Transition between excited states

In summary, the book says that the energy uncertainty for the photon emitted when an electron makes a transition between the two excited states is 1.31*10^-8 eV, while the uncertainty calculated using the uncertainty principle is 4.17*10^-8 eV.
  • #1
gildomar
99
2

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



[itex]\Delta[/itex]E[itex]\Delta[/itex]t[itex]\geq[/itex][itex]\frac{\hbar}{2}[/itex]

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?
 
Physics news on Phys.org
  • #2
Shouldn't it follow from

[itex]\Delta E \Delta t \geq \frac{\hbar}{2}[/itex]

that

[itex]\Delta E \geq \frac{\hbar}{2 \Delta t}[/itex]

?
 
  • #3
Whether adding or subtracting two quantities, the uncertainties add.
 
  • #4
@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.
 
  • #5
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.
 
  • #6
Thanks gneill! At least now I know I'm not going crazy.
 
  • #7
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 ??
 
  • #8
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!
 
  • #9
This thread is eight years old!
 
  • Wow
  • Haha
Likes Steve4Physics and Delta2

What is the transition between excited states?

The transition between excited states refers to the movement of an electron from a higher energy state to a lower energy state within an atom or molecule. This process is accompanied by the release of electromagnetic radiation, such as light.

What causes the transition between excited states?

The transition between excited states is caused by the absorption or emission of energy by an atom or molecule. This energy can come from a variety of sources, such as heat, light, or chemical reactions.

How is the transition between excited states measured?

The transition between excited states is measured using spectroscopy techniques. This involves shining light of different wavelengths onto a sample and measuring the intensity of the light that is absorbed or emitted by the sample. The resulting spectrum can provide information about the energy levels and transitions within the sample.

What factors affect the transition between excited states?

The transition between excited states can be affected by a variety of factors, including the structure and composition of the atom or molecule, the presence of other atoms or molecules nearby, and external factors such as temperature and pressure.

What are some real-world applications of the transition between excited states?

The transition between excited states is a fundamental process in many fields of science, including chemistry, physics, and astronomy. It is used in applications such as fluorescent lighting, lasers, and medical imaging techniques like MRI. Understanding and controlling this process also plays a crucial role in many technological advancements, such as solar panels and LED screens.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
836
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
10K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
Back
Top