1 00:00:01,040 --> 00:00:03,380 The following content is provided under a Creative 2 00:00:03,380 --> 00:00:04,770 Commons license. 3 00:00:04,770 --> 00:00:06,980 Your support will help MIT OpenCourseWare 4 00:00:06,980 --> 00:00:11,070 continue to offer high quality educational resources for free. 5 00:00:11,070 --> 00:00:13,640 To make a donation or to view additional materials 6 00:00:13,640 --> 00:00:17,600 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,600 --> 00:00:18,480 at ocw.mit.edu. 8 00:00:22,360 --> 00:00:25,040 MICHAEL SHORT: You guys asked to do some numerical examples 9 00:00:25,040 --> 00:00:26,873 of the stuff that we've been learning today, 10 00:00:26,873 --> 00:00:28,470 and I've got a fun one for you. 11 00:00:28,470 --> 00:00:31,700 A very real one, because it happens all the time 12 00:00:31,700 --> 00:00:33,930 in reactors all over the world. 13 00:00:33,930 --> 00:00:37,970 So to set the stage for this, hey, suppose I had-- 14 00:00:37,970 --> 00:00:42,560 just suppose, I had a radioactive cobalt-60 source, 15 00:00:42,560 --> 00:00:46,640 that was calibrated in March, 2011 16 00:00:46,640 --> 00:00:49,430 to be approximately one microcurie. 17 00:00:49,430 --> 00:00:51,747 And it's now October 2016. 18 00:00:51,747 --> 00:00:53,330 So let's start off with the easy part. 19 00:00:53,330 --> 00:00:57,890 Let's say I posed the question, how active was this source 20 00:00:57,890 --> 00:01:00,720 actually when it was made? 21 00:01:00,720 --> 00:01:02,570 Because it just says-- 22 00:01:02,570 --> 00:01:03,770 what does it say-- 23 00:01:03,770 --> 00:01:06,460 1 microcurie. 24 00:01:06,460 --> 00:01:15,920 Supposedly, this was 1 microcurie on March, 2011. 25 00:01:15,920 --> 00:01:19,850 What does that mean the actual activity could be? 26 00:01:19,850 --> 00:01:23,550 There's not a lot of confidence in that number. 27 00:01:23,550 --> 00:01:26,830 Remember significant figures from high school chemistry 28 00:01:26,830 --> 00:01:28,498 and physics. 29 00:01:28,498 --> 00:01:30,290 This is when they're important, because you 30 00:01:30,290 --> 00:01:31,790 got to know what you're buying. 31 00:01:31,790 --> 00:01:34,160 So what is the actual activity of this source 32 00:01:34,160 --> 00:01:35,120 when it was calibrated? 33 00:01:35,120 --> 00:01:36,590 What is our uncertainty on this? 34 00:01:43,100 --> 00:01:44,960 Anyone remember this from sig figs? 35 00:01:44,960 --> 00:01:46,252 I hope so. 36 00:01:46,252 --> 00:01:47,850 If not, I'll refresh your memory. 37 00:01:47,850 --> 00:01:51,060 It's plus or minus the next decimal point. 38 00:01:51,060 --> 00:01:57,180 So really, this was 1 plus or minus 0.5 microcuries. 39 00:01:57,180 --> 00:01:59,100 So you might have a 1/2 microcurie source. 40 00:01:59,100 --> 00:02:01,650 You might have a 1 and 1/2 microcurie source. 41 00:02:01,650 --> 00:02:03,355 They specifically decided to leave out 42 00:02:03,355 --> 00:02:04,980 the second decimal point so they're not 43 00:02:04,980 --> 00:02:07,860 liable for a source that's out of that calibration level. 44 00:02:07,860 --> 00:02:11,610 So we supposedly have a 1 microcurie source. 45 00:02:11,610 --> 00:02:14,610 And I've measured it on-- let's say it's now-- 46 00:02:14,610 --> 00:02:16,680 October, 2016. 47 00:02:16,680 --> 00:02:20,230 Suppose I went and made a measurement, 48 00:02:20,230 --> 00:02:23,890 and it was 0.52 microcuries. 49 00:02:23,890 --> 00:02:26,830 And we want to know, how active was this source actually 50 00:02:26,830 --> 00:02:29,497 when we got it? 51 00:02:29,497 --> 00:02:30,330 Where would I begin? 52 00:02:34,503 --> 00:02:35,902 AUDIENCE: Look up the half-life. 53 00:02:35,902 --> 00:02:37,360 MICHAEL SHORT: OK, yeah, that's why 54 00:02:37,360 --> 00:02:40,040 I've got the table of nuclides right here. 55 00:02:40,040 --> 00:02:44,010 So let's look up the half-life of cobalt-60. 56 00:02:44,010 --> 00:02:47,290 And there it is right there, 1,925 days. 57 00:02:47,290 --> 00:02:58,140 So we know that the half-live for cobalt-60 is 1,925.4 days, 58 00:02:58,140 --> 00:02:59,150 which equals-- 59 00:02:59,150 --> 00:03:02,610 and I pre-did the math out so I wouldn't spend lots of time 60 00:03:02,610 --> 00:03:05,010 on the calculator. 61 00:03:05,010 --> 00:03:09,130 Activity, how many seconds is that? 62 00:03:09,130 --> 00:03:15,000 1.66 times 10 to the eighth seconds. 63 00:03:15,000 --> 00:03:17,920 And we also remember that activity equation, 64 00:03:17,920 --> 00:03:21,250 where the activity as a function of time 65 00:03:21,250 --> 00:03:26,970 equals the original activity times e to the minus lambda t. 66 00:03:26,970 --> 00:03:29,910 The only thing missing here is the initial activity 67 00:03:29,910 --> 00:03:33,160 and lambda, so how do we find lambda? 68 00:03:33,160 --> 00:03:34,808 Anyone remember that expression? 69 00:03:34,808 --> 00:03:37,108 AUDIENCE: Log of 2 over half-life. 70 00:03:37,108 --> 00:03:37,900 MICHAEL SHORT: Yep. 71 00:03:37,900 --> 00:03:42,520 And we know that lambda equals log of 2 72 00:03:42,520 --> 00:03:46,910 over the half-life, which in this case 73 00:03:46,910 --> 00:03:55,630 is about 0.693 over 1.66 times 10 74 00:03:55,630 --> 00:04:04,050 to the eighth, which equals 4.17 times 10 to the minus 9 75 00:04:04,050 --> 00:04:06,980 per second. 76 00:04:06,980 --> 00:04:08,320 And so now it's pretty easy. 77 00:04:08,320 --> 00:04:10,000 We know what lambda is. 78 00:04:10,000 --> 00:04:11,560 We know what our current A is. 79 00:04:11,560 --> 00:04:16,000 So we can say that our original activity is simply 80 00:04:16,000 --> 00:04:22,640 our current activity divided by e to the minus lambda t. 81 00:04:22,640 --> 00:04:25,480 What's t? 82 00:04:25,480 --> 00:04:28,090 Well, I made an approximation here. 83 00:04:28,090 --> 00:04:30,640 It's been about five years and seven months. 84 00:04:30,640 --> 00:04:34,700 I assumed that a month has 30 days, on average, 85 00:04:34,700 --> 00:04:37,840 which comes out to t-- 86 00:04:37,840 --> 00:04:41,220 I remember calculating this already-- 87 00:04:41,220 --> 00:04:46,730 1.74 times 10 to the eighth seconds. 88 00:04:46,730 --> 00:04:50,090 So we plug in that t right here, plug in that lambda 89 00:04:50,090 --> 00:04:55,190 right there, and we get our initial activity-- 90 00:04:55,190 --> 00:04:58,860 what did I get?-- 91 00:04:58,860 --> 00:05:04,450 was 1.07 microcuries. 92 00:05:04,450 --> 00:05:07,570 Now hopefully, this result is fairly intuitive. 93 00:05:07,570 --> 00:05:10,300 Because the half-life of cobalt-57 94 00:05:10,300 --> 00:05:14,050 is 1,925 days, which is about 5 and 1/4 years. 95 00:05:14,050 --> 00:05:17,620 And it's just over 5 and 1/4 years since this source 96 00:05:17,620 --> 00:05:18,940 was calibrated. 97 00:05:18,940 --> 00:05:22,510 And we have just under half of the original activity. 98 00:05:22,510 --> 00:05:25,800 So hopefully that's an intuitive numerical example. 99 00:05:25,800 --> 00:05:30,260 Now, let's say, how many atoms of cobalt-60 did we have? 100 00:05:30,260 --> 00:05:33,277 Or better yet, what was the mass of cobalt-60? 101 00:05:33,277 --> 00:05:34,360 Where would I go for that? 102 00:05:41,066 --> 00:05:41,958 I'll give you a hint. 103 00:05:41,958 --> 00:05:42,750 It's on the screen. 104 00:05:46,884 --> 00:05:48,820 AUDIENCE: Are you saying the mass 105 00:05:48,820 --> 00:05:50,870 that we have in the pellet? 106 00:05:50,870 --> 00:05:54,200 MICHAEL SHORT: Yeah, so for this plastic check source 107 00:05:54,200 --> 00:05:55,690 right here, what's the actual mass 108 00:05:55,690 --> 00:05:58,310 of cobalt-60 that we put in in the beginning, 109 00:05:58,310 --> 00:06:00,740 again, supposing I had one of these right in front of you? 110 00:06:03,795 --> 00:06:05,580 AUDIENCE: Look at the binding energy. 111 00:06:05,580 --> 00:06:07,830 MICHAEL SHORT: You could look at binding energy, which 112 00:06:07,830 --> 00:06:10,140 is a form of mass. 113 00:06:10,140 --> 00:06:14,690 Or luckily, we've got the atomic mass right up there in AMU. 114 00:06:14,690 --> 00:06:17,840 So again, this is a quick review of high school chemistry. 115 00:06:17,840 --> 00:06:21,250 We need to find out how many atoms we have in this pellet. 116 00:06:21,250 --> 00:06:23,010 And once we know the number of atoms-- 117 00:06:23,010 --> 00:06:26,710 and we have the molar mass up there in either AMU per atom 118 00:06:26,710 --> 00:06:30,070 or same thing as moles per gram, pretty much-- 119 00:06:30,070 --> 00:06:33,430 we'll know what the mass of cobalt-60 was. 120 00:06:33,430 --> 00:06:35,770 So one important conversion factor to note 121 00:06:35,770 --> 00:06:43,240 is that 1 curie of radiation is 3.7 times 122 00:06:43,240 --> 00:06:46,770 10 to the 10 becquerel. 123 00:06:46,770 --> 00:06:51,550 And remember that 1 becquerel is 1 disintegration per second. 124 00:07:02,760 --> 00:07:07,620 So if we know the initial activity of our material, 125 00:07:07,620 --> 00:07:11,070 using our decay constant, we should know the number of atoms 126 00:07:11,070 --> 00:07:12,610 that we had right there. 127 00:07:12,610 --> 00:07:16,810 So we know our initial activity, A0 is 1.07 microcurie. 128 00:07:16,810 --> 00:07:18,690 And anyone remember, what's the relation 129 00:07:18,690 --> 00:07:22,780 between the activity and the number of atoms present? 130 00:07:22,780 --> 00:07:23,866 Just yell it out. 131 00:07:27,514 --> 00:07:29,170 I hope you'd know that by now. 132 00:07:32,443 --> 00:07:33,860 Does this look familiar to anyone? 133 00:07:37,540 --> 00:07:40,660 Where the activity is directly proportional to the number 134 00:07:40,660 --> 00:07:44,390 of atoms there, times it's decay constant. 135 00:07:44,390 --> 00:07:50,160 So since we now know A0, we can find N0, because now we 136 00:07:50,160 --> 00:07:52,320 know lambda as well. 137 00:07:52,320 --> 00:07:56,000 So the number of atoms we had at the beginning 138 00:07:56,000 --> 00:08:00,250 is just, our activity over lambda. 139 00:08:00,250 --> 00:08:06,410 And our activity is 1.07 microcuries. 140 00:08:06,410 --> 00:08:08,450 Let's convert up to curies. 141 00:08:08,450 --> 00:08:16,210 So we know that there is 1 curie in 10 to the 6 microcuries. 142 00:08:16,210 --> 00:08:19,150 And we'll use that conversion factor right there, 143 00:08:19,150 --> 00:08:29,710 times 3.7, times 10 to the 10 becquerel per 1 curie, 144 00:08:29,710 --> 00:08:37,320 divided by our decay constant, 4.17 times 10 to the minus 9 145 00:08:37,320 --> 00:08:38,760 per second. 