1 00:00:04,746 --> 00:00:06,120 TINA CHEN: In this video, we will 2 00:00:06,120 --> 00:00:09,870 be exploring the effect of bond energies on vacancy diffusion 3 00:00:09,870 --> 00:00:11,880 using a Mathematica simulation. 4 00:00:11,880 --> 00:00:15,150 Even the most carefully produced materials have defects. 5 00:00:15,150 --> 00:00:17,730 These defects often affect the physical properties 6 00:00:17,730 --> 00:00:18,690 of the material. 7 00:00:18,690 --> 00:00:21,270 One type of defect is a vacancy in which 8 00:00:21,270 --> 00:00:24,270 an atom or molecule is missing from a point in the lattice. 9 00:00:24,270 --> 00:00:27,090 Vacancies can diffuse or move around. 10 00:00:27,090 --> 00:00:28,800 In this video, we will investigate 11 00:00:28,800 --> 00:00:32,430 diffusion of a single vacancy in a binary AB alloy. 12 00:00:32,430 --> 00:00:34,620 To simplify the simulation process, 13 00:00:34,620 --> 00:00:37,320 this AB alloy will have a two-dimensional, square 14 00:00:37,320 --> 00:00:38,160 lattice. 15 00:00:38,160 --> 00:00:40,470 In order to simulate the movement of a vacancy, 16 00:00:40,470 --> 00:00:43,200 we must consider, on the atomic level, the interaction 17 00:00:43,200 --> 00:00:45,930 between A and B. Specifically, we 18 00:00:45,930 --> 00:00:47,790 will look at the bond energies. 19 00:00:47,790 --> 00:00:50,370 Here, we have our AB alloy square lattice 20 00:00:50,370 --> 00:00:55,020 with randomly placed atoms A and B. If we have a vacancy here, 21 00:00:55,020 --> 00:00:58,770 then the vacancy can potentially jump to one of four places. 22 00:00:58,770 --> 00:01:00,600 We look at the energy required to jump 23 00:01:00,600 --> 00:01:02,320 to each of the four places. 24 00:01:02,320 --> 00:01:04,830 If the vacancy wants to jump to this lattice point, 25 00:01:04,830 --> 00:01:06,540 then it has to break each of the bonds 26 00:01:06,540 --> 00:01:08,580 connecting the atom to other atoms. 27 00:01:08,580 --> 00:01:11,220 The same is true for the vacancy to jump to any 28 00:01:11,220 --> 00:01:12,660 of the other lattice points. 29 00:01:12,660 --> 00:01:15,060 It has to break the bonds connecting 30 00:01:15,060 --> 00:01:17,190 the atoms to other atoms. 31 00:01:17,190 --> 00:01:19,080 The frequency with which the vacancy 32 00:01:19,080 --> 00:01:21,660 jumps to any of the four neighboring lattice points 33 00:01:21,660 --> 00:01:25,050 is proportional to E to the bond energy. 34 00:01:25,050 --> 00:01:27,090 The specific equation relating these two 35 00:01:27,090 --> 00:01:28,740 is the Arrhensius equation, which 36 00:01:28,740 --> 00:01:31,970 shows that the frequency is also proportional to V, 37 00:01:31,970 --> 00:01:35,100 the Debye frequency, which is a property of the material, 38 00:01:35,100 --> 00:01:36,030 and the temperature. 39 00:01:36,030 --> 00:01:38,070 For our simulation, we will only look 40 00:01:38,070 --> 00:01:40,770 at how the bond energy affects vacancy diffusion. 41 00:01:40,770 --> 00:01:43,530 Since we only have two types of atoms in this alloy, 42 00:01:43,530 --> 00:01:45,930 there will only be three types of bonds-- 43 00:01:45,930 --> 00:01:49,890 AA, BB, and AB with corresponding bond energies, 44 00:01:49,890 --> 00:01:52,770 VAA, VBB, and VAB. 45 00:01:52,770 --> 00:01:55,860 For our simulation, we will start with a 2D square lattice 46 00:01:55,860 --> 00:01:57,840 with randomly placed A and B atoms. 