This is College Physics Answers with Shaun Dychko. The uncertainty in the decay energy of some nuclear state is 2 electron volts. And so that uncertainty multiplied by the uncertainty in time has to be greater than or equal to Planck's constant over 4<i>π</i>; this is the Heisenberg uncertainty principle. And that means that <i>Δt</i> then after we multiply both sides by 1 over <i>ΔE</i> here, is Planck's constant over 4<i>π</i> times <i>ΔE</i>. So the minimum uncertainty in time is 4.14 times 10 to the minus 15 electron volt seconds divided by 4<i>π</i> times 2 electron volts and this works out to 0.16 femtoseconds, is the minimum uncertainty in the time of this decay.