146 00:08:38,760 --> 00:08:41,520 Let's check our units to make sure everything comes out. 147 00:08:44,720 --> 00:08:47,970 I brought a canceling color. 148 00:08:47,970 --> 00:08:51,440 We have microcuries on the top, microcuries on the bottom, 149 00:08:51,440 --> 00:08:54,420 curies on the top, curies on the bottom. 150 00:08:54,420 --> 00:08:55,860 And we have a becquerels, which is 151 00:08:55,860 --> 00:08:58,260 a disintegrations per second. 152 00:08:58,260 --> 00:09:01,450 And we have a per second down on the bottom. 153 00:09:01,450 --> 00:09:03,720 So the per second goes away and the becquerels just 154 00:09:03,720 --> 00:09:06,675 becomes atoms. 155 00:09:09,260 --> 00:09:13,600 And so we actually get, N0 is-- 156 00:09:13,600 --> 00:09:15,310 I think it's something times 10 to 12th. 157 00:09:15,310 --> 00:09:19,930 Yeah, 9.5 times 10 to the 12th atoms. 158 00:09:22,700 --> 00:09:24,250 That's the way it looks. 159 00:09:24,250 --> 00:09:27,790 So the final step, how do we go from number of atoms to mass? 160 00:09:27,790 --> 00:09:29,730 Can anyone tell me? 161 00:09:29,730 --> 00:09:33,530 AUDIENCE: Convert it to Avogrado's number and moles. 162 00:09:33,530 --> 00:09:34,760 MICHAEL SHORT: Yep. 163 00:09:34,760 --> 00:09:36,920 So the last thing we'll do is we'll 164 00:09:36,920 --> 00:09:44,080 say we have 9.5 times 10 to the 12th atoms, 165 00:09:44,080 --> 00:09:49,300 times Avogadro's number, which is 1 mole in every 6 times 10 166 00:09:49,300 --> 00:09:50,605 to the 23rd atoms. 167 00:09:53,732 --> 00:09:55,190 Then we go to the table of nuclides 168 00:09:55,190 --> 00:09:57,308 to get its atomic mass. 169 00:09:57,308 --> 00:09:58,850 This is one of those situations where 170 00:09:58,850 --> 00:10:00,850 you don't need to take the eighth decimal place, 171 00:10:00,850 --> 00:10:04,070 because you're not converting from mass to energy. 172 00:10:04,070 --> 00:10:05,330 You're just getting mass. 173 00:10:05,330 --> 00:10:08,120 So if you might wonder, where did all the decimal points go, 174 00:10:08,120 --> 00:10:10,520 think of what type of calculation we're doing. 175 00:10:10,520 --> 00:10:14,007 If we started off with a one significant digit number, 176 00:10:14,007 --> 00:10:16,340 do we really care about keeping the eighth decimal place 177 00:10:16,340 --> 00:10:18,050 in the rest of everything? 178 00:10:18,050 --> 00:10:20,480 No, definitely not. 179 00:10:20,480 --> 00:10:24,080 We're not turning AMU into MEV, in which case 180 00:10:24,080 --> 00:10:27,417 the sixth decimal point could put you off by half an MEV, 181 00:10:27,417 --> 00:10:29,000 or like the rest mass of the electron. 182 00:10:29,000 --> 00:10:30,690 We're just getting masses here. 183 00:10:30,690 --> 00:10:33,530 So let's just call that 59.9-- 184 00:10:33,530 --> 00:10:36,340 that's enough for me-- 185 00:10:36,340 --> 00:10:43,310 59.9 grams in 1 mole of cobalt-60. 186 00:10:43,310 --> 00:10:49,950 And just to confirm, we have atoms here, atoms there, 187 00:10:49,950 --> 00:10:52,540 moles there, moles there. 188 00:10:52,540 --> 00:10:54,840 We should get a mass in grams. 189 00:10:54,840 --> 00:11:05,150 And this came out to 0.95 nanograms of cobalt-60. 190 00:11:05,150 --> 00:11:08,910 Not a lot of cobalt-60 can pack quite a wallop 191 00:11:08,910 --> 00:11:10,570 in terms of activity. 192 00:11:10,570 --> 00:11:12,570 So even though this has a pretty long half-life, 193 00:11:12,570 --> 00:11:16,230 as far as isotopes go, it takes very little of it 194 00:11:16,230 --> 00:11:18,540 to have quite a bit of activity, certainly enough 195 00:11:18,540 --> 00:11:21,705 for our fairly inefficient handmade Geiger counter 196 00:11:21,705 --> 00:11:22,830 to measure what's going on. 197 00:11:25,340 --> 00:11:28,130 Pretty neat, huh? 198 00:11:28,130 --> 00:11:30,530 Cool. 199 00:11:30,530 --> 00:11:33,620 Now, let's say I ask you another question, a simpler question. 200 00:11:33,620 --> 00:11:35,570 How many disintegrations per second 201 00:11:35,570 --> 00:11:38,060 are coming out of that cobalt-60 source? 202 00:11:38,060 --> 00:11:41,690 Let's say we wanted to find the efficiency of our Geiger 203 00:11:41,690 --> 00:11:42,680 counter. 204 00:11:42,680 --> 00:11:45,132 You'd have to know how many counts you measure. 205 00:11:45,132 --> 00:11:47,090 You'd have to know how far away your source is. 206 00:11:47,090 --> 00:11:51,200 And you'd have to know how many disintegrations per second 207 00:11:51,200 --> 00:11:51,890 are happening. 208 00:11:51,890 --> 00:11:55,840 So how would I get to the number of disintegrations per second? 209 00:11:55,840 --> 00:11:57,560 AUDIENCE: Convert it to becquerels. 210 00:11:57,560 --> 00:11:58,727 MICHAEL SHORT: That's right. 211 00:11:58,727 --> 00:12:00,680 We'll just take our current activity, 212 00:12:00,680 --> 00:12:04,430 which is 0.52 microcuries. 213 00:12:04,430 --> 00:12:11,420 And so we'll say, 0.52 times 10 to the minus 6 curies, 214 00:12:11,420 --> 00:12:23,230 times 3.7, times 10 to the 10 becquerel in 1 curie is-- 215 00:12:23,230 --> 00:12:24,880 well, we have curies here. 216 00:12:24,880 --> 00:12:27,260 We have curies there. 217 00:12:27,260 --> 00:12:30,955 And this comes out to-- what was our current activity. 218 00:12:30,955 --> 00:12:33,230 It was in the tens of thousands. 219 00:12:33,230 --> 00:12:37,910 About 19,000 becquerels. 220 00:12:37,910 --> 00:12:39,880 So we know that, right now, this source 221 00:12:39,880 --> 00:12:44,480 is giving off about 19,000 disintegrations per second. 222 00:12:44,480 --> 00:12:46,340 Is that how many gamma rays it's giving off? 223 00:12:46,340 --> 00:12:48,593 Do we know that yet? 224 00:12:48,593 --> 00:12:49,922 AUDIENCE: No. 225 00:12:49,922 --> 00:12:50,880 MICHAEL SHORT: Why not? 226 00:12:50,880 --> 00:12:51,797 I heard a lot of no's. 227 00:12:54,700 --> 00:12:57,864 What other information do we need to know? 228 00:12:57,864 --> 00:12:59,700 AUDIENCE: Type of radiation. 229 00:12:59,700 --> 00:13:00,620 MICHAEL SHORT: Sure. 230 00:13:00,620 --> 00:13:03,980 Luckily we've got that right here. 231 00:13:03,980 --> 00:13:05,890 So if you look here, the mode of decay 232 00:13:05,890 --> 00:13:08,590 is beta decay to nickel-60. 233 00:13:08,590 --> 00:13:12,130 So cobalt-60 is actually primarily a beta source. 234 00:13:12,130 --> 00:13:16,610 However, it's used for its characteristic gamma rays. 235 00:13:16,610 --> 00:13:18,870 So let's take a quick look. 236 00:13:18,870 --> 00:13:22,920 We look at its decay diagram, it's pretty simple. 237 00:13:22,920 --> 00:13:27,870 And let's just say, somewhere near 100% of the time, 238 00:13:27,870 --> 00:13:29,970 it beta decays up to this energy level 239 00:13:29,970 --> 00:13:33,150 and can undergo any number of transitions like this. 240 00:13:33,150 --> 00:13:34,620 The only two we really tend to see 241 00:13:34,620 --> 00:13:40,020 is, like that one and that one are the most likely ones. 242 00:13:40,020 --> 00:13:42,750 So on average, each disintegration 243 00:13:42,750 --> 00:13:46,590 of a cobalt-60 atom is going to produce two highly energetic 244 00:13:46,590 --> 00:13:48,250 gamma rays. 245 00:13:48,250 --> 00:13:51,240 So actually, what you'd have to know for this source is, 246 00:13:51,240 --> 00:13:56,550 despite it being 19,000 becquerels of cobalt-60, 247 00:13:56,550 --> 00:13:59,970 it's giving off 38,000 gamma rays per second. 248 00:13:59,970 --> 00:14:01,890 So that way your source calculations 249 00:14:01,890 --> 00:14:04,528 wouldn't be off by a factor of 2. 250 00:14:04,528 --> 00:14:06,570 Then if you know the distance between your source 251 00:14:06,570 --> 00:14:09,720 and your detector, and you know how many of those gamma 252 00:14:09,720 --> 00:14:13,440 rays are going through the detector itself-- 253 00:14:13,440 --> 00:14:15,870 which we can count calculate with a solid angle formula, 254 00:14:15,870 --> 00:14:17,472 which I'll give you a little later-- 255 00:14:17,472 --> 00:14:18,930 you'll know how many of them should 256 00:14:18,930 --> 00:14:23,340 interact in here, once you learn photon nuclear interactions. 257 00:14:23,340 --> 00:14:25,920 And you'll know how many of them actually get captured. 258 00:14:25,920 --> 00:14:28,500 And that's how you can get the efficiency of the detector. 259 00:14:28,500 --> 00:14:30,540 So we've kind of filled in half the puzzle. 260 00:14:30,540 --> 00:14:32,790 You now know, for a fixed source, 261 00:14:32,790 --> 00:14:35,910 how many atoms there are, how many disintegrations there are 262 00:14:35,910 --> 00:14:38,442 and how many gamma rays you expect it to give off. 263 00:14:38,442 --> 00:14:39,900 And then later in the course, we'll 264 00:14:39,900 --> 00:14:41,750 tell you how to figure out how many of them 265 00:14:41,750 --> 00:14:44,460 make it into the detector and how many of those 266 00:14:44,460 --> 00:14:48,380 should interact in the detector. 267 00:14:48,380 --> 00:14:50,350 Is everyone clear on what we talked about here? 268 00:14:50,350 --> 00:14:50,850 Yep. 269 00:14:50,850 --> 00:14:54,800 AUDIENCE: Where did you get the number for 38,000 gamma rays? 270 00:14:54,800 --> 00:14:58,730 MICHAEL SHORT: Double the number of cobalt disintegrations. 271 00:14:58,730 --> 00:15:00,777 AUDIENCE: Because we assume it drops 2? 272 00:15:00,777 --> 00:15:01,610 MICHAEL SHORT: Yeah. 273 00:15:01,610 --> 00:15:05,720 So I'm ignoring the 0.2% and 0.6% 274 00:15:05,720 --> 00:15:08,060 decays because they're extremely unlikely. 275 00:15:08,060 --> 00:15:09,800 And having looked this up ahead of time, 276 00:15:09,800 --> 00:15:13,970 I know that this transition and that transition 277 00:15:13,970 --> 00:15:16,123 are by far the most likely ones. 278 00:15:16,123 --> 00:15:17,290 And if you don't know that-- 279 00:15:17,290 --> 00:15:19,040 AUDIENCE: Yeah, does it say it down there? 280 00:15:19,040 --> 00:15:21,730 MICHAEL SHORT: It sure do. 281 00:15:21,730 --> 00:15:25,980 Those right there, the 99.9-something intensity ones, 282 00:15:25,980 --> 00:15:26,680 it's all there. 283 00:15:26,680 --> 00:15:28,990 So I had some questions on Piazza. 284 00:15:28,990 --> 00:15:32,590 What happens when we can't read the pixelated decay diagram? 