47 00:01:57,840 --> 00:02:00,540 We will see the ratio of the atoms is 1 to 1. 48 00:02:00,540 --> 00:02:02,940 We will also have a randomly placed vacancy. 49 00:02:02,940 --> 00:02:05,220 We will then simulate a single jump of the vacancy 50 00:02:05,220 --> 00:02:07,020 by comparing the frequencies with which 51 00:02:07,020 --> 00:02:09,630 a vacancy will jump to the four neighboring lattice points. 52 00:02:09,630 --> 00:02:12,000 We will consider only the case that the vacancy jumps. 53 00:02:12,000 --> 00:02:14,130 That is, it does not stay at it's lattice point, 54 00:02:14,130 --> 00:02:16,440 and it can only move to one of the four neighboring lattice 55 00:02:16,440 --> 00:02:16,860 points. 56 00:02:16,860 --> 00:02:18,930 Then we can calculate the probability with which 57 00:02:18,930 --> 00:02:20,430 any of the four jumps occurs. 58 00:02:20,430 --> 00:02:22,500 This probability will be the jump frequency 59 00:02:22,500 --> 00:02:25,050 for a specific lattice point over the sum of all the jump 60 00:02:25,050 --> 00:02:25,697 frequencies. 61 00:02:25,697 --> 00:02:27,780 Note that we could have considered the possibility 62 00:02:27,780 --> 00:02:30,410 that the vacancy did not jump, but then the value of the Debye 63 00:02:30,410 --> 00:02:31,680 frequency would be needed. 64 00:02:31,680 --> 00:02:33,420 And when we use the probabilities instead 65 00:02:33,420 --> 00:02:35,850 of the actual jump frequency, the Debye frequency 66 00:02:35,850 --> 00:02:38,940 cancels out, and we can consider a general binary alloy. 67 00:02:38,940 --> 00:02:41,820 Now that we have the probabilities P1, P2, P3 68 00:02:41,820 --> 00:02:44,310 and P4 of jumping to the respective lattice points, 69 00:02:44,310 --> 00:02:46,470 we can simulate what the vacancy at this lattice 70 00:02:46,470 --> 00:02:48,660 point surrounded by these atoms will do. 71 00:02:48,660 --> 00:02:51,480 First, we pick a random number from 0 to 1. 72 00:02:51,480 --> 00:02:53,370 The value of the random number will determine 73 00:02:53,370 --> 00:02:54,600 where the vacancy will go. 74 00:02:54,600 --> 00:02:57,250 From 0 to P1, the vacancy will jump to the first lattice 75 00:02:57,250 --> 00:02:57,750 point. 76 00:02:57,750 --> 00:03:00,810 From P1 to P1 plus P2, the vacancy 77 00:03:00,810 --> 00:03:03,090 will jump to the second lattice point and so on. 78 00:03:03,090 --> 00:03:04,980 This is the areas for which the vacancy 79 00:03:04,980 --> 00:03:07,500 will jump to the third and fourth lattice points. 80 00:03:07,500 --> 00:03:09,750 The number picked by our random number generator 81 00:03:09,750 --> 00:03:11,760 will fall in one of these four areas. 82 00:03:11,760 --> 00:03:13,770 And the area that number falls in 83 00:03:13,770 --> 00:03:15,930 is the lattice point the vacancy will move to. 84 00:03:15,930 --> 00:03:18,000 The vacancy then jumps to the lattice point 85 00:03:18,000 --> 00:03:19,650 indicated by the random number. 86 00:03:19,650 --> 00:03:21,570 In this way, we are using random samples 87 00:03:21,570 --> 00:03:24,210 to simulate the randomness of the vacancy moving, given then 88 00:03:24,210 --> 00:03:27,060 calculated probabilities to one of the four lattice points. 