WEBVTT 00:00:00.030 --> 00:00:02.400 The following content is provided under a Creative 00:00:02.400 --> 00:00:03.820 Commons license. 00:00:03.820 --> 00:00:06.850 Your support will help MIT Open Courseware continue to 00:00:06.850 --> 00:00:10.520 offer high-quality educational resources for free. 00:00:10.520 --> 00:00:13.390 To make a donation or view additional materials from 00:00:13.390 --> 00:00:17.490 hundreds of MIT courses, visit MIT Open Courseware at 00:00:17.490 --> 00:00:18.740 ocw.mit.edu. 00:00:21.560 --> 00:00:22.310 Hi I'm Sal. 00:00:22.310 --> 00:00:24.980 Today we're going to solve problem number 6 of 00:00:24.980 --> 00:00:27.320 exam 1 of fall 2009. 00:00:27.320 --> 00:00:30.040 Now before you attempt a problem, there are a lot of 00:00:30.040 --> 00:00:33.680 things that you should know, that is background material 00:00:33.680 --> 00:00:35.740 that's going to help you solve the problem. 00:00:35.740 --> 00:00:38.060 Especially during an exam, given that this was the 00:00:38.060 --> 00:00:39.560 hardest on this exam. 00:00:39.560 --> 00:00:41.540 So you should be able to finish this within 00:00:41.540 --> 00:00:43.740 15 minutes or so. 00:00:43.740 --> 00:00:47.520 So before you start the problem you should know what 00:00:47.520 --> 00:00:49.800 the energy is of a charged particle in an electric field, 00:00:49.800 --> 00:00:51.830 which is given by this equation. 00:00:51.830 --> 00:00:55.640 The conversion from an electron volt to a joule, 00:00:55.640 --> 00:00:58.810 which is just a conversion of energy and this should be 00:00:58.810 --> 00:01:03.350 given to you on your table of constants. 00:01:03.350 --> 00:01:06.970 Also the energy associated with an emission spectrum, 00:01:06.970 --> 00:01:10.610 which is given by this equation where k is actually 00:01:10.610 --> 00:01:13.740 the ionization energy of hydrogen, which is 13.6 00:01:13.740 --> 00:01:18.140 electron volts and the z squared is the atomic number 00:01:18.140 --> 00:01:22.320 of your atom of interest. And then the n f and the n sub i 00:01:22.320 --> 00:01:24.340 are your transition states. 00:01:24.340 --> 00:01:27.360 The DeBroglie wavelength, which is given by lambda, is 00:01:27.360 --> 00:01:31.370 equal to a Planck's Constant divided by the momentum. 00:01:31.370 --> 00:01:33.430 And always remember that you need to conserve energy. 00:01:33.430 --> 00:01:35.950 This is what's going to get you the full 00:01:35.950 --> 00:01:37.400 points on the problem. 00:01:37.400 --> 00:01:41.940 So the problem, it's best to solve if you draw a little 00:01:41.940 --> 00:01:44.690 image as you read the problem. 00:01:44.690 --> 00:01:49.170 So the problem reads, atoms of ionized helium gas, He plus, 00:01:49.170 --> 00:01:52.350 are struck by electrons in a gas discharge tube operating 00:01:52.350 --> 00:01:54.040 with the potential difference between the 00:01:54.040 --> 00:01:57.060 electrodes set at 8.8 volts. 00:01:57.060 --> 00:01:59.580 So I'm going to go ahead and start drawing an image. 00:02:05.910 --> 00:02:13.070 So let's say I have two plates where the voltage between 00:02:13.070 --> 00:02:21.930 these two is set at 8.88 volts. 