285 00:15:32,590 --> 00:15:34,270 That's just there for fun. 286 00:15:34,270 --> 00:15:36,760 The actual table that will tell you all the information you 287 00:15:36,760 --> 00:15:40,110 need to know is right below. 288 00:15:40,110 --> 00:15:43,110 And I just decided, all right, forget the ones that are 2e 289 00:15:43,110 --> 00:15:48,210 to the minus 6 likely or point 0.007, whatever, too many 0's. 290 00:15:48,210 --> 00:15:50,700 So you can say that, on average, to probably two 291 00:15:50,700 --> 00:15:54,240 or three significant digits, that gamma ray, 292 00:15:54,240 --> 00:15:57,140 that gamma ray are the only ones you tend to see. 293 00:15:59,750 --> 00:16:03,373 Is everyone clear on how I made that determination? 294 00:16:03,373 --> 00:16:06,000 Cool. 295 00:16:06,000 --> 00:16:06,500 Yes. 296 00:16:06,500 --> 00:16:10,260 AUDIENCE: So are disintegrations just a beta, 297 00:16:10,260 --> 00:16:14,333 and then for each one of those, it's two gamma rays? 298 00:16:14,333 --> 00:16:15,500 MICHAEL SHORT: That's right. 299 00:16:15,500 --> 00:16:17,330 The number of disintegrations is the number 300 00:16:17,330 --> 00:16:19,740 of atoms that leave this position. 301 00:16:19,740 --> 00:16:22,520 So let's take a crazier example. 302 00:16:22,520 --> 00:16:27,310 I've already had you look at americium-241. 303 00:16:27,310 --> 00:16:31,270 Sort of bring home the message that the number of americium 304 00:16:31,270 --> 00:16:34,450 disintegrations does not necessarily 305 00:16:34,450 --> 00:16:40,400 equal the number of gamma rays that you will expect to see. 306 00:16:40,400 --> 00:16:42,990 Because any one of these is possible. 307 00:16:42,990 --> 00:16:45,710 There is an 84% likely one. 308 00:16:45,710 --> 00:16:48,290 It looks like it goes to the third level from the bottom, 309 00:16:48,290 --> 00:16:50,900 because it's just the third number I'm seeing here. 310 00:16:50,900 --> 00:16:53,312 If you don't believe that, then check 311 00:16:53,312 --> 00:16:55,520 how much you have to zoom out to see what's going on. 312 00:16:58,890 --> 00:17:00,610 Let's see. 313 00:17:00,610 --> 00:17:02,190 Intensity, yep. 314 00:17:02,190 --> 00:17:04,950 The third most energetic alpha ray, 315 00:17:04,950 --> 00:17:09,390 so the third from the bottom, is indeed the 84% likely one. 316 00:17:09,390 --> 00:17:11,730 Then there's another one that's 13% likely, 317 00:17:11,730 --> 00:17:14,527 so we can't discount that either. 318 00:17:14,527 --> 00:17:16,319 So you're going to hit anything from, like, 319 00:17:16,319 --> 00:17:19,560 the second to the fourth energy level 320 00:17:19,560 --> 00:17:24,780 and any number of those gamma cascades that come off there. 321 00:17:24,780 --> 00:17:28,740 So the intensity of your source in becquerels or curies 322 00:17:28,740 --> 00:17:30,270 does not immediately tell you how 323 00:17:30,270 --> 00:17:33,060 many gamma rays or other disintegration products 324 00:17:33,060 --> 00:17:35,850 you should expect. 325 00:17:35,850 --> 00:17:40,080 There are some pretty simple ones, like the dysprosium 326 00:17:40,080 --> 00:17:41,070 one that we saw. 327 00:17:41,070 --> 00:17:43,603 Was it 151? 328 00:17:43,603 --> 00:17:44,520 I've been here before. 329 00:17:48,140 --> 00:17:49,110 It's not purple. 330 00:17:49,110 --> 00:17:51,240 No, that's a complicated one. 331 00:17:51,240 --> 00:17:54,270 All right, I don't remember which dysprosium isotope just 332 00:17:54,270 --> 00:17:56,320 had a single decay. 333 00:17:56,320 --> 00:17:58,020 But the only time that you would get 334 00:17:58,020 --> 00:18:02,010 one particle per disintegration is if your decay diagram 335 00:18:02,010 --> 00:18:04,750 looked like this. 336 00:18:04,750 --> 00:18:07,110 That's your parent and that's your daughter. 337 00:18:07,110 --> 00:18:10,390 That's the only time you can expect one thing. 338 00:18:10,390 --> 00:18:13,220 If you have a more complex decay diagram than that, 339 00:18:13,220 --> 00:18:16,570 you'll be looking at more than one quantum of radiation 340 00:18:16,570 --> 00:18:20,480 of some form per disintegration. 341 00:18:20,480 --> 00:18:24,716 So is that unclear to anybody? 342 00:18:24,716 --> 00:18:26,640 Cool. 343 00:18:26,640 --> 00:18:30,540 Now let's get into a more fun problem. 344 00:18:30,540 --> 00:18:32,690 And I'll leave this up here, since we're going 345 00:18:32,690 --> 00:18:35,100 to refer to it a fair bit. 346 00:18:35,100 --> 00:18:38,970 So this is a good example of activity, half-life, mass, 347 00:18:38,970 --> 00:18:40,540 number of atom calculations. 348 00:18:40,540 --> 00:18:42,570 We're going to use this information 349 00:18:42,570 --> 00:18:46,150 to answer a more interesting question. 350 00:18:46,150 --> 00:18:46,910 How was this made? 351 00:18:49,800 --> 00:18:55,640 Anyone have any idea what sort of intentional nuclear reaction 352 00:18:55,640 --> 00:18:58,860 could have produced cobalt-60? 353 00:18:58,860 --> 00:19:02,530 And I'll go back to cobalt-60 so we can get its proton number. 354 00:19:02,530 --> 00:19:03,148 There it is. 355 00:19:03,148 --> 00:19:04,440 Can you say it a little louder? 356 00:19:04,440 --> 00:19:06,870 AUDIENCE: [INAUDIBLE] Oh, I said neutron bombardment 357 00:19:06,870 --> 00:19:08,710 of cobalt-59. 358 00:19:08,710 --> 00:19:12,220 MICHAEL SHORT: Sure, you can have a neutron bombardment. 359 00:19:12,220 --> 00:19:20,840 So you can have, cobalt-59 absorbs a neutron, 360 00:19:20,840 --> 00:19:24,520 becomes cobalt-60 with some half-life. 361 00:19:24,520 --> 00:19:28,515 And then we'll decay with some decay constant lambda, 362 00:19:28,515 --> 00:19:29,015 and become-- 363 00:19:32,810 --> 00:19:34,620 in this case, it undergoes beta decay-- 364 00:19:37,680 --> 00:19:38,460 to nickel-60. 365 00:19:38,460 --> 00:19:41,785 Let me make sure I got the proton number right. 366 00:19:41,785 --> 00:19:43,675 I'm going from memory here. 367 00:19:43,675 --> 00:19:45,414 Hooray. 368 00:19:45,414 --> 00:19:50,320 OK, how do we set up the series radioactive 369 00:19:50,320 --> 00:19:53,290 decay and production equations to describe this phenomenon? 370 00:19:53,290 --> 00:19:57,170 So we have a physical picture of what's going on here. 371 00:19:57,170 --> 00:19:59,535 How do we construct the differential equation model? 372 00:20:02,720 --> 00:20:05,030 The same way we've been doing it Friday and Tuesday. 373 00:20:05,030 --> 00:20:07,830 So someone, kick me off. 374 00:20:07,830 --> 00:20:08,856 What do we start with? 375 00:20:08,856 --> 00:20:13,640 AUDIENCE: So the production of cobalt-59 is 0. 376 00:20:13,640 --> 00:20:20,750 Destruction would be sigma times flux times the amount 377 00:20:20,750 --> 00:20:22,540 of cobalt. 378 00:20:22,540 --> 00:20:25,670 MICHAEL SHORT: OK, let's make a couple of designations. 379 00:20:25,670 --> 00:20:27,830 Let's call cobalt-59-- 380 00:20:27,830 --> 00:20:30,500 I'll use another color for this, just so it's clear. 381 00:20:30,500 --> 00:20:33,360 We'll call cobalt-59 N1. 382 00:20:33,360 --> 00:20:36,280 We'll call cobalt-60 N2. 383 00:20:36,280 --> 00:20:43,650 We'll call nickel-60 N3, just to keep our notation straight. 384 00:20:43,650 --> 00:20:48,567 And then continuing our notation pile here, 385 00:20:48,567 --> 00:20:50,150 you mentioned that there's going to be 386 00:20:50,150 --> 00:20:54,620 some sigma, some microscopic cross-section times some flux, 387 00:20:54,620 --> 00:20:56,780 the number of neutrons whizzing about the reactor, 388 00:20:56,780 --> 00:20:59,900 times the amount of cobalt-60. 389 00:20:59,900 --> 00:21:06,070 So let's stick our DN, Dt, equals no production 390 00:21:06,070 --> 00:21:08,440 minus some destruction. 391 00:21:08,440 --> 00:21:12,410 Does this look eerily familiar to any of you? 392 00:21:12,410 --> 00:21:15,120 Forget the fact that it's a sigma and a flux. 393 00:21:15,120 --> 00:21:16,370 Just treat those as constants. 394 00:21:16,370 --> 00:21:20,220 What does this look exactly like? 395 00:21:20,220 --> 00:21:21,560 AUDIENCE: The minus lambda D. 396 00:21:21,560 --> 00:21:22,880 MICHAEL SHORT: Exactly. 397 00:21:22,880 --> 00:21:30,950 It looks just like dN1 dt equals minus lambda and 1. 398 00:21:30,950 --> 00:21:32,540 Yeah, that's exactly it. 399 00:21:32,540 --> 00:21:41,760 This, in effect, is like your artificially-induced decay. 400 00:21:44,400 --> 00:21:48,210 So the probability that one atom of N1 absorbs a neutron, 401 00:21:48,210 --> 00:21:50,250 times the number of neutrons are there, 402 00:21:50,250 --> 00:21:53,130 gives you some rate at which these atoms are destroyed. 403 00:21:53,130 --> 00:21:56,100 Just like a lambda gives you the rate at which 404 00:21:56,100 --> 00:21:58,620 those atoms naturally self-destruct. 405 00:21:58,620 --> 00:22:03,360 So you can think of the sigma times flux like a lambda. 406 00:22:03,360 --> 00:22:05,970 Mathematically, they're treated identically. 407 00:22:05,970 --> 00:22:09,390 The only difference is, we're imposing this neutron flux. 408 00:22:09,390 --> 00:22:12,302 So it's like an artificial lambda, 409 00:22:12,302 --> 00:22:14,510 which means solving the equations and setting them up 410 00:22:14,510 --> 00:22:16,590 is exactly the same. 411 00:22:16,590 --> 00:22:17,480 So how about N2? 412 00:22:17,480 --> 00:22:22,140 What's the production and destruction rate of cobalt-60? 413 00:22:22,140 --> 00:22:27,678 First of all, someone yell it out, what's the creation rate? 414 00:22:27,678 --> 00:22:30,480 AUDIENCE: Lambda N1. 415 00:22:30,480 --> 00:22:31,680 MICHAEL SHORT: Which lambda? 416 00:22:31,680 --> 00:22:33,600 AUDIENCE: Oh, lambda 1 I mean. 417 00:22:33,600 --> 00:22:35,940 MICHAEL SHORT: So what is lambda 1, this one? 418 00:22:35,940 --> 00:22:37,240 AUDIENCE: Sigma, yeah. 419 00:22:37,240 --> 00:22:39,617 MICHAEL SHORT: OK, so let's keep with these notations. 420 00:22:39,617 --> 00:22:41,700 I'm going to say this is like a lambda artificial, 421 00:22:41,700 --> 00:22:47,840 but I'm going to keep with sigma flux and 1. 422 00:22:47,840 --> 00:22:50,100 And just so we keep our notation straight, 423 00:22:50,100 --> 00:22:51,830 I want to be able to cleanly separate 424 00:22:51,830 --> 00:22:54,920 the natural and the artificial production and destruction. 425 00:22:54,920 --> 00:22:58,300 And what's the destruction rate? 426 00:22:58,300 --> 00:23:00,372 AUDIENCE: Sigma phi N2. 