# Quick Uncertainty Problem - atom energy levels

1. Jan 14, 2012

### binbagsss

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 any one 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 ... =]

2. Jan 17, 2012

### BruceW

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. Jan 17, 2012

### binbagsss

ahh thanks, so to clarify my method used for part b is correct or ?

4. Jan 17, 2012

### BruceW

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. Jan 17, 2012

### vela

Staff Emeritus
Consider
$$E + \Delta E = \frac{hc}{\lambda + \Delta \lambda}$$and solve for ΔE to first order in Δλ.

6. Jan 17, 2012

### BruceW

that's pretty clever, vela. Thanks for jumping in, I did not have the right answer. binbagsss - listen to vela!