00:02:21.930 --> 00:02:25.690 And what's happening here is that you're accelerating 00:02:25.690 --> 00:02:26.750 electrons through here. 00:02:26.750 --> 00:02:29.660 So you can imagine that-- 00:02:29.660 --> 00:02:30.992 say this is positively charges and this 00:02:30.992 --> 00:02:32.570 is negatively charged-- 00:02:32.570 --> 00:02:35.910 that an electron starts from your negative plate and then 00:02:35.910 --> 00:02:39.750 it accelerates toward your positive plate and you can 00:02:39.750 --> 00:02:42.940 imagine your plate having hole or something small where your 00:02:42.940 --> 00:02:46.550 electron can then exit and then be in free space under 00:02:46.550 --> 00:02:50.890 the influence of no potential. 00:02:50.890 --> 00:02:57.330 So then these electrons are actually being aimed towards 00:02:57.330 --> 00:02:59.350 your helium plus ions. 00:02:59.350 --> 00:03:01.580 So that's your target. 00:03:01.580 --> 00:03:05.770 Now if I continue reading the problem, it says that the 00:03:05.770 --> 00:03:07.920 emmision spectrum includes the line associated with the 00:03:07.920 --> 00:03:11.310 transition from n equals 3 to n equals 2. 00:03:11.310 --> 00:03:13.890 Calculate the minimum value of the DeBroglie wavelength of 00:03:13.890 --> 00:03:16.480 scattered electrons that have collided with helium plus and 00:03:16.480 --> 00:03:19.050 generated this line in the emission spectrum. 00:03:19.050 --> 00:03:21.950 So it's an energy conservation problem. 00:03:21.950 --> 00:03:26.010 And all this tells you is that, you have an electron 00:03:26.010 --> 00:03:28.780 that is being accelerated through your potential. 00:03:28.780 --> 00:03:32.600 The electron is going to strike a helium plus atom and 00:03:32.600 --> 00:03:35.960 upon that you detect an emission spectrum. 00:03:35.960 --> 00:03:36.910 Well what does that tell you? 00:03:36.910 --> 00:03:41.360 That tells you that that electron actually excited an 00:03:41.360 --> 00:03:42.950 electron in the atom. 00:03:42.950 --> 00:03:45.710 The electron had to transition to a higher state to absorb 00:03:45.710 --> 00:03:49.340 energy and then decay and in the process of decaying it 00:03:49.340 --> 00:03:52.800 emits a photon, which is what you detect. 00:03:52.800 --> 00:03:54.630 So this information is given to you. 00:03:54.630 --> 00:03:57.415 But the problem asks to find the value of the DeBroglie 00:03:57.415 --> 00:03:59.150 wavelength. 00:03:59.150 --> 00:04:03.220 So since this an energy conservation problem, and this 00:04:03.220 --> 00:04:07.510 is the very first step that happens and I know that my 00:04:07.510 --> 00:04:09.530 energy equation for a charged particle is 00:04:09.530 --> 00:04:13.220 simply q times the voltage. 00:04:13.220 --> 00:04:19.600 So I know that q, which is a certain value of charge, the 00:04:19.600 --> 00:04:26.400 product of these two when I'm multiplying by 8.8 volts, this 00:04:26.400 --> 00:04:29.730 yields 8.88 electron volts. 00:04:29.730 --> 00:04:32.120 So it's a new unit of energy where we just saw the 00:04:32.120 --> 00:04:33.140 conversion. 00:04:33.140 --> 00:04:36.255 And the reason why this takes this shape is because one 00:04:36.255 --> 00:04:40.880 electron has the charge of 1.6 times 10 to the negative 19 00:04:40.880 --> 00:04:42.835 coulombs and that's where the conversion from electrical 00:04:42.835 --> 00:04:45.100 volts to joules come from. 00:04:45.100 --> 00:04:50.260 So I'm going to go ahead and call this e sub i initial. 00:04:50.260 --> 00:04:53.740 Because this is our initial energy that we're given. 