427 00:23:00,372 --> 00:23:04,175 MICHAEL SHORT: So you said sigma phi N2. 428 00:23:04,175 --> 00:23:06,050 Is that the physical picture we have up here? 429 00:23:11,440 --> 00:23:12,880 Not quite, because like we can see 430 00:23:12,880 --> 00:23:18,240 here, cobalt-60 self-destructs on its own, right? 431 00:23:18,240 --> 00:23:20,380 Yeah. 432 00:23:20,380 --> 00:23:23,740 However, that actually is a correct term to put in. 433 00:23:23,740 --> 00:23:26,430 The other one that we're missing would be minus lambda N2. 434 00:23:30,890 --> 00:23:32,120 So you know what? 435 00:23:32,120 --> 00:23:35,330 Let's escalate this a bit into reality and say, 436 00:23:35,330 --> 00:23:36,290 we're going to do this. 437 00:23:36,290 --> 00:23:38,623 The actual equation is not going to be that much harder. 438 00:23:38,623 --> 00:23:39,890 So thanks, Sean. 439 00:23:39,890 --> 00:23:42,120 Let's do this. 440 00:23:42,120 --> 00:23:42,620 Yeah. 441 00:23:42,620 --> 00:23:44,370 AUDIENCE: Would those sigmas be different? 442 00:23:44,370 --> 00:23:45,850 MICHAEL SHORT: Yes, good question. 443 00:23:45,850 --> 00:23:47,290 That's what I was getting to. 444 00:23:47,290 --> 00:23:49,900 So there is an absorption cross-section 445 00:23:49,900 --> 00:23:53,320 for every reaction for every isotope. 446 00:23:53,320 --> 00:23:56,080 And now I'd like to show you guys where to find them. 447 00:23:56,080 --> 00:23:58,810 On the 22.01 site, there is a link 448 00:23:58,810 --> 00:24:00,640 to what's called the Janis Database, which 449 00:24:00,640 --> 00:24:04,460 is tabulated and plottable cross-sections of all kinds. 450 00:24:04,460 --> 00:24:06,820 So every single database that we know of, 451 00:24:06,820 --> 00:24:09,640 it's updated continuously by the OECD, 452 00:24:09,640 --> 00:24:12,593 so you can trust that the data is updated. 453 00:24:12,593 --> 00:24:15,010 I won't say accurate, because some of these cross-sections 454 00:24:15,010 --> 00:24:16,960 are not very well known. 455 00:24:16,960 --> 00:24:19,310 So I've already started it up. 456 00:24:19,310 --> 00:24:22,520 In this case, we don't have neutron capture data 457 00:24:22,520 --> 00:24:24,170 for cobalt-60. 458 00:24:24,170 --> 00:24:26,450 Let's keep it in the symbols for now. 459 00:24:26,450 --> 00:24:29,300 And then later on, we're just going to say, it's probably 0. 460 00:24:29,300 --> 00:24:31,460 But let's keep it together. 461 00:24:31,460 --> 00:24:35,600 So for cobalt-59, double-click on that, and you 462 00:24:35,600 --> 00:24:40,190 are presented with an enormous host of possible reactions. 463 00:24:40,190 --> 00:24:42,990 Like right here, you have N comma total, 464 00:24:42,990 --> 00:24:47,030 which is the total cross-section for all interactions 465 00:24:47,030 --> 00:24:50,460 of neutrons with cobalt-59. 466 00:24:50,460 --> 00:24:51,890 And you can see that this varies, 467 00:24:51,890 --> 00:24:55,580 but the same sort of shape that I was haphazardly drawing. 468 00:24:55,580 --> 00:24:56,330 It's low. 469 00:24:56,330 --> 00:24:58,310 There are some resonances. 470 00:24:58,310 --> 00:25:01,250 And then it continuously increases at lower energies. 471 00:25:01,250 --> 00:25:03,050 Is this the right cross-section to use? 472 00:25:07,290 --> 00:25:09,390 If it accounts for every possible interaction 473 00:25:09,390 --> 00:25:12,880 of a neutron with cobalt-59. 474 00:25:12,880 --> 00:25:13,380 No. 475 00:25:13,380 --> 00:25:16,260 So what other reactions, besides the one that we want, which is 476 00:25:16,260 --> 00:25:20,178 capture, could this account for? 477 00:25:20,178 --> 00:25:21,130 AUDIENCE: Scattering. 478 00:25:21,130 --> 00:25:21,606 MICHAEL SHORT: Yep. 479 00:25:21,606 --> 00:25:22,560 AUDIENCE: Fission. 480 00:25:22,560 --> 00:25:26,960 MICHAEL SHORT: Scattering, fission, n2n production. 481 00:25:26,960 --> 00:25:28,760 Sometimes one neutron goes in and two 482 00:25:28,760 --> 00:25:31,070 come out, like for beryllium, like we 483 00:25:31,070 --> 00:25:32,720 talked about at the beginning of class. 484 00:25:32,720 --> 00:25:35,840 So you've got to know to choose the right cross-section. 485 00:25:35,840 --> 00:25:38,450 And in nuclear reaction parlance-- 486 00:25:38,450 --> 00:25:40,040 that's shorthand right there-- 487 00:25:40,040 --> 00:25:41,180 that's N comma total. 488 00:25:41,180 --> 00:25:43,460 That accounts for elastic scattering, 489 00:25:43,460 --> 00:25:47,690 inelastic scattering to any energy level, capture, fission, 490 00:25:47,690 --> 00:25:52,550 n2n reactions, sometimes proton release, sometimes exploding, 491 00:25:52,550 --> 00:25:54,830 whatever nuclei do. 492 00:25:54,830 --> 00:25:57,380 So let's look at our other choices. 493 00:25:57,380 --> 00:25:59,150 I'll shrink that. 494 00:25:59,150 --> 00:26:02,920 We have the elastic cross-section. 495 00:26:02,920 --> 00:26:04,840 Compare that with a total, and you 496 00:26:04,840 --> 00:26:07,600 get a general idea of how much the total 497 00:26:07,600 --> 00:26:10,790 at elastic cross-sections actually matter. 498 00:26:10,790 --> 00:26:14,622 So a lot of those resonances are elastic cross-section 499 00:26:14,622 --> 00:26:16,330 resonances, but there are other reactions 500 00:26:16,330 --> 00:26:18,788 that are responsible for a lot of the other craziness going 501 00:26:18,788 --> 00:26:20,600 on. 502 00:26:20,600 --> 00:26:25,001 So let's unselect that. 503 00:26:25,001 --> 00:26:25,950 Oh, I'm sorry. 504 00:26:25,950 --> 00:26:31,210 I meant MT 1 Let's compare MT 1 and MT 2. 505 00:26:31,210 --> 00:26:32,545 So MT 2, elastic. 506 00:26:35,880 --> 00:26:36,990 There we go. 507 00:26:36,990 --> 00:26:39,620 OK, that's what I was hoping for. 508 00:26:39,620 --> 00:26:42,350 So right now, the red one is the total cross-section, 509 00:26:42,350 --> 00:26:46,095 and the green one is the elastic cross-section. 510 00:26:46,095 --> 00:26:47,720 And you can see that, at high energies, 511 00:26:47,720 --> 00:26:49,370 the total cross-section is mostly 512 00:26:49,370 --> 00:26:51,050 the elastic cross-section. 513 00:26:51,050 --> 00:26:54,790 But at low energies, especially right around here 514 00:26:54,790 --> 00:26:57,330 at the thermal energy of neutrons in a reactor, 515 00:26:57,330 --> 00:26:59,100 there's something else responsible. 516 00:26:59,100 --> 00:27:01,470 That's probably what we're going after. 517 00:27:01,470 --> 00:27:06,762 So let's keep looking through this Janis Database. 518 00:27:06,762 --> 00:27:09,080 Hey, there's the n2n reaction, if you guys 519 00:27:09,080 --> 00:27:11,710 want to see how likely this is. 520 00:27:11,710 --> 00:27:13,180 It's another one of those reactions 521 00:27:13,180 --> 00:27:17,560 that-- look at that-- it's 0 until you get to 11 MEV. 522 00:27:17,560 --> 00:27:19,570 So what do you guys think the q-value 523 00:27:19,570 --> 00:27:24,068 for n2n production of cobalt-59 is? 524 00:27:24,068 --> 00:27:25,543 AUDIENCE: Very negative. 525 00:27:25,543 --> 00:27:26,710 MICHAEL SHORT: How negative? 526 00:27:26,710 --> 00:27:28,674 It's on the graph. 527 00:27:28,674 --> 00:27:29,570 AUDIENCE: 10. 528 00:27:29,570 --> 00:27:32,160 MICHAEL SHORT: Yeah, negative looks like 10, or maybe 10 529 00:27:32,160 --> 00:27:33,510 and 1/2, MEV. 530 00:27:33,510 --> 00:27:36,210 AUDIENCE: 0.454. 531 00:27:36,210 --> 00:27:38,280 MICHAEL SHORT: Oh, hey, awesome. 532 00:27:38,280 --> 00:27:38,780 [LAUGHTER] 533 00:27:38,780 --> 00:27:40,680 AUDIENCE: I think it moves it around. 534 00:27:40,680 --> 00:27:41,972 MICHAEL SHORT: So there you go. 535 00:27:41,972 --> 00:27:44,460 Yeah, indeed, it's very energetically unlikely 536 00:27:44,460 --> 00:27:46,650 to fire in one neutrons and get two. 537 00:27:46,650 --> 00:27:50,580 But if you have a 10.454 MEV neutron, 538 00:27:50,580 --> 00:27:52,700 you can make it happen. 539 00:27:52,700 --> 00:27:54,470 Pretty cool, huh? 540 00:27:54,470 --> 00:27:56,870 That's not the reaction we're going for. 541 00:27:56,870 --> 00:28:01,370 What we want-- let's see if I can find it. 542 00:28:01,370 --> 00:28:04,520 Proton plus neutron, neutron plus deuteron, all 543 00:28:04,520 --> 00:28:06,320 the inelastic energy levels. 544 00:28:06,320 --> 00:28:09,510 There it is, capture. 545 00:28:09,510 --> 00:28:13,810 The gamma reaction here is what's referred to as capture. 546 00:28:13,810 --> 00:28:16,660 And there we go, a nice normal-looking cross-section. 547 00:28:16,660 --> 00:28:20,990 So for cobalt-59, if we go down to about 0.025 548 00:28:20,990 --> 00:28:26,830 EV here, read it off, it's about 20 barns. 549 00:28:26,830 --> 00:28:29,440 Because you guys asked for a numerical example. 550 00:28:29,440 --> 00:28:36,670 So let's say that our capture cross-section for cobalt-59 551 00:28:36,670 --> 00:28:39,640 is about 20 barns, which is to say 552 00:28:39,640 --> 00:28:47,160 20 times 10 to the minus 24th centimeter squared. 553 00:28:47,160 --> 00:28:50,645 Let's also put up our capture one for cobalt-60. 554 00:28:55,430 --> 00:28:58,403 So let's go back to our table, take a look at cobalt-60. 555 00:28:58,403 --> 00:29:00,320 I think I know what the answer is going to be, 556 00:29:00,320 --> 00:29:03,980 which is, we don't know. 557 00:29:03,980 --> 00:29:05,870 Not in this database, unfortunately. 558 00:29:05,870 --> 00:29:08,330 So for symbolism, let's keep it there. 559 00:29:08,330 --> 00:29:10,850 But we're going to say, well, we don't know. 560 00:29:14,220 --> 00:29:15,147 Yeah. 561 00:29:15,147 --> 00:29:17,230 So let's designate these different cross-sections. 562 00:29:17,230 --> 00:29:25,780 Let's call it sigma-59 and sigma-60. 563 00:29:25,780 --> 00:29:28,930 So those will be our two cross-sections. 564 00:29:28,930 --> 00:29:32,445 We'll just call this one sigma-59, 565 00:29:32,445 --> 00:29:34,360 and call this one sigma-60. 566 00:29:39,050 --> 00:29:41,450 And we already know the lambda for cobalt-60. 567 00:29:41,450 --> 00:29:46,160 So let's say the lambda for 60 cobalt, from this stuff 568 00:29:46,160 --> 00:29:53,690 up here, 4.17 times 10 to the minus 9 per second. 569 00:29:53,690 --> 00:29:55,430 Let's just refer to that as lambda 570 00:29:55,430 --> 00:29:58,220 for ease of writing things down. 571 00:30:00,840 --> 00:30:04,985 So we've got a complete set of reactions for a dN2 dt. 572 00:30:04,985 --> 00:30:05,610 What about dN3? 