- $32.9 \textrm{ keV}$
- $\dfrac{\Delta E}{E_o} = 6.44 \%$

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This is College Physics Answers with Shaun Dychko. The lifetime of an unstable nucleus is 10 to the minus 20 seconds and we are asked to figure out, what is the minimum possible uncertainty in its energy? So the Heisenberg uncertainty principle says that the uncertainty in energy times the uncertainty in time has to be more than or equal to Planck's constant over 4<i>π</i>. And so we'll divide both sides by <i>Δt</i> and we get the uncertainty in energy then is at a minimum Planck's constant over 4<i>πΔt</i>. And so to minimize this fraction, we should maximum <i>Δt</i> and take the largest number possible that we can put there which will be the measurement of 10 to the minus 20 seconds. If we knew what the uncertainty in this measurement was or an estimate of the uncertainty, we could plug that in instead. But since we are not given that information, we can just take the entire reading to be the uncertainty such that we can maximize the denominator and therefore minimize this fraction and therefore find the minimum uncertainty in the energy. So we have 4.14 times 10 to the minus 15 electron volt seconds is Planck's constant and we'll divide by 4<i>π</i> times 1 times 10 to the minus 20 seconds giving us 32.9 kiloelectron volts is the minimum uncertainty in the energy. And we are gonna compare that energy to the rest energy of an electron which is 511 kiloelectron volts. So 32.9 over 511 is 6.44 percent.