00:04:53.740 --> 00:04:55.570 So we just start with this energy, nothing else. 00:04:55.570 --> 00:04:57.370 Nothing else was given from the problem. 00:04:57.370 --> 00:04:59.140 So we want to go ahead and conserve this. 00:04:59.140 --> 00:05:03.690 So our final state, the combination of whatever we are 00:05:03.690 --> 00:05:06.500 solving, should not have a higher energy than what 00:05:06.500 --> 00:05:08.670 initially we started with. 00:05:08.670 --> 00:05:13.780 So because the problem says that we have an emission 00:05:13.780 --> 00:05:19.370 spectrum, I'm going to go ahead and draw another image. 00:05:19.370 --> 00:05:24.560 I'll label this one, initial and I'll 00:05:24.560 --> 00:05:26.190 label this one, final. 00:05:29.660 --> 00:05:37.690 So my final image, you can imagine a helium plus atom. 00:05:37.690 --> 00:05:44.990 And the electron strikes the helium plus atom and we detect 00:05:44.990 --> 00:05:51.620 a photon and you're getting a scattered electron. 00:05:51.620 --> 00:05:54.280 That's what the problem says. 00:05:54.280 --> 00:05:56.740 Scattered electron. 00:05:56.740 --> 00:06:03.050 So what this tells me is that this energy now is going into 00:06:03.050 --> 00:06:05.340 creating a photon and scattering an 00:06:05.340 --> 00:06:07.510 electron from your atom. 00:06:07.510 --> 00:06:12.180 So if I was to equate those two or what not I can already 00:06:12.180 --> 00:06:18.360 say that, well in my final state I detect an emission or 00:06:18.360 --> 00:06:19.450 an emission spectrum. 00:06:19.450 --> 00:06:21.670 And I know what the transition says. 00:06:21.670 --> 00:06:23.570 It goes from n equals 3 to n equals 2. 00:06:23.570 --> 00:06:25.100 That's what the problem tells us. 00:06:25.100 --> 00:06:29.680 So if I take my equation, my delta e of my emission 00:06:29.680 --> 00:06:35.730 spectrum, is simply going to be negative k, which is 13.6 00:06:35.730 --> 00:06:36.980 electron volts. 00:06:41.030 --> 00:06:43.400 And I'm going to multiply by the atomic 00:06:43.400 --> 00:06:44.670 number, squared now. 00:06:44.670 --> 00:06:47.400 This is where people also make mistakes during the exam. 00:06:47.400 --> 00:06:49.770 The atomic number is not 1. 00:06:49.770 --> 00:06:52.000 It's actually 2 because we're talking about helium. 00:06:52.000 --> 00:06:58.200 And the atomic number is not the number of electrons or 00:06:58.200 --> 00:07:02.440 values in terms of an ion, but it's actually the number of 00:07:02.440 --> 00:07:05.140 protons that you have. So helium has 2 protons. 00:07:05.140 --> 00:07:07.630 So I can multiply this by 2. 00:07:07.630 --> 00:07:10.120 And I'm going to square it. 00:07:10.120 --> 00:07:16.160 And I know that I'm going from n and equals 3 to n equals 2. 00:07:16.160 --> 00:07:21.600 So if I plug this into my equation I have 1 over 3 00:07:21.600 --> 00:07:27.300 squared minus 1 over 2 squared, which is 9 and 4. 00:07:27.300 --> 00:07:31.530 And if you plug this in, and assuming that your calculator 00:07:31.530 --> 00:07:37.920 works and is correct, you get a value of 7.56 electron volts 00:07:37.920 --> 00:07:42.230 because that was the unit that I was using. 00:07:42.230 --> 00:07:46.270 Everything else is unitless, so my energy is now 7.56 00:07:46.270 --> 00:07:48.050 electron volts for my transition. 00:07:48.050 --> 00:07:50.800 Which covers this part for my photon. 