89 00:03:27,060 --> 00:03:28,860 To complete the simulation, we then 90 00:03:28,860 --> 00:03:30,900 repeat the computation of the probabilities 91 00:03:30,900 --> 00:03:33,240 of jumping to the neighboring atoms, the random number 92 00:03:33,240 --> 00:03:35,490 sampling that chooses where the vacancy moves, 93 00:03:35,490 --> 00:03:37,730 and actually moving the vacancy. 94 00:03:37,730 --> 00:03:40,560 First, we will look at the case when atoms of the same type 95 00:03:40,560 --> 00:03:43,860 have lower bond energies than atoms of different types. 96 00:03:43,860 --> 00:03:47,100 The animation allows us to visualize how the vacancy will 97 00:03:47,100 --> 00:03:48,990 move within the lattice. 98 00:03:48,990 --> 00:03:51,930 We can wait and see as T gets larger, 99 00:03:51,930 --> 00:03:54,090 something seems to occur. 100 00:03:54,090 --> 00:03:56,430 And that seems to be the agglomeration 101 00:03:56,430 --> 00:03:58,290 of the same type of atom. 102 00:03:58,290 --> 00:04:01,110 For instance, we can see the accumulation of red atoms 103 00:04:01,110 --> 00:04:01,890 in this area. 104 00:04:05,030 --> 00:04:07,190 We can let the simulation run or we 105 00:04:07,190 --> 00:04:11,090 can see the beginning and end points after 10,000 steps. 106 00:04:11,090 --> 00:04:14,660 We see here that originally the lattice was pretty random. 107 00:04:14,660 --> 00:04:17,690 But after 10,000 steps of the vacancy, 108 00:04:17,690 --> 00:04:21,709 we can see accumulation of red atoms here and blue atoms here. 109 00:04:21,709 --> 00:04:25,040 Next, we will look at the case when atoms of the same type 110 00:04:25,040 --> 00:04:28,760 have higher bond energies and atoms of different types 111 00:04:28,760 --> 00:04:30,710 have lower bond energies. 112 00:04:30,710 --> 00:04:34,905 Again, we animate and see what happens. 113 00:04:43,190 --> 00:04:46,220 Now, atoms of the same type seem to be going away 114 00:04:46,220 --> 00:04:47,100 from each other. 115 00:04:47,100 --> 00:04:50,090 And instead, order seems to be appearing in this system. 116 00:04:50,090 --> 00:04:54,410 We see here red, blue, red, blue, red, blue, red, blue, 117 00:04:54,410 --> 00:04:55,500 red, blue. 118 00:04:55,500 --> 00:04:58,430 Again, we can look at the initial and final products. 119 00:04:58,430 --> 00:05:01,490 And we can see the initial lattice was quite random 120 00:05:01,490 --> 00:05:03,860 and the end lattice is very ordered. 121 00:05:03,860 --> 00:05:05,330 It's almost completely ordered. 122 00:05:05,330 --> 00:05:06,800 Finally, we will look at the case 123 00:05:06,800 --> 00:05:09,470 when atoms of the same type and atoms of different type 124 00:05:09,470 --> 00:05:11,150 have similar bond energies. 125 00:05:11,150 --> 00:05:14,450 We can animate, and we can wait a while, 126 00:05:14,450 --> 00:05:16,850 but nothing particularly special will happen. 127 00:05:19,610 --> 00:05:21,290 Again, our end products can show us 128 00:05:21,290 --> 00:05:23,390 what will happen after 10,000 steps. 129 00:05:23,390 --> 00:05:26,900 And we see that there is not much significant going on. 130 00:05:26,900 --> 00:05:30,020 The beginning and end lattices are both rather random. 131 00:05:30,020 --> 00:05:31,670 So how does a lattice decide which 132 00:05:31,670 --> 00:05:33,980 form to approach as T gets very large? 