Question #87972
An electron in the n=2 state of hydrogen remains there on the average about 10^-8s before making a transition to n=1 state the uncertainty in the energy of an n=2 state is
Ans: 4.1×10^-7 ev
1
Expert's answer
2019-04-15T10:17:04-0400

The uncertainty in energy is connected with the lifetime of the corresponding excited state via the following expression:


ΔEΔt=\Delta E \cdot \Delta t = \hbar

Hence:


ΔE=Δt=6.631034108J=6.631026J\Delta E= \frac{\hbar}{\Delta t} = \frac{6.63 \cdot 10^{-34}}{10^{-8}} \, J \, = 6.63 \cdot 10^{-26} \, J

As long as


1eV=1.61019J,1 \, eV = 1.6 \cdot 10^{-19} \, J,

we obtain


ΔE=6.6310261.610194.1107eV\Delta E= \frac{6.63 \cdot 10^{-26}}{1.6 \cdot 10^{-19}} \approx 4.1 \cdot 10^{-7} \, eV

Answer: 4.1*10-7 eV.



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