Is this the correct way to estimate the natural lifetime of an atomic state?

In summary, the conversation was about estimating the natural lifetime of an atomic state with a dominant decay mode that produces an emission line of wavelength 6 x 10^-7 m and natural width 10^-13 m. The attempt at a solution involved using the equations ΔEΔT ≥ ħ/2 and E = hc/λ, but the mistake was using ΔE = Δλ instead of ΔE = hc/λ^2 * Δλ. After correcting the mistake, the estimated natural lifetime was approximately 1.59 x 10^-16 s.
  • #1
doomhalo
2
0
Hi there,

I've just been having a little trouble with this short question from a past exam paper...

Homework Statement


"An atomic state has a dominant decay mode which produces an emission line of wavelength [itex] 6 \times 10^{-7} m [/itex] and natural width [itex] 10^{-13} m [/itex]. Estimate it's natural lifetime.

Homework Equations


[tex] \Delta E \Delta T \geq \frac{\hbar}{2} [/tex]
[tex] E = \frac{hc}{\lambda} [/tex]

The Attempt at a Solution



[tex] \tau \approx \frac{\hbar}{2E} [/tex]
[tex] \tau \approx \frac{\lambda}{4\pi c} \approx 1.59 \times 10^{-16} [/tex]

I was just wondering if this seemed right? I'm concerned that I've not used the natural width provided in the question but I'm not sure whether it's a case of I've a) Missed a relevant equation, or b) I've used the equations I do have wrong.

Thank you in advance!
 
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  • #2
The mistake here is that you used:[tex]\Delta E = \Delta \lambda[/tex]

What is ##\Delta E## in terms of ##\lambda## and ##\Delta \lambda##?
 
  • #3
unscientific said:
The mistake here is that you used:[tex]\Delta E = \Delta \lambda[/tex]

What is ##\Delta E## in terms of ##\lambda## and ##\Delta \lambda##?

Ah I see, so
[tex] \Delta E = \frac{hc}{\lambda^2} \Delta\lambda [/tex]

And then

[tex] \tau \approx \frac{\lambda^2}{4\pi c \Delta\lambda} [/tex]

?

Thank you very much!
 
  • #4
Yes that's right. That gives a reasonable answer. (0.3s I think)
 
  • #5


Hi there,

Your solution is correct. The natural width is not needed in this case because it is already taken into account in the uncertainty principle equation, which you used to calculate the natural lifetime. The natural width represents the spread of energies in the emission line, and it is related to the natural lifetime through the Heisenberg uncertainty principle. So, your solution is correct and you have not missed any relevant equation. Good job!
 

1. What is atomic state natural lifetime?

Atomic state natural lifetime refers to the amount of time it takes for an excited atom to return to its ground state, releasing the excess energy in the form of electromagnetic radiation.

2. How is atomic state natural lifetime measured?

Atomic state natural lifetime is typically measured using spectroscopic techniques, where the decay of the excited state is observed and the lifetime is calculated based on the rate of decay.

3. What factors can affect the atomic state natural lifetime?

The atomic state natural lifetime can be affected by external factors such as temperature, pressure, and electric or magnetic fields. It can also be influenced by the electronic structure of the atom and the specific energy level of the excited state.

4. Why is the atomic state natural lifetime important in atomic physics?

The atomic state natural lifetime is important because it provides information about the stability of an excited state and the transitions between energy levels. It is also essential for understanding and predicting atomic and molecular spectra.

5. Can the atomic state natural lifetime be manipulated?

Yes, the atomic state natural lifetime can be manipulated through various techniques such as laser cooling, which can control the energy levels of atoms and increase their stability. This is useful for applications such as atomic clocks and precision measurements.

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