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(A) According to the Heisenberg uncertainty principle,the uncertainty in energy $\Delta E$ and the uncertainty in time $\Delta t$ are related by the e

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$6.56 	imes 10^{-8} \, eV$

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(A) According to the Heisenberg uncertainty principle,the uncertainty in energy $\Delta E$ and the uncertainty in time $\Delta t$ are related by the expression: $\Delta E \cdot \Delta t \approx \frac{h}{2\pi}$.Given,the average lifetime of the excited state is $\Delta t = 10^{-8} \, s$.Substituting the values,we get $\Delta E = \frac{h}{2\pi \cdot \Delta t}$.Using $h = 6.626 	imes 10^{-34} \, J \cdot s$,we have $\Delta E = \frac{6.626 	imes 10^{-34}}{2 	imes 3.14159 	imes 10^{-8}} \approx 1.054 	imes 10^{-26} \, J$.To convert this energy into electron-volts $(eV)$,we divide by the charge of an electron $(1.6 	imes 10^{-19} \, C)$:$\Delta E = \frac{1.054 	imes 10^{-26}}{1.6 	imes 10^{-19}} \approx 6.59 	imes 10^{-8} \, eV$.Rounding to the nearest provided option,the uncertainty in energy is $6.56 	imes 10^{-8} \, eV$.

$A$ sodium atom is in one of the states labeled 'Lowest excited levels'. It remains in that state for an average time of $10^{-8} \, s$,before it makes a transition back to a ground state. What is the uncertainty in energy of that excited state?

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