00:07:50.800 --> 00:07:53.900 So now we're left with what the energy of this is. 00:07:53.900 --> 00:07:56.920 So what is the energy of the scattered electron? 00:07:56.920 --> 00:08:01.730 Well if I equate the initial to the final I know that the 00:08:01.730 --> 00:08:04.950 final is a composition of the transition 00:08:04.950 --> 00:08:06.120 in my scatter state. 00:08:06.120 --> 00:08:16.110 So I know that e initial, which is 8.88 electron volts, 00:08:16.110 --> 00:08:21.370 this equals to the energy for the emission spectrum, which 00:08:21.370 --> 00:08:26.610 is 7.56 electron volts. 00:08:26.610 --> 00:08:30.270 So I'll put the units in square brackets so you don't 00:08:30.270 --> 00:08:31.980 get confused. 00:08:31.980 --> 00:08:43.850 And then plus the energy of the scattered electron. 00:08:43.850 --> 00:08:46.120 So if I subtract the two, I get the energy of 00:08:46.120 --> 00:08:48.290 my scattered electron. 00:08:48.290 --> 00:08:51.870 And that pretty much gives me a value of-- 00:08:51.870 --> 00:08:54.080 after I subtract that-- 00:08:54.080 --> 00:09:01.390 e of scattered electron is equal to-- 00:09:01.390 --> 00:09:06.090 well the distance between these two is actually 1.32 00:09:06.090 --> 00:09:07.340 electron volts. 00:09:10.060 --> 00:09:12.330 But don't stop here because the problem didn't ask you to 00:09:12.330 --> 00:09:14.690 figure out the energy of the scattered electron. 00:09:14.690 --> 00:09:16.150 It actually told us to figure out what the DeBroglie 00:09:16.150 --> 00:09:19.330 wavelength is of this scattered electron. 00:09:19.330 --> 00:09:23.870 Now the fact that the electron is scattered, to me that means 00:09:23.870 --> 00:09:26.780 that the only energy that this particle 00:09:26.780 --> 00:09:28.140 has is kinetic energy. 00:09:28.140 --> 00:09:31.970 So it's a classical form of energy. 00:09:31.970 --> 00:09:37.176 So I can go ahead and I can equate this to just 1/2 mass 00:09:37.176 --> 00:09:40.880 of the electrons times my velocity squared-- whatever 00:09:40.880 --> 00:09:42.010 velocity it has. 00:09:42.010 --> 00:09:45.160 Now I'd don't need to figure out what the velocity is. 00:09:45.160 --> 00:09:47.960 This is just important to figure out what the DeBroglie 00:09:47.960 --> 00:09:48.750 wavelength is. 00:09:48.750 --> 00:09:51.840 So this is good. 00:09:51.840 --> 00:09:56.050 Now if we look at what our DeBroglie wavelength is, we 00:09:56.050 --> 00:10:02.460 know that lambda is equal to Planck's 00:10:02.460 --> 00:10:06.640 Constant divided by p. 00:10:06.640 --> 00:10:15.140 And Planck's Constant has a value of 6.6 times 10 to the 00:10:15.140 --> 00:10:21.450 minus 34 with units of joules seconds. 00:10:21.450 --> 00:10:22.460 So this is important too. 00:10:22.460 --> 00:10:24.670 Always keep track of your units because dimensional 00:10:24.670 --> 00:10:27.420 analysis will help you a lot when solving these problems. A 00:10:27.420 --> 00:10:30.620 lot of the times you're given energy values in electron 00:10:30.620 --> 00:10:35.050 volts but your solution will have a constant that doesn't 00:10:35.050 --> 00:10:37.500 have an electron volt, that will have a joule, so you're 00:10:37.500 --> 00:10:40.520 forced to make that conversion or else your problem is going 00:10:40.520 --> 00:10:42.700 to be wrong. 00:10:42.700 --> 00:10:45.820 OK so with that in mind, we know what h is. 00:10:45.820 --> 00:10:46.800 But what about p? 00:10:46.800 --> 00:10:47.980 Well p is momentum. 