Transition between excited states

AI Thread Summary
The discussion revolves around calculating the uncertainty in energy for a photon emitted during an electron transition between two excited states, with lifetimes of 1.2 x 10^-8 sec and 2.3 x 10^-8 sec. The uncertainty principle is applied, yielding uncertainties of 2.74 x 10^-8 eV and 1.43 x 10^-8 eV for the respective states. A discrepancy arises when the calculated energy uncertainty of 1.31 x 10^-8 eV differs from the book's answer of 4.17 x 10^-8 eV, leading to confusion about whether to add or subtract the uncertainties. It is clarified that uncertainties should be added, as they follow statistical rules similar to those for combining measurements. The conversation concludes with a consensus on the correct approach to handling uncertainties in this context.
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Homework Statement



An atom in an excited state has a lifetime of 1.2 x 10 -8 sec; in a second excited state the
lifetime is 2.3 x 10 -8 sec. What is the uncertainty in energy for the photon emitted when
an electron makes a transition between these two levels?

Homework Equations



\DeltaE\Deltat\geq\frac{\hbar}{2}

The Attempt at a Solution



So I just found the uncertainty in energies for the two excited states using the uncertainty principle, getting 2.74*10^-8 eV for the 1.2*10^-8 sec state, and 1.43*10^-8 eV for the 2.3*10^-8 sec state. And figured that the uncertainty in energy would just be the difference in the energies, giving 1.31*10^-8 eV. But the book gives an answer of 4.17*10^-8 eV, which I noticed is what you get if you add the energies instead.

So is the book wrong, or is there some weird thing about the uncertainties combining such that I have to add them instead?
 
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Shouldn't it follow from

\Delta E \Delta t \geq \frac{\hbar}{2}

that

\Delta E \geq \frac{\hbar}{2 \Delta t}

?
 
Whether adding or subtracting two quantities, the uncertainties add.
 
@Basic_Physics: Yes, that's how I got the energies.

@gneill: That's a rule from statistics? Cause the book was a bit sparse on that point.
 
gildomar said:
@gneill: That's a rule from statistics? Cause the book was a bit sparse on that point.
Yup. Consider that subtraction is just adding the negative of one of the values. The uncertainty in the negative value is the same as for the positive value. So in terms of uncertainty, addition and subtraction are the same.
 
Thanks gneill! At least now I know I'm not going crazy.
 
gildomar said:
Thanks gneill! At least now I know I'm not going crazy
I'm kinda confused here, what have you resolved at?? Is your value correct or wrong ??
 
A good way to realize that you don't subtract the uncertainties is to imagine that the two uncertainies happen to be equal.

Subtracting would give a result of zero uncertainty, which (I hope you see) makes no sense!
 
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