573 00:30:11,450 --> 00:30:13,197 What's the production rate of N3? 574 00:30:18,921 --> 00:30:20,180 AUDIENCE: Lambda 575 00:30:20,180 --> 00:30:21,430 MICHAEL SHORT: Yep, lambda N2. 576 00:30:24,920 --> 00:30:25,597 Anything else? 577 00:30:25,597 --> 00:30:26,930 What about the destruction rate? 578 00:30:32,120 --> 00:30:32,620 Is what? 579 00:30:32,620 --> 00:30:34,408 AUDIENCE: [INAUDIBLE] 580 00:30:34,408 --> 00:30:35,950 MICHAEL SHORT: It's a stable isotope. 581 00:30:35,950 --> 00:30:37,300 AUDIENCE: We don't know. 582 00:30:37,300 --> 00:30:38,290 MICHAEL SHORT: But you were on to it, Sean. 583 00:30:38,290 --> 00:30:40,840 So what would you add, based on what you added to N2? 584 00:30:45,750 --> 00:30:51,160 AUDIENCE: Sigma whatever, phi N3. 585 00:30:51,160 --> 00:30:53,600 MICHAEL SHORT: Yeah, there's going to be some new sigma-- 586 00:30:53,600 --> 00:30:58,460 let's call it sigma nickel-60, phi N3. 587 00:31:02,080 --> 00:31:04,580 AUDIENCE: Yeah, is the sigma going to be 0 this time around, 588 00:31:04,580 --> 00:31:06,000 or is it actually going to be-- 589 00:31:06,000 --> 00:31:08,170 MICHAEL SHORT: Let's find out. 590 00:31:08,170 --> 00:31:09,280 Let's go to our tables. 591 00:31:09,280 --> 00:31:12,760 There, there's data for nickel-60. 592 00:31:12,760 --> 00:31:16,050 Let's look up its absorption cross-section. 593 00:31:16,050 --> 00:31:19,950 So we'll scroll down to our Z gamma, 594 00:31:19,950 --> 00:31:23,650 our capture cross-section, plot it. 595 00:31:23,650 --> 00:31:28,200 Take a look at around 1e minus 8. 596 00:31:28,200 --> 00:31:31,460 And it's like 2 barns-- 597 00:31:31,460 --> 00:31:32,751 not negligible. 598 00:31:35,520 --> 00:31:40,365 So our capture cross-section for nickel-60-- 599 00:31:40,365 --> 00:31:42,100 I keep overwriting myself, and then I 600 00:31:42,100 --> 00:31:45,580 remember it's a blackboard. 601 00:31:45,580 --> 00:31:53,150 2 barns, which is 2 times 10 to the minus 24th cm squared. 602 00:31:53,150 --> 00:31:57,820 So we can just refer to that as sigma nickel. 603 00:31:57,820 --> 00:32:00,460 For the purposes of this problem, 604 00:32:00,460 --> 00:32:03,640 we don't particularly care how much nickel-60 we're making. 605 00:32:03,640 --> 00:32:06,390 Nickel-60 is a stable isotope of nickel. 606 00:32:06,390 --> 00:32:08,320 Eh, forget it. 607 00:32:08,320 --> 00:32:11,610 So for the purposes of this problem, 608 00:32:11,610 --> 00:32:13,677 forget the N3 equation. 609 00:32:13,677 --> 00:32:15,760 We don't care how much stable nickel-60 that we're 610 00:32:15,760 --> 00:32:17,760 making, because it's a lot cheaper to get it out 611 00:32:17,760 --> 00:32:19,950 of the ground, probably something like 10 612 00:32:19,950 --> 00:32:21,260 orders of magnitude cheaper. 613 00:32:21,260 --> 00:32:21,890 Yep. 614 00:32:21,890 --> 00:32:24,182 AUDIENCE: Where did you pick that incident energy from, 615 00:32:24,182 --> 00:32:26,120 the MEV [INAUDIBLE] 1 times 10/8? 616 00:32:26,120 --> 00:32:29,100 MICHAEL SHORT: Yeah, so the incident energy I picked-- 617 00:32:29,100 --> 00:32:30,990 because one thing we had gone over 618 00:32:30,990 --> 00:32:35,730 is that the thermal energy of a neutron 619 00:32:35,730 --> 00:32:43,450 is around 0.025 EV, which is equal to 2.5 times 620 00:32:43,450 --> 00:32:47,770 10 to the minus eighth MEV. 621 00:32:47,770 --> 00:32:51,340 So I took that value, about 2.5 times 10 622 00:32:51,340 --> 00:32:55,850 to the minus eighth MEV, so around here, and just went up. 623 00:32:55,850 --> 00:33:00,150 And on a log scale, it looks to be closest to 2 barns. 624 00:33:00,150 --> 00:33:04,830 You can always get the actual value. 625 00:33:04,830 --> 00:33:07,280 So if you want to zoom right in, I'm 626 00:33:07,280 --> 00:33:12,162 going to keep zooming into the 10 to the minus eighth region. 627 00:33:15,060 --> 00:33:16,850 Maybe not. 628 00:33:16,850 --> 00:33:21,660 You can set your bounds accordingly. 629 00:33:21,660 --> 00:33:25,380 Oh, 0 or negative values-- ah, OK, whatever. 630 00:33:25,380 --> 00:33:27,030 Let's just read off the graph for now. 631 00:33:27,030 --> 00:33:29,340 You can use this tool to get the actual value. 632 00:33:29,340 --> 00:33:31,530 But for problems like this, I think 633 00:33:31,530 --> 00:33:33,930 estimating it from the graph is going to be fine. 634 00:33:33,930 --> 00:33:35,370 When you get into 22.05 and you're 635 00:33:35,370 --> 00:33:38,280 like, what's the actual flux in the reactor to within 1% 636 00:33:38,280 --> 00:33:40,140 or something, estimating from the graph 637 00:33:40,140 --> 00:33:42,270 is no longer allowable. 638 00:33:42,270 --> 00:33:45,100 There are tabulated values of these things. 639 00:33:47,840 --> 00:33:50,690 Oh, yeah, so you can actually set the plot settings, 640 00:33:50,690 --> 00:33:53,650 get the tables. 641 00:33:53,650 --> 00:33:56,260 Oh, tabulated-- there we go. 642 00:33:56,260 --> 00:33:59,200 So you can read off values of the cross-sections 643 00:33:59,200 --> 00:34:00,393 from a table like that. 644 00:34:00,393 --> 00:34:01,810 But for now, since we want to make 645 00:34:01,810 --> 00:34:03,670 sure to get this problem done, let's 646 00:34:03,670 --> 00:34:05,450 just stick with the graph. 647 00:34:05,450 --> 00:34:08,130 So we don't care about the N3 equation. 648 00:34:08,130 --> 00:34:14,179 We're also going to say, well, let's not ignore sigma-60 yet. 649 00:34:14,179 --> 00:34:15,980 So how do we solve this set of equations? 650 00:34:18,620 --> 00:34:20,120 Let's make a little separation here. 651 00:34:20,120 --> 00:34:23,880 First of all, the easy one-- 652 00:34:23,880 --> 00:34:25,320 what's N1 as a function of time? 653 00:34:28,666 --> 00:34:35,952 AUDIENCE: e to the minus sigma-59, phi. 654 00:34:35,952 --> 00:34:42,560 MICHAEL SHORT: e to the minus sigma-59, phi, and what else? 655 00:34:42,560 --> 00:34:44,590 AUDIENCE: N0. 656 00:34:44,590 --> 00:34:48,554 MICHAEL SHORT: There is an N1 0, and there's a t. 657 00:34:48,554 --> 00:34:49,929 Because as you're burning it out, 658 00:34:49,929 --> 00:34:51,550 it matters how much time you have. 659 00:34:51,550 --> 00:34:56,420 Doesn't this look eerily similar to an N equals N0 e 660 00:34:56,420 --> 00:34:59,820 to the minus lambda t? 661 00:34:59,820 --> 00:35:03,360 Again, it's like exponential artificial decay, 662 00:35:03,360 --> 00:35:06,120 because we're burning those things out. 663 00:35:06,120 --> 00:35:08,880 For a fixed amount of neutrons going in, 664 00:35:08,880 --> 00:35:11,940 the amount that we burn is proportional to the amount 665 00:35:11,940 --> 00:35:12,930 that is there. 666 00:35:12,930 --> 00:35:14,340 So that's the burn rate. 667 00:35:14,340 --> 00:35:16,050 That's the amount that's there. 668 00:35:16,050 --> 00:35:18,370 This is our artificial lambda. 669 00:35:18,370 --> 00:35:19,990 So that equation's easy. 670 00:35:19,990 --> 00:35:21,550 What we really want to know is, what 671 00:35:21,550 --> 00:35:26,360 is N2 as a function of time? 672 00:35:26,360 --> 00:35:28,410 That's the $60 million dollar question. 673 00:35:28,410 --> 00:35:32,500 And I'm not exaggerating there, because cobalt-60 is expensive. 674 00:35:32,500 --> 00:35:34,860 So the question I'm posing to you guys on the homework-- 675 00:35:34,860 --> 00:35:37,250 so for those of you who have started problem set 4, 676 00:35:37,250 --> 00:35:39,292 I'm going to be swapping out the noodle scratcher 677 00:35:39,292 --> 00:35:43,080 problem to this problem that we're going to begin together 678 00:35:43,080 --> 00:35:44,080 in class. 679 00:35:44,080 --> 00:35:46,140 And you guys are going to finish on the homework. 680 00:35:46,140 --> 00:35:48,480 So I'm kind of giving you help on the homework. 681 00:35:48,480 --> 00:35:51,690 What is N2, the amount of cobalt-60 in your reactor, 682 00:35:51,690 --> 00:35:53,100 as a function of time? 683 00:35:53,100 --> 00:35:58,500 And what is your profit for running the reactor 684 00:35:58,500 --> 00:36:00,150 as a function of time? 685 00:36:00,150 --> 00:36:03,930 Assuming a few things-- so let's set up some parameters. 686 00:36:03,930 --> 00:36:06,120 I'm going to say that the neutron flux is 687 00:36:06,120 --> 00:36:09,950 the same as that in the MIT reactor, which is about 10 688 00:36:09,950 --> 00:36:15,440 to the 14th neutrons per centimeter squared per second. 689 00:36:15,440 --> 00:36:17,210 We already have all of our lambdas. 690 00:36:17,210 --> 00:36:19,190 We have all of our sigmas. 691 00:36:19,190 --> 00:36:24,950 I'm going to say that the cost of running the reactor is-- 692 00:36:24,950 --> 00:36:27,170 and I have a quote on this-- 693 00:36:27,170 --> 00:36:37,630 $1,000 per day, which is the same as $0.01 per second. 694 00:36:37,630 --> 00:36:40,430 Not a bad rate to stick something in the reactor, 695 00:36:40,430 --> 00:36:42,160 right? 696 00:36:42,160 --> 00:36:43,228 It's not bad at all. 697 00:36:43,228 --> 00:36:44,770 If you had to build your own reactor, 698 00:36:44,770 --> 00:36:47,020 your daily operating cost would actually be $1 million 699 00:36:47,020 --> 00:36:50,230 a day for a commercial power plant. 700 00:36:50,230 --> 00:36:53,380 So every time a plant goes down, you lose $1 million a day, 701 00:36:53,380 --> 00:36:56,740 plus the lost electricity or whatever that you have to buy. 702 00:36:56,740 --> 00:36:58,210 So let's put that in there. 703 00:36:58,210 --> 00:37:04,270 And from the cost of this hypothetical cobalt-60 source, 704 00:37:04,270 --> 00:37:15,420 we know that cobalt-60 runs about $100 per microcurie, 705 00:37:15,420 --> 00:37:19,110 because these sources run for about $100. 706 00:37:19,110 --> 00:37:21,870 And so the eventual problem that we're going to set up here 707 00:37:21,870 --> 00:37:23,912 and you guys are going to solve on the homework-- 708 00:37:23,912 --> 00:37:26,520 and we can keep going a little bit on Friday if you want-- 709 00:37:26,520 --> 00:37:30,660 is, at what point, at what t do you shut off your reactor 710 00:37:30,660 --> 00:37:34,050 and extract your cobalt-60 to maximize your profit? 711 00:37:34,050 --> 00:37:35,640 And this is an actual value judgment 712 00:37:35,640 --> 00:37:38,670 that folks that make cobalt-60 have to make. 713 00:37:38,670 --> 00:37:41,040 How long do you keep your nickel target in there 714 00:37:41,040 --> 00:37:43,410 and not hit diminishing returns? 