133 00:05:33,980 --> 00:05:35,870 The answer is, the strength of the bonds 134 00:05:35,870 --> 00:05:39,350 between A and A, between B and B, and between A and B, 135 00:05:39,350 --> 00:05:40,590 as we said before. 136 00:05:40,590 --> 00:05:42,170 That is, do atoms of the same type 137 00:05:42,170 --> 00:05:43,580 like or dislike each other? 138 00:05:43,580 --> 00:05:46,250 Based on the answer, atoms are the same type will 139 00:05:46,250 --> 00:05:50,210 either conglomerate, as in the first case, into regions of A 140 00:05:50,210 --> 00:05:54,680 and regions of B or atoms will rearrange into an ordered form, 141 00:05:54,680 --> 00:05:56,390 as in the second case, to be closer 142 00:05:56,390 --> 00:05:57,980 to the type of atom it likes. 143 00:05:57,980 --> 00:06:00,680 The affinity of an atom type for other atom types 144 00:06:00,680 --> 00:06:02,189 is given by the bond energy. 145 00:06:02,189 --> 00:06:03,980 And the type of lattice that will be formed 146 00:06:03,980 --> 00:06:06,380 is given by the effective interchange energy, 147 00:06:06,380 --> 00:06:10,760 which is given by VAA plus VBB minus 2VAB. 148 00:06:10,760 --> 00:06:12,650 If epsilon is greater than 0, then 149 00:06:12,650 --> 00:06:14,480 atoms of one type like the other type 150 00:06:14,480 --> 00:06:16,760 more than they like each other, so the vacancy 151 00:06:16,760 --> 00:06:19,520 will allow the crystal to obtain an ordered lattice. 152 00:06:19,520 --> 00:06:22,550 If epsilon is less than 0, then atoms of the same type 153 00:06:22,550 --> 00:06:25,160 like each other more than they like atoms of the other type, 154 00:06:25,160 --> 00:06:27,284 and they will conglomerate by the vacancy diffusion 155 00:06:27,284 --> 00:06:28,100 mechanism. 156 00:06:28,100 --> 00:06:30,230 If epsilon equals 0, then disordered, 157 00:06:30,230 --> 00:06:31,670 solid solution occurs. 158 00:06:31,670 --> 00:06:35,680 This is because neither the VAA, VBB, nor the VAB phases 159 00:06:35,680 --> 00:06:36,410 are favored. 160 00:06:36,410 --> 00:06:38,480 Instead, they maintain a random order. 161 00:06:38,480 --> 00:06:40,940 To create the simulation, me use the type of modeling 162 00:06:40,940 --> 00:06:42,560 called Monte Carlo simulation. 163 00:06:42,560 --> 00:06:44,600 The Monte Carlo method is simply a class 164 00:06:44,600 --> 00:06:47,810 of computational algorithms that use repeated random sampling 165 00:06:47,810 --> 00:06:48,800 for simulation. 166 00:06:48,800 --> 00:06:51,140 In this case, we used random sampling in addition 167 00:06:51,140 --> 00:06:53,210 to the probabilities to determine how likely 168 00:06:53,210 --> 00:06:55,940 a destination for the vacancy each lattice point is. 169 00:06:55,940 --> 00:06:58,820 Now take the model of the single vacancy diffusion 170 00:06:58,820 --> 00:07:00,950 and imagine many, many more vacancies. 171 00:07:00,950 --> 00:07:03,050 We can see now why defects can cause 172 00:07:03,050 --> 00:07:05,060 changes in the physical characteristics 173 00:07:05,060 --> 00:07:07,650 and even structure, in this case, of a material. 174 00:07:07,650 --> 00:07:10,550 I'd like to thank professors Craig Carter, [INAUDIBLE],, 175 00:07:10,550 --> 00:07:12,620 and Dwayne Johnson for their help and advice 176 00:07:12,620 --> 00:07:15,100 with the coding in this video.