A sodium atom is in one of the states labeled | vedclass

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(A) The uncertainty principle for energy and time is given by $\Delta E \cdot \Delta t \approx \frac{h}{2\pi}$.Given the average lifetime of the excit

Vedclass · Accepted Answer

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

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(A) The uncertainty principle for energy and time is given by $\Delta E \cdot \Delta t \approx \frac{h}{2\pi}$.Given the average lifetime of the excited state is $\Delta t = 10^{-8} \ s$.The uncertainty in energy is $\Delta E = \frac{h}{2\pi \Delta t}$.Substituting the values: $h = 6.626 	imes 10^{-34} \ J \cdot s$,$\pi = 3.14$,and $\Delta t = 10^{-8} \ s$.$\Delta E = \frac{6.626 	imes 10^{-34}}{2 	imes 3.14 	imes 10^{-8}} \approx 1.055 	imes 10^{-26} \ J$.To convert this energy into electron-volts $(eV)$,divide by $1.6 	imes 10^{-19} \ J/eV$:$\Delta E = \frac{1.055 	imes 10^{-26}}{1.6 	imes 10^{-19}} \approx 6.59 	imes 10^{-8} \ eV$.Rounding to the nearest provided option,we get $6.56 	imes 10^{-8} \ eV$.

$A$ 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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