715 00:37:43,410 --> 00:37:45,360 Because you're always going to be, 716 00:37:45,360 --> 00:37:49,680 let's say, increasing the amount of cobalt-60 717 00:37:49,680 --> 00:37:54,270 that you make if your source is basically undefeatable, 718 00:37:54,270 --> 00:37:57,270 until you reach some certain half-life criterion. 719 00:37:57,270 --> 00:38:00,580 But it might not make financial sense to do so. 720 00:38:00,580 --> 00:38:02,550 So let's start getting the solution to N2. 721 00:38:05,650 --> 00:38:06,910 Let's see. 722 00:38:06,910 --> 00:38:09,860 So I'm going to rewrite our N2 equation 723 00:38:09,860 --> 00:38:13,370 and we can start solving it. 724 00:38:13,370 --> 00:38:19,070 I'm sorry, that's a dN2 dt, equals 725 00:38:19,070 --> 00:38:33,800 sigma-59, flux, N1 minus lambda N2 minus sigma-60 flux N2. 726 00:38:33,800 --> 00:38:37,490 So how do we go about solving this differential equation? 727 00:38:37,490 --> 00:38:39,290 AUDIENCE: Integrating factor. 728 00:38:39,290 --> 00:38:41,248 MICHAEL SHORT: Yep, the old integrating factor. 729 00:38:41,248 --> 00:38:42,748 First, we want to get rid of the N1, 730 00:38:42,748 --> 00:38:44,100 because that's another variable. 731 00:38:44,100 --> 00:38:53,140 And we've already decided right here 732 00:38:53,140 --> 00:39:03,980 that N1 is N1 0 times e to the minus sigma-59 flux t. 733 00:39:03,980 --> 00:39:07,810 And we haven't specified, what's N1 0? 734 00:39:07,810 --> 00:39:08,930 Let's do that now. 735 00:39:08,930 --> 00:39:14,830 The last number we'll put in is, we started with a 100 gram 736 00:39:14,830 --> 00:39:19,300 source of cobalt-59. 737 00:39:19,300 --> 00:39:21,700 When we write it in isotope parlance, it sounds exotic. 738 00:39:21,700 --> 00:39:24,452 But that's actually the only stable isotope of cobalt. 739 00:39:24,452 --> 00:39:26,410 So that's just a lump of cobalt from the ground 740 00:39:26,410 --> 00:39:27,118 that we stick in. 741 00:39:30,170 --> 00:39:32,090 So we know what N1 0 is. 742 00:39:32,090 --> 00:39:35,770 So we can now rewrite this equation as-- 743 00:39:35,770 --> 00:39:39,190 let's just go with N2 prime for shorthand. 744 00:39:39,190 --> 00:39:42,190 And we'll put everything on one side of the equation. 745 00:39:42,190 --> 00:39:53,730 So we'll have, plus lambda plus sigma-60 phi N2, 746 00:39:53,730 --> 00:40:06,590 minus sigma 59 phi times e to the minus sigma-59 phi t, 747 00:40:06,590 --> 00:40:08,820 equals 0. 748 00:40:08,820 --> 00:40:11,160 So what is our integrating factor here? 749 00:40:15,260 --> 00:40:18,587 AUDIENCE: e to the lambda plus sigma-60 phi t. 750 00:40:18,587 --> 00:40:19,420 MICHAEL SHORT: Yeah. 751 00:40:19,420 --> 00:40:22,210 So it is, e to the integral of whatever 752 00:40:22,210 --> 00:40:33,810 is in front of our N2, lambda plus sigma-60 phi dt, 753 00:40:33,810 --> 00:40:40,650 which equals e to the lambda plus sigma-60 phi t. 754 00:40:40,650 --> 00:40:42,930 So we now multiply every term in this equation 755 00:40:42,930 --> 00:40:47,170 by our mu, our integrating factor. 756 00:40:47,170 --> 00:40:52,510 So let's say we have N2 prime times-- 757 00:40:52,510 --> 00:40:56,700 I'm just going to use mu for shorthand, since it's going 758 00:40:56,700 --> 00:40:58,770 to take a long time to write. 759 00:40:58,770 --> 00:41:05,050 Plus-- let's see, that right there is like mu prime, 760 00:41:05,050 --> 00:41:09,580 because if we take the integral of mu times N2-- 761 00:41:09,580 --> 00:41:11,080 let's see. 762 00:41:11,080 --> 00:41:12,700 Oh, yeah, this is right. 763 00:41:12,700 --> 00:41:22,960 So we have lambda plus sigma-60 N2 times mu, 764 00:41:22,960 --> 00:41:35,790 minus mu times sigma-59 phi e to the minus sigma-59 phi t, 765 00:41:35,790 --> 00:41:38,280 equals 0. 766 00:41:38,280 --> 00:41:43,530 This stuff right here is like the an expanded product rule, 767 00:41:43,530 --> 00:41:45,250 so we can write it more simply. 768 00:41:45,250 --> 00:41:51,895 So we can say, N2 times mu prime-- 769 00:41:54,780 --> 00:42:03,530 let's see-- equals mu sigma-59 phi e to the minus sigma 770 00:42:03,530 --> 00:42:08,730 59 phi times t. 771 00:42:08,730 --> 00:42:12,315 So next we integrate both sides. 772 00:42:16,160 --> 00:42:22,500 And we get N2 times mu equals-- 773 00:42:22,500 --> 00:42:25,090 let's see. 774 00:42:25,090 --> 00:42:26,610 Let's expand everything out now. 775 00:42:29,160 --> 00:42:37,890 So that stuff would be sigma-59 phi 776 00:42:37,890 --> 00:42:45,630 e to the lambda plus sigma-60 phi, 777 00:42:45,630 --> 00:42:53,230 minus sigma-59 phi times t. 778 00:42:53,230 --> 00:42:54,850 So if we integrate all of that, we're 779 00:42:54,850 --> 00:43:01,335 going to get sigma-59 phi. 780 00:43:01,335 --> 00:43:05,300 I think there's an N0 missing here, isn't there? 781 00:43:05,300 --> 00:43:05,825 Let's see. 782 00:43:08,600 --> 00:43:11,820 There should be an N1 0 missing here. 783 00:43:11,820 --> 00:43:12,320 Yep. 784 00:43:15,200 --> 00:43:19,380 There's an N1 0 that I dropped for some reason. 785 00:43:19,380 --> 00:43:22,170 Let's stick that back in-- 786 00:43:22,170 --> 00:43:27,600 N1 0, and N1 0. 787 00:43:31,020 --> 00:43:40,900 N1 0 over that stuff, lambda plus sigma-60 phi 788 00:43:40,900 --> 00:43:50,100 minus sigma-59 phi, times whatever is left, 789 00:43:50,100 --> 00:44:05,570 e to the lambda plus sigma-60 phi, minus sigma-59 phi t, 790 00:44:05,570 --> 00:44:08,800 plus C. 791 00:44:08,800 --> 00:44:13,252 So now we can say, what's our integration constant C? 792 00:44:13,252 --> 00:44:14,710 The last thing we haven't specified 793 00:44:14,710 --> 00:44:16,240 is our initial condition. 794 00:44:16,240 --> 00:44:24,070 So let's assume that when we started our reactor, 795 00:44:24,070 --> 00:44:26,300 there was no cobalt-60. 796 00:44:26,300 --> 00:44:29,110 That makes for the simplest initial condition. 797 00:44:29,110 --> 00:44:31,400 So we can substitute that in here. 798 00:44:31,400 --> 00:44:33,740 So at t equals 0, N equals 0. 799 00:44:33,740 --> 00:44:36,530 So we've got get 0 equals-- 800 00:44:36,530 --> 00:44:45,110 if t is 0, then that just becomes sigma-59 phi N10 801 00:44:45,110 --> 00:44:57,050 over lambda plus sigma-60 phi, minus sigma-59 phi, 802 00:44:57,050 --> 00:45:02,500 plus C. And so that makes things pretty easy, 803 00:45:02,500 --> 00:45:09,900 because we know C equals minus sigma-59 phi N1 0, 804 00:45:09,900 --> 00:45:13,290 over that stuff that I keep saying over and over again. 805 00:45:20,222 --> 00:45:22,180 And then we've pretty much solved the equation. 806 00:45:22,180 --> 00:45:25,210 The last thing we have to do is divide by mu, 807 00:45:25,210 --> 00:45:26,950 and we'll end up with the same solution 808 00:45:26,950 --> 00:45:29,290 that we got on Friday and the same solution 809 00:45:29,290 --> 00:45:30,820 that we got on Tuesday. 810 00:45:30,820 --> 00:45:33,430 So I realized, the second after I said it last time, that, 811 00:45:33,430 --> 00:45:36,640 oh, we can't just absorb some e to the something 812 00:45:36,640 --> 00:45:39,190 t into our integration constant C because there's 813 00:45:39,190 --> 00:45:40,660 a variable t in it. 814 00:45:40,660 --> 00:45:43,420 So I would say, look at this derivation 815 00:45:43,420 --> 00:45:45,340 to know the whole solution. 816 00:45:45,340 --> 00:45:47,770 And so finally we end up with-- 817 00:45:47,770 --> 00:45:51,776 anyone mind if I erase a little bit of this stuff up top? 818 00:45:51,776 --> 00:45:54,810 OK, I'll erase the decay diagram because we're not 819 00:45:54,810 --> 00:45:57,090 using that anymore. 820 00:45:57,090 --> 00:45:59,838 And hopefully everybody knows our conversion factor. 821 00:46:02,650 --> 00:46:10,260 So the end solution, N2 of t, would look like, 822 00:46:10,260 --> 00:46:20,910 sigma-59 phi N1 0, over lambda plus sigma-60 phi, 823 00:46:20,910 --> 00:46:30,300 minus sigma-59 phi, times e to the minus-- 824 00:46:30,300 --> 00:46:33,990 what's lambda 2 in this case? 825 00:46:33,990 --> 00:46:39,660 Lambda plus sigma-60 phi t, minus e 826 00:46:39,660 --> 00:46:45,411 to the minus sigma-59 phi t. 827 00:46:48,100 --> 00:46:51,860 And that right there is our full equation for N2. 828 00:46:51,860 --> 00:46:53,800 Now you guys said, let's make this numerical. 829 00:46:53,800 --> 00:46:59,568 OK, we have every numerical value already chosen. 830 00:46:59,568 --> 00:47:01,610 I've already plugged these into the Desmos thing, 831 00:47:01,610 --> 00:47:05,500 so you can see generally how this goes. 832 00:47:05,500 --> 00:47:07,060 So we've solved it theoretically, 833 00:47:07,060 --> 00:47:10,090 so now let's make this numerical and make some sort of a value 834 00:47:10,090 --> 00:47:11,290 judgment, right? 835 00:47:11,290 --> 00:47:15,410 We know sigma-59, because we just looked that up. 836 00:47:15,410 --> 00:47:16,700 We know phi. 837 00:47:16,700 --> 00:47:23,240 We impose that as 10 to the 14th neutrons per second. 838 00:47:23,240 --> 00:47:25,210 There it is. 839 00:47:25,210 --> 00:47:28,060 We know our lambda. 840 00:47:28,060 --> 00:47:31,060 We don't know our sigma-60, so we're just 841 00:47:31,060 --> 00:47:33,140 going to forget that for now. 842 00:47:33,140 --> 00:47:37,360 But the point is, for everything except time and N2, 843 00:47:37,360 --> 00:47:40,100 we have numerical constants for this. 844 00:47:40,100 --> 00:47:44,290 So once we plug it all in, I modified the Desmos example 845 00:47:44,290 --> 00:47:46,870 from last time to have the actual unit. 846 00:47:46,870 --> 00:47:53,130 So you can see that our fake L1, our lambda-- 847 00:47:53,130 --> 00:47:58,780 we'll call it lambda 1-- equals sigma-59 times 848 00:47:58,780 --> 00:48:02,620 phi, which is 20 barns. 849 00:48:02,620 --> 00:48:09,160 20 times 10 to the minus 24th centimeters squared, times 10 850 00:48:09,160 --> 00:48:14,230 to the 14th neutrons per centimeter squared per second. 851 00:48:14,230 --> 00:48:18,160 And the centimeter squareds cancel. 852 00:48:18,160 --> 00:48:21,810 That becomes, to the minus 10. 853 00:48:21,810 --> 00:48:29,770 And we get that our lambda 1 is like 2 times 10 to the minus 9 854 00:48:29,770 --> 00:48:32,570 per second. 855 00:48:32,570 --> 00:48:37,190 Our lambda 2, well, we already have that, 4.17 times 10 856 00:48:37,190 --> 00:48:39,120 to the minus 9 per second. 