WEBVTT 00:00:00.000 --> 00:00:02.840 This is College Physics Answers with Shaun Dychko. 00:00:03.340 --> 00:00:06.490 Formula [27.3] tells us 00:00:06.490 --> 00:00:11.970 what the distance between slits and the diffraction grating must be 00:00:11.970 --> 00:00:15.460 to have a maximum at a certain angle, 00:00:15.460 --> 00:00:17.710 given a particular wavelength of light 00:00:17.790 --> 00:00:20.220 and m is the order of the maximum 00:00:20.220 --> 00:00:23.410 so each wavelength will give many maxima 00:00:23.410 --> 00:00:27.810 and we want to find the first one in part (a), in which case, m is 1. 00:00:28.130 --> 00:00:31.160 And we are told that the angle to this first maximum is 20 degrees 00:00:31.160 --> 00:00:34.480 and so what is d, the distance between the slits 00:00:34.480 --> 00:00:38.250 and we divide it by sin Θ on both sides to solve for d. 00:00:39.080 --> 00:00:41.390 Now the wavelength is something we have to figure out 00:00:41.390 --> 00:00:44.050 using equation [30.13] 00:00:44.050 --> 00:00:47.720 and we are asked in part (a) to use the first wavelength 00:00:47.720 --> 00:00:50.200 in the Balmer series for hydrogen. 00:00:51.250 --> 00:00:53.300 So we have the reciprocal of that wavelength 00:00:53.300 --> 00:00:57.490 is Rydberg's constant times 1 over the final energy level squared 00:00:57.490 --> 00:01:00.730 minus 1 over the initial energy level squared. 00:01:01.530 --> 00:01:05.070 And for the Balmer series, 00:01:05.070 --> 00:01:07.020 the final is 2 00:01:07.020 --> 00:01:10.520 and the initial has to be more than the final in order for there to be a 00:01:10.520 --> 00:01:14.190 drop in energy level of the electron 00:01:14.190 --> 00:01:17.440 and therefore an emission of some photon of some energy. 00:01:18.220 --> 00:01:19.950 So n i has to be more than n f 00:01:19.950 --> 00:01:21.580 and so if we want the first wavelength, 00:01:21.580 --> 00:01:25.470 we'll take the first number greater than n f which is 3. 00:01:25.850 --> 00:01:29.450 So we plug in the number 3 here for n i. 00:01:29.980 --> 00:01:31.940 Now I solved for wavelength by 00:01:31.940 --> 00:01:33.940 taking the reciprocal of both sides which means 00:01:34.410 --> 00:01:36.690 raise both sides to the exponent negative 1 00:01:36.690 --> 00:01:39.880 and this right-hand side, I just left it as wrapped in brackets 00:01:39.880 --> 00:01:42.430 with an exponent negative 1 because 00:01:42.430 --> 00:01:45.660 you know, getting a common denominator here and then flipping the fraction 00:01:45.890 --> 00:01:47.380 is more work than it's worth probably. 00:01:47.380 --> 00:01:50.430 So let's just do it like this. 00:01:50.800 --> 00:01:54.610 So we have the first wavelength in the Balmer series is Rydberg's constant— 00:01:54.610 --> 00:01:57.800 1.097 times 10 to the 7 reciprocal meters— 00:01:58.270 --> 00:02:01.000 times 1 over the final energy level squared 00:02:01.000 --> 00:02:03.780 minus 1 over the initial energy level squared 00:02:03.780 --> 00:02:08.650 and all to the negative 1 giving us 656.34 nanometers. 00:02:08.890 --> 00:02:12.040 So that's the wavelength we plug in to this formula here 00:02:13.290 --> 00:02:15.090 to get the distance between the slits; 00:02:15.220 --> 00:02:20.010 the order is 1—it's the first maximum—times its wavelength divided by sin of 20 degrees 00:02:20.010 --> 00:02:22.730 which is 1.92 micrometers. 00:02:24.200 --> 00:02:27.690 Now in part (b), we are asked to figure out 00:02:28.480 --> 00:02:29.970 what will the angle be 00:02:30.860 --> 00:02:35.080 for the fourth line in the Balmer series? 00:02:35.310 --> 00:02:38.780 So n i has to be greater than n f; 00:02:38.780 --> 00:02:43.400 n f is definitely 2 because of the word 'Balmer', Balmer series 00:02:43.400 --> 00:02:47.970 and so n i can be 3, 4, 5, 6 and so on. 00:02:48.260 --> 00:02:51.810 And we want the fourth wavelength so that means n i is 6; 00:02:52.730 --> 00:02:55.900 number 3 would be the first, 4, 5, 6, 6 is the fourth. 00:02:55.900 --> 00:02:58.330 So that's why we plug in the number 6 there. 00:02:59.310 --> 00:03:02.990 And we rearranged this formula to solve for Θ now 00:03:02.990 --> 00:03:06.350 so Θ is the inverse sign of the order times the wavelength divided by 00:03:06.370 --> 00:03:07.770 the distance between the slits. 00:03:08.610 --> 00:03:10.840 And so we find out what the wavelength is 00:03:10.840 --> 00:03:13.960 by going Rydberg's constant times 1 over 2 squared minus 00:03:13.960 --> 00:03:16.130 1 over 6 squared and then all to the negative 1 00:03:16.130 --> 00:03:18.300 which is 410.21 nanometers. 00:03:19.600 --> 00:03:23.460 And then that wavelength is substituted in here 00:03:24.130 --> 00:03:27.610 and we are multiplying by 1 cause it's the first order maximum 00:03:27.610 --> 00:03:30.040 for this fourth wavelength in the Balmer series 00:03:30.200 --> 00:03:33.160 and dividing by the separation 00:03:33.160 --> 00:03:35.980 between the slits that we calculated in part (a); 00:03:36.980 --> 00:03:40.250 that number's written here times 10 to the minus 6 instead of 00:03:40.250 --> 00:03:42.250 the prefix 'micro' 00:03:42.650 --> 00:03:45.250 and then we are taking the inverse sign of all that to get the angle 00:03:45.250 --> 00:03:47.060 which is 12.3 degrees. 00:03:49.420 --> 00:03:52.480 And then part (c) says, 00:03:53.420 --> 00:03:57.500 at what angle will you find the second order maximum for 00:03:57.500 --> 00:04:01.120 the first wavelength in the Balmer series? 00:04:01.400 --> 00:04:02.970 So that means m is 2 00:04:02.980 --> 00:04:06.880 and the wavelength is what we calculated up here— 00:04:06.880 --> 00:04:09.620 first wavelength in the Balmer series— 00:04:11.100 --> 00:04:13.790 and then dividing by the separation between the slits, 00:04:13.790 --> 00:04:16.930 taking the inverse sign of that and we get 43.2 degrees.