right unit out or else you 00:15:59.080 --> 00:16:00.930 will get not Kelvin-- 00:16:00.930 --> 00:16:03.180 you would get something different here, which wouldn't 00:16:03.180 --> 00:16:06.850 make any sense given the question that was asked. 00:16:06.850 --> 00:16:09.600 So with that, again I advise that-- always read 00:16:09.600 --> 00:16:10.760 the problem in detail. 00:16:10.760 --> 00:16:14.760 Look at the units that you're working with and make the 00:16:14.760 --> 00:16:17.050 appropriate conversions because everything should be 00:16:17.050 --> 00:16:19.280 on your table of contents.