857 00:48:39,120 --> 00:48:44,860 So this is one of those cases where lambda 1 approximately 858 00:48:44,860 --> 00:48:47,197 equals lambda 2. 859 00:48:47,197 --> 00:48:48,780 So just like you see in the book, when 860 00:48:48,780 --> 00:48:52,630 you plug in all the numbers, you get a very similar equation. 861 00:48:52,630 --> 00:48:57,400 Which is to say, there's going to be some maximum of cobalt-60 862 00:48:57,400 --> 00:48:57,900 produced. 863 00:48:57,900 --> 00:48:59,820 And in this case, the x-axis I have 864 00:48:59,820 --> 00:49:01,770 in seconds, because that's the units we're 865 00:49:01,770 --> 00:49:03,210 using for everything. 866 00:49:03,210 --> 00:49:05,080 The y-axis is number of atoms. 867 00:49:05,080 --> 00:49:09,210 So right there, 6 times 10 to the 23rd, that's one mole. 868 00:49:09,210 --> 00:49:13,050 So at most, you can make up about 1/3 869 00:49:13,050 --> 00:49:21,540 of a mole of cobalt-60 out of 100 moles of cobalt-59, which 870 00:49:21,540 --> 00:49:22,920 means you're never actually going 871 00:49:22,920 --> 00:49:25,380 to have one mole of cobalt-60. 872 00:49:25,380 --> 00:49:30,150 Because of the way that our natural and artificial decay 873 00:49:30,150 --> 00:49:32,700 constants work out, because they're fairly equal, 874 00:49:32,700 --> 00:49:35,580 you're never going to be able to convert and harvest it all. 875 00:49:35,580 --> 00:49:38,520 That's the numerical output of this. 876 00:49:38,520 --> 00:49:40,710 Now I have another question for you. 877 00:49:40,710 --> 00:49:44,490 Is the top of this curve necessarily the profit point 878 00:49:44,490 --> 00:49:45,770 for this reactor? 879 00:49:45,770 --> 00:49:47,090 No, good answer. 880 00:49:47,090 --> 00:49:47,840 Why do you say no? 881 00:49:47,840 --> 00:49:50,780 AUDIENCE: Well, you have to write another equation 882 00:49:50,780 --> 00:49:54,985 for the costs and the profits to maximize both of them. 883 00:49:54,985 --> 00:49:56,040 MICHAEL SHORT: Exactly. 884 00:49:56,040 --> 00:49:58,430 That's what you guys are going to do on the homework. 885 00:49:58,430 --> 00:50:01,340 I think we've done the hard part together here. 886 00:50:01,340 --> 00:50:03,110 And so now I want you guys to decide, 887 00:50:03,110 --> 00:50:06,020 given those profit parameters-- and I will write them down 888 00:50:06,020 --> 00:50:07,280 on the Pset-- 889 00:50:07,280 --> 00:50:09,650 how long do you run your reactor to maximize 890 00:50:09,650 --> 00:50:11,977 your cobalt-60 profit? 891 00:50:11,977 --> 00:50:13,560 So this is one of those examples where 892 00:50:13,560 --> 00:50:18,220 we did the whole theoretical derivation, 893 00:50:18,220 --> 00:50:21,340 we decided, yes, let's escalate the situation for reality. 894 00:50:21,340 --> 00:50:23,260 Everything works out just fine. 895 00:50:23,260 --> 00:50:26,900 The final answer, well, you just tack 896 00:50:26,900 --> 00:50:29,240 on this extra artificial bit of decay 897 00:50:29,240 --> 00:50:30,920 from the cobalt-60 being in the reactor. 898 00:50:30,920 --> 00:50:34,800 But the form of the equation is exactly the same. 899 00:50:34,800 --> 00:50:37,760 There's just a couple other constants in it for reality. 900 00:50:37,760 --> 00:50:40,820 Then if you plug in all the numbers for the constants, 901 00:50:40,820 --> 00:50:44,270 so you pretend like that stuff is lambda 2 and that stuff 902 00:50:44,270 --> 00:50:50,120 is lambda 1, there's lambda 1 again, there's lambda 2, 903 00:50:50,120 --> 00:50:51,980 there's lambda 1, and it's exactly 904 00:50:51,980 --> 00:50:56,665 the same equational form as the original solution that we had. 905 00:50:56,665 --> 00:50:58,040 When you plug in all the numbers, 906 00:50:58,040 --> 00:50:59,957 you get something remarkably similar to what's 907 00:50:59,957 --> 00:51:03,620 in the book, just scaled for actual units of atoms 908 00:51:03,620 --> 00:51:04,660 and seconds. 909 00:51:04,660 --> 00:51:05,900 Is there a question? 910 00:51:05,900 --> 00:51:06,485 Yeah. 911 00:51:06,485 --> 00:51:07,860 AUDIENCE: On the homework, do you 912 00:51:07,860 --> 00:51:10,237 want us to just assume that sigma-60 is 0? 913 00:51:10,237 --> 00:51:11,070 MICHAEL SHORT: Sure. 914 00:51:11,070 --> 00:51:12,080 AUDIENCE: And have that all cancel out? 915 00:51:12,080 --> 00:51:14,080 MICHAEL SHORT: If you can find it, that's great. 916 00:51:14,080 --> 00:51:16,160 But I couldn't find it that easily. 917 00:51:16,160 --> 00:51:18,470 You can hunt through the different databases in Janis 918 00:51:18,470 --> 00:51:19,220 to try to find it. 919 00:51:19,220 --> 00:51:21,170 But I'm not going to penalize you if you can't find it, 920 00:51:21,170 --> 00:51:22,087 if I couldn't find it. 921 00:51:24,895 --> 00:51:26,270 So yeah, a lot of the homework is 922 00:51:26,270 --> 00:51:28,730 going to be redoing this derivation for yourself. 923 00:51:28,730 --> 00:51:30,230 Because I want to make sure that you 924 00:51:30,230 --> 00:51:36,740 can go from a set of equations that 925 00:51:36,740 --> 00:51:39,320 models an actual physical solution. 926 00:51:39,320 --> 00:51:41,990 And I guarantee you you'll have another physical solution 927 00:51:41,990 --> 00:51:44,750 on the exam. 928 00:51:44,750 --> 00:51:47,960 Solve them using your knowledge of 1803, 929 00:51:47,960 --> 00:51:50,840 get some sort of a solution, which looks crazy, 930 00:51:50,840 --> 00:51:53,600 but it comes from straightforward math. 931 00:51:53,600 --> 00:51:55,250 Then plug in some realistic numbers 932 00:51:55,250 --> 00:51:56,720 and answer an actual question. 933 00:51:56,720 --> 00:51:58,220 How long should you run your reactor 934 00:51:58,220 --> 00:52:01,062 to maximize your profit? 935 00:52:01,062 --> 00:52:02,290 So it's kind of neat. 936 00:52:02,290 --> 00:52:03,955 We've been here one month together, 937 00:52:03,955 --> 00:52:06,490 and you can already start answering these value judgment 938 00:52:06,490 --> 00:52:09,670 questions about running a reactor. 939 00:52:09,670 --> 00:52:12,880 And so again, I don't know who did, but I'm glad you asked, 940 00:52:12,880 --> 00:52:14,620 is this field mostly simulation? 941 00:52:14,620 --> 00:52:16,090 And the answer is no. 942 00:52:16,090 --> 00:52:18,670 You actually have to use math to make value judgments if you 943 00:52:18,670 --> 00:52:20,020 want to go and make isotopes. 944 00:52:20,020 --> 00:52:22,732 And then you go and make the isotopes. 945 00:52:22,732 --> 00:52:24,190 And you sell them to people like me 946 00:52:24,190 --> 00:52:29,390 so I can bring them into class and scare unwitting members 947 00:52:29,390 --> 00:52:30,960 of the public, theoretically. 948 00:52:30,960 --> 00:52:34,190 Yeah, hypothetical source indeed. 949 00:52:34,190 --> 00:52:38,410 So any questions on what we did here, from start to finish? 950 00:52:38,410 --> 00:52:39,250 Yep. 951 00:52:39,250 --> 00:52:44,250 AUDIENCE: On the equation right in the middle there, 952 00:52:44,250 --> 00:52:47,090 where it says-- in parenthesis, it has lambda plus sigma-60-- 953 00:52:47,090 --> 00:52:48,090 MICHAEL SHORT: This one? 954 00:52:48,090 --> 00:52:49,080 AUDIENCE: Yeah. 955 00:52:49,080 --> 00:52:50,657 Was the phi drop just like a-- 956 00:52:50,657 --> 00:52:52,490 MICHAEL SHORT: Oh, yeah, that was a mistake. 957 00:52:52,490 --> 00:52:53,032 AUDIENCE: OK. 958 00:52:53,032 --> 00:52:55,007 I can just pull it back later. 959 00:52:55,007 --> 00:52:56,590 MICHAEL SHORT: Yeah, it probably means 960 00:52:56,590 --> 00:52:58,420 I was talking and thinking at the same time 961 00:52:58,420 --> 00:52:59,462 and forgot to write that. 962 00:52:59,462 --> 00:53:01,840 But indeed, it's back everywhere else. 963 00:53:01,840 --> 00:53:03,080 Thank you. 964 00:53:03,080 --> 00:53:03,580 Yep. 965 00:53:03,580 --> 00:53:06,490 AUDIENCE: So the lambda in the final N2 function equation, 966 00:53:06,490 --> 00:53:08,750 is that lambda 1, the theoretical lambda 1, 967 00:53:08,750 --> 00:53:10,890 fake lambda 1, or is that lambda 2? 968 00:53:10,890 --> 00:53:14,520 MICHAEL SHORT: This lambda right here 969 00:53:14,520 --> 00:53:16,740 is the actual lambda for cobalt-60. 970 00:53:16,740 --> 00:53:17,550 Yep. 971 00:53:17,550 --> 00:53:18,990 This right here is just an analogy 972 00:53:18,990 --> 00:53:23,670 I'm drawing to say that, it's almost like that stuff 973 00:53:23,670 --> 00:53:25,230 is lambda 1. 974 00:53:25,230 --> 00:53:29,550 That's our original artificial decay constant. 975 00:53:29,550 --> 00:53:32,973 And this stuff here is like our lambda 2, 976 00:53:32,973 --> 00:53:34,890 because there's natural decay and then there's 977 00:53:34,890 --> 00:53:37,904 reactor-induced destruction. 978 00:53:37,904 --> 00:53:38,780 Yep. 979 00:53:38,780 --> 00:53:41,756 AUDIENCE: What is that factor of dividing by lambda [INAUDIBLE].. 980 00:53:41,756 --> 00:53:43,740 Where does that come from? 981 00:53:43,740 --> 00:53:47,598 MICHAEL SHORT: That comes from this solution right here. 982 00:53:47,598 --> 00:53:49,140 There's another interesting bit, too. 983 00:53:49,140 --> 00:53:51,600 Did we necessarily say-- 984 00:53:51,600 --> 00:53:53,130 let's see. 985 00:53:53,130 --> 00:53:55,620 Did we necessarily say that-- 986 00:53:55,620 --> 00:53:58,350 no, never mind, that's fine. 987 00:53:58,350 --> 00:54:01,470 That comes from-- let's trace it through. 988 00:54:01,470 --> 00:54:06,030 So mu contains-- yeah, it comes from C. That's right. 989 00:54:06,030 --> 00:54:12,300 So we had it over here, because that's 990 00:54:12,300 --> 00:54:14,370 part of our solution for-- 991 00:54:14,370 --> 00:54:16,770 let's see. 992 00:54:16,770 --> 00:54:18,280 Where would we trace it back to? 993 00:54:18,280 --> 00:54:20,520 It starts off here in the differential equation. 994 00:54:20,520 --> 00:54:22,870 It starts off here as well. 995 00:54:22,870 --> 00:54:25,110 So that's part of what's inside mu. 996 00:54:25,110 --> 00:54:27,030 OK, that's where it came from. 997 00:54:27,030 --> 00:54:29,465 So a mu contains this stuff. 998 00:54:29,465 --> 00:54:29,965 Yep. 999 00:54:29,965 --> 00:54:33,230 AUDIENCE: And then once you integrate, 1000 00:54:33,230 --> 00:54:35,100 it just becomes e to the that. 1001 00:54:35,100 --> 00:54:38,760 It doesn't actually become lambda plus 65. 