WEBVTT 00:00:00.000 --> 00:00:02.840 This is College Physics Answers with Shaun Dychko. 00:00:03.140 --> 00:00:07.280 A 50 kilovolt x-ray tube accelerates electrons 00:00:07.485 --> 00:00:12.594 and the maximum energy of the photon emitted by this x-ray tube 00:00:12.590 --> 00:00:15.897 will be equal to the kinetic energy of the electrons. 00:00:16.411 --> 00:00:18.560 And that's gonna be the electron charge 00:00:18.560 --> 00:00:21.302 multiplied by the voltage through which it is accelerated. 00:00:22.160 --> 00:00:25.040 And so we can equate the energy which we can write as 00:00:25.040 --> 00:00:28.068 Planck's constant times speed of light divided by lambda 00:00:28.060 --> 00:00:31.531 and you can equate that to qV. 00:00:33.085 --> 00:00:37.177 And let's solve this for lambda by taking the reciprocal of both sides 00:00:37.691 --> 00:00:40.251 so this λ over hc equals 1 over qV 00:00:40.251 --> 00:00:42.250 and then multiply both sides by hc. 00:00:43.200 --> 00:00:47.691 So we have the shortest possible wavelength is hc over qV. 00:00:47.690 --> 00:00:51.794 So that's 6.626 times 10 to the minus 34 joule seconds—Planck's constant— 00:00:51.790 --> 00:00:54.274 times speed of light divided by elementary charge 00:00:54.270 --> 00:00:56.377 times 50 times 10 to the 3 volts 00:00:56.370 --> 00:01:01.657 giving 2.48 times 10 to the minus 11 meters is the shortest possible wavelength. 00:01:03.417 --> 00:01:06.925 And then in part (b), we calculate the energy of this wavelength 00:01:06.920 --> 00:01:09.702 and that's hc divided by λ 00:01:09.700 --> 00:01:14.445 and hc we are going to take to be 1240 electron volt nanometers 00:01:14.440 --> 00:01:16.628 and so we'll divide by this wavelength 00:01:16.620 --> 00:01:18.617 also expressed in units of nanometers 00:01:18.610 --> 00:01:22.365 so that these nanometers cancel and we'll get an answer in electron volts. 00:01:22.360 --> 00:01:26.788 So this written in nanometers is 0.02480 nanometers 00:01:27.108 --> 00:01:30.697 and this works out to 50.0 kiloelectron volts. 00:01:31.440 --> 00:01:33.977 And in part (c), we talk about 00:01:33.970 --> 00:01:37.897 how this corresponds to this 00:01:37.890 --> 00:01:41.634 this is the voltage of the x-ray tube— 50 kilovolts— 00:01:41.630 --> 00:01:45.668 and this is the maximum energy photon emitted 00:01:46.171 --> 00:01:50.480 and the maximum photon energy 00:01:50.480 --> 00:01:52.857 is the kinetic energy of the electron 00:01:52.850 --> 00:01:57.120 which is q times V and this is the elementary charge 00:01:57.360 --> 00:02:01.714 of an electron multiplied by the voltage so that makes electron volts 00:02:01.897 --> 00:02:06.331 and therefore, we see that the maximum photon energy is 00:02:06.330 --> 00:02:09.737 the x-ray tube voltage written in units of electron volts. 00:02:09.885 --> 00:02:12.491 So where we started with 50 kilovolts, 00:02:12.490 --> 00:02:15.622 we can write that as 50 kiloelectron volts 00:02:15.620 --> 00:02:18.365 for calculating the maximum photon energy.