Transition between excited states

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SUMMARY

The discussion centers on the calculation of uncertainty in energy for a photon emitted during an electron transition between two excited states with lifetimes of 1.2 x 10-8 seconds and 2.3 x 10-8 seconds. The uncertainty principle, expressed as ΔEΔt ≥ ħ/2, is applied to find uncertainties of 2.74 x 10-8 eV and 1.43 x 10-8 eV for the respective states. The confusion arises regarding whether to add or subtract these uncertainties, with the correct approach being to add them, resulting in a total uncertainty of 4.17 x 10-8 eV, as confirmed by forum participants.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically the uncertainty principle.
  • Familiarity with the concept of excited states in atomic physics.
  • Basic knowledge of energy quantization in photons.
  • Proficiency in performing calculations involving scientific notation.
NEXT STEPS
  • Study the implications of the uncertainty principle in quantum mechanics.
  • Learn about the statistical rules governing the addition of uncertainties.
  • Explore the concept of energy levels and transitions in quantum systems.
  • Investigate the relationship between lifetime and energy uncertainty in atomic physics.
USEFUL FOR

Students of quantum mechanics, physics educators, and researchers interested in atomic transitions and uncertainty calculations will benefit from this discussion.

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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