00:10:47.980 --> 00:10:51.520 And classical momentum is just mass times your velocity. 00:10:51.520 --> 00:10:57.390 So this is just mass times velocity, but this is the mass 00:10:57.390 --> 00:10:58.940 times our velocity. 00:10:58.940 --> 00:11:00.260 We know what the mass of the election is. 00:11:00.260 --> 00:11:02.250 That's given on our table of constants. 00:11:02.250 --> 00:11:04.970 But the velocity, we don't know what it is but we can 00:11:04.970 --> 00:11:07.850 grab it from our classical energy. 00:11:07.850 --> 00:11:11.250 So if I look at my energy-- 00:11:11.250 --> 00:11:16.200 from the side so I'm going to look at my energy-- 00:11:16.200 --> 00:11:18.450 I know that my energy-- 00:11:18.450 --> 00:11:21.984 I'm going to call this scat, for scattered-- 00:11:21.984 --> 00:11:27.810 the energy of my scattered electron is simply 1/2 the 00:11:27.810 --> 00:11:30.980 mass of the electron times our velocity squared. 00:11:30.980 --> 00:11:36.880 Now if I multiply both sides by 2m, I'm going to go ahead 00:11:36.880 --> 00:11:41.520 and get us to the point where I can get an equation that is 00:11:41.520 --> 00:11:42.960 just my mass times v. 00:11:42.960 --> 00:11:47.380 So I'm going to multiply both sides by 2m, So I get 2 mass 00:11:47.380 --> 00:11:55.360 of the electron, e scat of the electron, equals-- 00:11:55.360 --> 00:11:57.125 if I multiply this by 2m, the two cancel, 00:11:57.125 --> 00:11:58.820 so I get an m squared. 00:11:58.820 --> 00:12:00.790 m squared times v squared is essentially 00:12:00.790 --> 00:12:03.250 just m times v squared. 00:12:03.250 --> 00:12:08.630 And this becomes just mass of your electron v squared. 00:12:08.630 --> 00:12:16.580 Now if I take the square root of both sides, I end up 00:12:16.580 --> 00:12:18.510 getting a nice relation for just m times v. 00:12:18.510 --> 00:12:24.990 So I know that m times v now is just simply equal to this-- 00:12:24.990 --> 00:12:27.710 canceled, the square root cancels the square. 00:12:27.710 --> 00:12:33.050 The square root of 2 mass of the electron times the energy 00:12:33.050 --> 00:12:36.780 of the scattered electron. 00:12:36.780 --> 00:12:38.750 So if I go back to my DeBroglie 00:12:38.750 --> 00:12:40.550 wavelength, I have m v. 00:12:40.550 --> 00:12:43.830 So I can take that value as a function of energy and just 00:12:43.830 --> 00:12:45.820 substitute it into my equation. 00:12:45.820 --> 00:12:48.410 And that'll help us get the answer because we know what 00:12:48.410 --> 00:12:49.900 the energy of the scattered electron is. 00:12:49.900 --> 00:12:52.770 We calculated that on the first part. 00:12:52.770 --> 00:12:56.380 So with that in mind we'll come over here. 00:12:56.380 --> 00:13:01.590 Again this is 6.6 times 10 to the minus 34. 00:13:01.590 --> 00:13:05.760 And the units are joules seconds. 00:13:05.760 --> 00:13:15.800 And down here I have the square root of 2 times my mass 00:13:15.800 --> 00:13:18.480 of the electron, which is on your table of constants. 00:13:18.480 --> 00:13:24.620 And it's just 9.11 times 10 to the minus 31. 00:13:24.620 --> 00:13:30.070 And this has units of kilograms. And the energy of 00:13:30.070 --> 00:13:38.460 my scattered electron, which is 1.3 times 1.32 and this 00:13:38.460 --> 00:13:41.640 units of electron volts. 00:13:41.640 --> 00:13:44.610 Now this has units of electron volts. 