WEBVTT 00:00:00.000 --> 00:00:02.840 This is College Physics Answers with Shaun Dychko. 00:00:03.245 --> 00:00:06.308 The uncertainty in the decay energy of some nuclear state 00:00:06.308 --> 00:00:08.300 is 2 electron volts. 00:00:08.582 --> 00:00:11.874 And so that uncertainty multiplied by the uncertainty in time 00:00:11.870 --> 00:00:14.560 has to be greater than or equal to Planck's constant over 4π; 00:00:14.590 --> 00:00:16.731 this is the Heisenberg uncertainty principle. 00:00:16.971 --> 00:00:19.348 And that means that Δt then 00:00:19.714 --> 00:00:23.634 after we multiply both sides by 1 over ΔE here, 00:00:23.817 --> 00:00:26.400 is Planck's constant over 4π times ΔE. 00:00:26.400 --> 00:00:28.605 So the minimum uncertainty in time 00:00:28.600 --> 00:00:32.182 is 4.14 times 10 to the minus 15 electron volt seconds 00:00:32.180 --> 00:00:35.131 divided by 4π times 2 electron volts 00:00:35.302 --> 00:00:39.485 and this works out to 0.16 femtoseconds, 00:00:39.480 --> 00:00:43.028 is the minimum uncertainty in the time of this decay.