Transition between excited states

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Homework Help Overview

The problem involves an atom transitioning between two excited states, each with a specified lifetime, and seeks to determine the uncertainty in energy for the emitted photon during this transition. The context is rooted in quantum mechanics and the uncertainty principle.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the uncertainty principle to find the uncertainty in energy for each excited state and question how these uncertainties should be combined. There is also exploration of whether the book's answer is correct and how statistical rules apply to uncertainties.

Discussion Status

Participants are actively engaging with the problem, questioning the method of combining uncertainties and discussing the implications of adding versus subtracting uncertainties. Some guidance has been provided regarding the statistical treatment of uncertainties, but no consensus has been reached on the correct approach to the problem.

Contextual Notes

There is mention of the book being sparse on details regarding the combination of uncertainties, which may be affecting participants' understanding. The discussion also reflects on the age of the thread, indicating a long-standing interest in the topic.

gildomar
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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!
 
This thread is eight years old!
 
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