1002 00:54:38,760 --> 00:54:41,910 MICHAEL SHORT: It becomes e to the lambda plus 60 phi, 1003 00:54:41,910 --> 00:54:43,230 times t. 1004 00:54:43,230 --> 00:54:46,710 AUDIENCE: Yeah, but when we do the integration of that, 1005 00:54:46,710 --> 00:54:49,868 we don't get any factors of lambda plus 60 phi coming down. 1006 00:54:49,868 --> 00:54:51,160 MICHAEL SHORT: We do, actually. 1007 00:54:51,160 --> 00:54:53,280 There is a mu stuck in right here. 1008 00:54:53,280 --> 00:54:56,100 And so I just wanted to say that, expanding this term 1009 00:54:56,100 --> 00:54:59,280 comes out to lambda plus sigma 60 phi, 1010 00:54:59,280 --> 00:55:02,700 minus lambda sigma-59 phi, times t. 1011 00:55:02,700 --> 00:55:05,260 So when you integrate this whole term-- 1012 00:55:05,260 --> 00:55:09,460 and again, there's an N0 that should be there. 1013 00:55:09,460 --> 00:55:12,730 You do bring this whole pile in front of the t down 1014 00:55:12,730 --> 00:55:14,500 on the bottom of the equation. 1015 00:55:14,500 --> 00:55:17,214 So that's where it comes from. 1016 00:55:17,214 --> 00:55:19,410 Yeah, cool. 1017 00:55:19,410 --> 00:55:22,680 I don't think there's any more missing terms. 1018 00:55:22,680 --> 00:55:24,900 OK, maybe time for a last question, because it is 10 1019 00:55:24,900 --> 00:55:27,053 o'clock. 1020 00:55:27,053 --> 00:55:28,470 AUDIENCE: This doesn't really have 1021 00:55:28,470 --> 00:55:31,215 to do with your derivation or anything. 1022 00:55:31,215 --> 00:55:33,340 So I'm pretty sure you also already explained this. 1023 00:55:33,340 --> 00:55:39,730 But why can you put a cross-section, 1024 00:55:39,730 --> 00:55:41,770 like that's a measure of probably in units, 1025 00:55:41,770 --> 00:55:44,530 of centimeters squared. 1026 00:55:44,530 --> 00:55:46,370 How does that [INAUDIBLE]? 1027 00:55:46,370 --> 00:55:49,170 MICHAEL SHORT: Ah, so the cross-section is almost like, 1028 00:55:49,170 --> 00:55:52,080 if you fire a neutron at an atom, 1029 00:55:52,080 --> 00:55:55,200 the bigger the atom appears to the neutron, the more likely 1030 00:55:55,200 --> 00:55:56,610 it's going to hit it. 1031 00:55:56,610 --> 00:55:58,805 So it's kind of a theoretical construct, 1032 00:55:58,805 --> 00:56:00,930 to say, if something has an enormous cross-section, 1033 00:56:00,930 --> 00:56:03,110 it's like shooting a bullet at a gigantic target, 1034 00:56:03,110 --> 00:56:06,108 with a high probability of impact or interaction. 1035 00:56:06,108 --> 00:56:07,650 Something with a small cross-section, 1036 00:56:07,650 --> 00:56:09,348 there's still only one atom in the way, 1037 00:56:09,348 --> 00:56:11,640 but it's like you're shooting a bullet at a tiny target 1038 00:56:11,640 --> 00:56:13,307 and have less of a chance of hitting it. 1039 00:56:16,231 --> 00:56:17,940 Does that makes sense? 1040 00:56:17,940 --> 00:56:18,500 Cool. 1041 00:56:18,500 --> 00:56:21,660 AUDIENCE: So is it just a theoretical construct, 1042 00:56:21,660 --> 00:56:25,104 or can you actually relate it to an actual physical 1043 00:56:25,104 --> 00:56:26,933 cross-sectional area? 1044 00:56:26,933 --> 00:56:28,350 MICHAEL SHORT: You can't relate it 1045 00:56:28,350 --> 00:56:30,850 to a physical cross-sectional area, as I know. 1046 00:56:30,850 --> 00:56:33,630 It's not like a certain nucleus has a larger 1047 00:56:33,630 --> 00:56:35,063 cross-sectional area. 1048 00:56:35,063 --> 00:56:36,480 Otherwise, things like gadolinium, 1049 00:56:36,480 --> 00:56:38,490 which has a cross-section of 100,000 barns, 1050 00:56:38,490 --> 00:56:40,600 would just be a larger atom. 1051 00:56:40,600 --> 00:56:41,562 And it's not. 1052 00:56:41,562 --> 00:56:43,510 And yeah, Sean. 1053 00:56:43,510 --> 00:56:46,330 AUDIENCE: Are they determined only experimentally, 1054 00:56:46,330 --> 00:56:49,232 or do we know some of way to calculate it? 1055 00:56:49,232 --> 00:56:50,440 MICHAEL SHORT: Good question. 1056 00:56:50,440 --> 00:56:53,800 They can be theoretically calculated in some cases. 1057 00:56:53,800 --> 00:56:57,340 In the Yip book, Nuclear Radiation Interactions, 1058 00:56:57,340 --> 00:57:01,950 he does go over how to calculate those from quantum stuff. 1059 00:57:01,950 --> 00:57:04,380 And so you'll get a little bit of that in 22.02, 1060 00:57:04,380 --> 00:57:07,590 in terms of predicting the cross-section for hydrogen 1061 00:57:07,590 --> 00:57:09,060 and the cross-section for water. 1062 00:57:09,060 --> 00:57:12,030 And molecular water is not just the sum of its parts. 1063 00:57:12,030 --> 00:57:13,530 That's the kind of crazy part. 1064 00:57:13,530 --> 00:57:16,560 Cross-sections do change when you put atoms and molecules 1065 00:57:16,560 --> 00:57:19,770 together, just tend to be at lower energies, 1066 00:57:19,770 --> 00:57:22,230 around thermal energies and such. 1067 00:57:22,230 --> 00:57:23,760 Let's say, all of them probably can 1068 00:57:23,760 --> 00:57:26,940 be theoretically calculated, just not that easily. 1069 00:57:26,940 --> 00:57:31,800 But the really simple ones you can predict theoretically. 1070 00:57:31,800 --> 00:57:36,300 Predicting the resonances in those cross-sections, 1071 00:57:36,300 --> 00:57:39,190 that's tough. 1072 00:57:39,190 --> 00:57:43,377 Let's look at a simple cross-section, like hydrogen. 1073 00:57:43,377 --> 00:57:47,120 AUDIENCE: Can't you calculate it using simulations? 1074 00:57:47,120 --> 00:57:48,010 MICHAEL SHORT: Yes. 1075 00:57:48,010 --> 00:57:51,520 Like if you know, let's say, the full wave function for a given 1076 00:57:51,520 --> 00:57:55,440 atom or for all the electrons in an atom, you should be able to. 1077 00:57:55,440 --> 00:57:58,890 So let's do N total for hydrogen. 1078 00:57:58,890 --> 00:58:02,880 Much simpler, this is the kind of thing that can be predicted 1079 00:58:02,880 --> 00:58:03,970 from theory quite easily. 1080 00:58:03,970 --> 00:58:13,545 In fact, you will be doing this in 22.02. 1081 00:58:13,545 --> 00:58:14,920 The other one, no, I don't expect 1082 00:58:14,920 --> 00:58:16,870 you to be able to predict this. 1083 00:58:16,870 --> 00:58:18,850 But you will learn why the resonances are there 1084 00:58:18,850 --> 00:58:21,200 and why they take the shape that they do. 1085 00:58:21,200 --> 00:58:24,430 So last thing-- we did go like three minutes late, 1086 00:58:24,430 --> 00:58:26,230 but everyone's still here. 1087 00:58:26,230 --> 00:58:27,940 You can go if you have to, by the way. 1088 00:58:27,940 --> 00:58:29,530 I can't keep you here. 1089 00:58:29,530 --> 00:58:32,470 If you want to know, if you want to make this equation more 1090 00:58:32,470 --> 00:58:37,000 realistic and account for every possible energy in the reactor, 1091 00:58:37,000 --> 00:58:41,320 you can make these cross-sections 1092 00:58:41,320 --> 00:58:47,040 a function of energy, and integrate over the full energy 1093 00:58:47,040 --> 00:58:47,540 range. 1094 00:58:47,540 --> 00:58:49,320 And this is actually how it's done. 1095 00:58:49,320 --> 00:58:51,740 And you will do this in 22.05, where 1096 00:58:51,740 --> 00:58:54,880 you'll be able to take the energy-dependent cross-section 1097 00:58:54,880 --> 00:58:57,260 in tabulated or theoretical form, 1098 00:58:57,260 --> 00:58:59,510 and then integrate this whole equation, 1099 00:58:59,510 --> 00:59:04,360 and also account for the fact that the flux has an energy 1100 00:59:04,360 --> 00:59:04,900 component. 1101 00:59:04,900 --> 00:59:07,240 Usually, it looks something like-- in a light-water 1102 00:59:07,240 --> 00:59:10,918 reactor, if this is energy and this is flux, 1103 00:59:10,918 --> 00:59:12,460 there'll be a bit of a thermal spike. 1104 00:59:12,460 --> 00:59:14,650 There won't be much going on in the middle. 1105 00:59:14,650 --> 00:59:19,560 I'm sorry, a fast spike, and there will be a thermal spike. 1106 00:59:19,560 --> 00:59:22,080 And knowing how many neutrons are at every energy 1107 00:59:22,080 --> 00:59:25,500 level, what's the probability of every neutron at energy level 1108 00:59:25,500 --> 00:59:28,740 interacting, and what are the cross-sections at every energy 1109 00:59:28,740 --> 00:59:31,200 level integrated over the full energy range 1110 00:59:31,200 --> 00:59:34,230 is what gets you the accurate correct solution. 1111 00:59:34,230 --> 00:59:37,113 What we've done here is called the one-group approximation, 1112 00:59:37,113 --> 00:59:39,030 where we've assumed that all the neutrons have 1113 00:59:39,030 --> 00:59:42,030 the same energy, thermal energy, which 1114 00:59:42,030 --> 00:59:46,610 is an OK assumption for thermal light-water reactors. 1115 00:59:46,610 --> 00:59:49,300 And it'll get you a good estimate. 1116 00:59:49,300 --> 00:59:51,670 The more neutrons you have at different energies, 1117 00:59:51,670 --> 00:59:53,560 the less good that estimate becomes. 1118 00:59:53,560 --> 00:59:54,060 Yeah. 1119 00:59:54,060 --> 00:59:55,647 AUDIENCE: Wait, so that thermal energy 1120 00:59:55,647 --> 00:59:57,260 you gave us, like 0.02 [INAUDIBLE],, 1121 00:59:57,260 --> 01:00:00,506 that was estimated for the energy of the neutrons being 1122 01:00:00,506 --> 01:00:01,006 fired. 1123 01:00:02,865 --> 01:00:04,240 MICHAEL SHORT: Let's say you have 1124 01:00:04,240 --> 01:00:09,320 a neutron at about 298 Kelvin. 1125 01:00:09,320 --> 01:00:11,660 From the Maxwell-Boltzmann temperature distribution, 1126 01:00:11,660 --> 01:00:15,770 you can turn that temperature into an average kinetic energy. 1127 01:00:15,770 --> 01:00:19,400 And that average kinetic energy will give you a velocity. 1128 01:00:19,400 --> 01:00:25,360 And that velocity is around 2,200 meters per second. 1129 01:00:25,360 --> 01:00:29,630 And that average kinetic energy happens to be about 0.025 EV. 1130 01:00:29,630 --> 01:00:32,180 So thermalized neutrons, like the ones 1131 01:00:32,180 --> 01:00:35,270 flying about in the reactor, are moving quite slowly 1132 01:00:35,270 --> 01:00:38,652 at just 2,200 meters a second, compared 1133 01:00:38,652 --> 01:00:40,360 to the fast neutrons, which can be moving 1134 01:00:40,360 --> 01:00:43,820 closer to the speed of light, not that close, 1135 01:00:43,820 --> 01:00:45,163 but much, much, much closer. 1136 01:00:47,980 --> 01:00:48,480 Cool. 1137 01:00:48,480 --> 01:00:50,938 I'll take it as a good sign that you all voluntarily stayed 1138 01:00:50,938 --> 01:00:52,480 a little late. 1139 01:00:52,480 --> 01:00:55,350 So did you guys find this example useful?