Analogous to the uncertainty relation ΔpΔx > h/4π, there is an uncertainty relation for the time and energy, ΔEΔt > h/4π that stems from the methods
usually used to measure the energy. The uncertainty in the time, Δt, can be interpreted as a lifetime. The excited state of an atom responsible for the emission of a photon has an average lifetime of 10^-10 s. What is the corresponding uncertainty in the frequency associated with the emitted photon?

What I understood based on the uncertainty principle:
ΔEΔt > h/4π and that Δt=10^-10 s here in this problem.
I can use this to calculate ΔE.
However, my next step is something I am not sure about. I set ΔE=hΔv and solved for v but then got 7.96x10^8 ms-1 which is so big and doesn't make sense to me.

Can anyone explain what I did wrong, why, and give me some guidance on what I should do to get the uncertainty in frequency?

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Redbelly98
Staff Emeritus
Homework Helper
What I understood based on the uncertainty principle:
ΔEΔt > h/4π and that Δt=10^-10 s here in this problem.
I can use this to calculate ΔE.
However, my next step is something I am not sure about. I set ΔE=hΔv and solved for v but then got 7.96x10^8 ms-1 which is so big and doesn't make sense to me.

Can anyone explain what I did wrong, why, and give me some guidance on what I should do to get the uncertainty in frequency?
Do you mean for the units to be s-1 or ms-1?

Either way, consider that visible light has a frequency in the range of 1014-1015 Hz (or s-1). Compared to that, does your Δv value appear to be large or small?