Quick Uncertainty Problem - atom energy levels

In summary: The uncertainty in energy of that excited state is 2.105eV. The wavelength spread of the corresponding spectral line is 589.0nm.
  • #1
binbagsss
1,254
11
Two quick problems, not sure whether the 2 are related or not:

1) The question is:
A sodium atom in one of the states laveled 'lowest excited lvels' remains in that state, on average, for 1.6 x 10^-8 s before it makes a transition to the ground level, emitting a photon with wavelength 589.0nm and energy 2.105eV
a)What is the uncertainty in energy of that excited state?
b)What is the wavelength spread of the corresponding spectral line?

a) Using the average time the atom spends in the excited state as Δt, and applying the uncertainty principle, you are able to solve for ΔE
b) Here, I do not understand why you are not able to use : ΔE = hc/Δλ
, and solve for Δλ: (Would it be a coincidence that when using this my answer is a factor of 10^-16 out?

Rather you should use the fact that the fractional uncertainties of the photon are the same, and you are able to find the fractional uncertainty in a) as this is the lowest exctied level.
Is it because we are not after the corresponding Δλ of this energy state , but rather of this photon specifcially?

2) I am confused as to the correct way to regard the uncertainty in the time -Δt - when considering energy levels of an electron in an atom - so in the above question, the lifetime of the energy state was taken to be the Δt - but I do not fully understand why this is the case - I am picturing the lifetime of an energy level in the following way:
- Any energy level consists of +1 electron, the lifetime of this energy state therefore must correspond to only once all the electrons have dropped to ground level, and so to me this makes sense in the way that any particular electron could drop to ground anytime between t=0 and t=lifetime, (rather than the lifetime specifying the average time taken for anyone electron to drop down to ground level).
- Another way I am picturing this is, an energy level consists of 1 electron only, the electron's time taken to drop to ground level would follow a normal distribution and so to me, for the lifetime to correspond to Δt , the lifetime should be equal to the maximum time an electron takes to drop down to ground level, rather than the average time ...

I think the problem stems from the fact that I do not know what is defined as the lifetime of an energy level.

Many thanks ... =]
 
Physics news on Phys.org
  • #2
It looks like you are using ΔE Δt = h/2pi which is simply used to get an estimate of the order of magnitude. I'm not sure why you got the question wrong... maybe incorrect units?
 
  • #3
ahh thanks, so to clarify my method used for part b is correct or ?
 
  • #4
ΔE=hc/Δλ Would be my first guess as well, since I would assume that the uncertainty in the photon's energy is the same as the uncertainty in the energy of the atom before it ejected the photon. So I think you've got it right. But the given answer is different to your answer? maybe we were both wrong then..
 
  • #5
Consider
$$E + \Delta E = \frac{hc}{\lambda + \Delta \lambda}$$and solve for ΔE to first order in Δλ.
 
  • #6
that's pretty clever, vela. Thanks for jumping in, I did not have the right answer. binbagsss - listen to vela!
 

1. What is the Quick Uncertainty Problem?

The Quick Uncertainty Problem is a concept in quantum mechanics that involves the uncertainty principle and the energy levels of an atom. It refers to the challenge of determining the exact position and momentum of an electron in an atom, as these two properties cannot be known simultaneously with absolute precision.

2. How does the Quick Uncertainty Problem affect energy levels?

The Quick Uncertainty Problem has a significant impact on energy levels in atoms. According to the uncertainty principle, the more precisely we know the position of an electron, the less we know about its momentum, and vice versa. This means that we cannot know the exact energy level of an electron in an atom, only a range of possible energy values.

3. Can the Quick Uncertainty Problem be solved?

No, the Quick Uncertainty Problem cannot be solved in the traditional sense. The uncertainty principle is a fundamental aspect of quantum mechanics, and it is impossible to determine the exact position and momentum of an electron simultaneously. However, we can use mathematical models and calculations to predict the probability of finding an electron at a particular energy level.

4. How does the Quick Uncertainty Problem relate to atomic spectra?

The Quick Uncertainty Problem is closely related to atomic spectra, which is the set of distinct energy levels that an electron can occupy in an atom. The uncertainty principle dictates that we cannot know the exact energy level of an electron, so we can only observe the different energy levels it can occupy and their corresponding spectral lines.

5. What are the practical applications of the Quick Uncertainty Problem?

The Quick Uncertainty Problem has significant practical applications in fields such as materials science, chemistry, and technology. It helps us understand the behavior of atoms and molecules, which is essential for developing new materials and technologies. It also plays a crucial role in quantum computing, as it allows for the manipulation of energy levels to store and process information.

Similar threads

  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
Replies
1
Views
380
Replies
3
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
12
Views
1K
Replies
7
Views
1K
Replies
15
Views
1K
  • Introductory Physics Homework Help
2
Replies
35
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
16K
  • Quantum Physics
Replies
11
Views
892
Back
Top