00:13:44.610 --> 00:13:46.402 This has units of joules. 00:13:46.402 --> 00:13:48.010 This problem's a little bit-- 00:13:48.010 --> 00:13:50.230 now you're not going to get a good answer with that unless 00:13:50.230 --> 00:13:52.160 you make the conversion. 00:13:52.160 --> 00:13:54.880 I pointed it out, the bullet point in the beginning was 00:13:54.880 --> 00:13:59.540 that 1 electron volt is equal to 1.6 times 10 to minus 19 00:13:59.540 --> 00:14:03.300 joules, so if I simply multiply this by that 00:14:03.300 --> 00:14:11.540 conversion, just 1.6 times 10 to the minus 19, this has 00:14:11.540 --> 00:14:18.800 units now of joules per electron volt. 00:14:18.800 --> 00:14:23.740 So this has units of joules per electron volt. 00:14:23.740 --> 00:14:25.220 The equation turned out to be pretty long. 00:14:25.220 --> 00:14:27.780 But this cancels the electron volt and now 00:14:27.780 --> 00:14:29.320 has units of joules. 00:14:29.320 --> 00:14:32.600 Which if you go through the math, you're going to go ahead 00:14:32.600 --> 00:14:34.820 and at the end get a unit of just meters. 00:14:34.820 --> 00:14:37.940 Because joule is just kilogram, meters squared per 00:14:37.940 --> 00:14:39.380 second squared. 00:14:39.380 --> 00:14:44.260 So all that factors out and you end up getting that. 00:14:44.260 --> 00:14:48.780 If you go through the math and you get your math right, then 00:14:48.780 --> 00:14:54.610 this should yield a value of 1.06 times 10 00:14:54.610 --> 00:14:59.920 to the minus 9 meters. 00:14:59.920 --> 00:15:06.550 So this is the value that will get you the right 00:15:06.550 --> 00:15:07.660 answer on the exam. 00:15:07.660 --> 00:15:10.410 Again this is lambda for your DeBroglie wavelength because 00:15:10.410 --> 00:15:11.740 that's what the problem is. 00:15:11.740 --> 00:15:15.120 So the problem asked for the DeBroglie wavelength but it 00:15:15.120 --> 00:15:17.800 give you all this information to get to the point where you 00:15:17.800 --> 00:15:18.890 needed to solve. 00:15:18.890 --> 00:15:25.050 And by conserving energy and knowing the right conversion 00:15:25.050 --> 00:15:27.660 factors between energies you're able to get an answer, 00:15:27.660 --> 00:15:29.740 which is good. 00:15:29.740 --> 00:15:34.940 It's really easy to complicate things and get a 00:15:34.940 --> 00:15:36.450 wrong problem now. 00:15:36.450 --> 00:15:41.280 I remember from grading the exams, the most common error 00:15:41.280 --> 00:15:45.410 that people faced was actually letting the energy of the 00:15:45.410 --> 00:15:47.900 electron be the energy of a photon. 00:15:47.900 --> 00:15:52.320 Now that can't be because the energy of a photon is just 00:15:52.320 --> 00:15:54.950 your Planck's Constant times a frequency. 00:15:54.950 --> 00:15:59.110 And that's the energy for massless particles, your 00:15:59.110 --> 00:16:02.390 electron has mass so if it's moving it's not 00:16:02.390 --> 00:16:03.410 going to be h nu. 00:16:03.410 --> 00:16:05.480 The energy is going to be kinetic energy. 00:16:05.480 --> 00:16:06.370 1/2 mb squared. 00:16:06.370 --> 00:16:09.650 If it's not in an electric field, that is. 00:16:09.650 --> 00:16:14.160 So with that in mind just be confident when you solve these 00:16:14.160 --> 00:16:18.690 problems and make sure that you know exactly what energy 00:16:18.690 --> 00:16:21.820 equations to use to get the right answer.