Estimate the lifetime of the excited state that produced this line.

You can use the equation ##\Delta \lambda=\frac{c}{f}## to find the frequency, and then use the equation ##\tau=\frac{1}{f}## to find the lifetime of the excited state. However, be careful to convert the units properly so that your final answer is in seconds.
  • #1
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Homework Statement



An atomic spectrum contains a line with a wavelength centered at 460 nm. Careful measurements show the line is really spread out between 459 and 461 nm.

Estimate the lifetime of the excited state that produced this line.

Homework Equations



Change in Frequency = Speed of light/Change in Wavelength

Time = The inverse of frequency

The Attempt at a Solution



Change in wavelength = 461nm -459 nm = 2 nm or 2x10^-9 meters

Change in frequency = speed of light/change in wavelength = (3x10^8)/(2x10^-9) = 1.5x10^17 in units of s^-1

Life time = 1/(1.5x10^17) = 6.7*10^-18 seconds

But this is wrong and I don't know where to go from here. Please help!
 
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  • #2
Neither of your relevant equations is correct. For the first one, use ##\lambda_1 = c/f_1## and use ##\lambda_2 = c/f_2## and find what ##\lambda_1 - \lambda_2## equals. The second equation only applies if the time is the period of oscillation, which it isn't in this case.
 
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  • #3
I'd have had approached the problem differently but I might be totally wrong.
Let me write the following to vela (please do not read this OP!):
Hey vela :)
I'd have used HUP for that one. So instead of looking for the delta lambda, I'd have looked for delta E and then apply the HUP to get delta t.
What do you think about the approach?
 
  • #4
Yeah, that's correct.
 
  • #5


I would first like to commend you for attempting to solve this problem and using the appropriate equations. However, it is important to note that the given information is not sufficient to accurately estimate the lifetime of the excited state.

To accurately determine the lifetime of the excited state, we would need additional information such as the energy level of the excited state and the transition involved in the emission of the spectral line. This information is necessary because the lifetime of an excited state depends on the energy difference between the excited state and the ground state, and the strength of the transition.

Additionally, the spread of the line does not necessarily indicate a shorter lifetime. The line could also be broadened due to other factors such as Doppler broadening or pressure broadening. Therefore, without further information, it is not possible to accurately estimate the lifetime of the excited state that produced this line.

In conclusion, while your attempt at a solution is commendable, it is important to consider all relevant factors and gather all necessary information before making an estimation of the lifetime of an excited state.
 

1. How is the lifetime of an excited state estimated?

The lifetime of an excited state can be estimated using various techniques such as spectroscopy, fluorescence, and phosphorescence measurements.

2. What factors affect the lifetime of an excited state?

The lifetime of an excited state can be affected by factors such as the energy of the excited state, the nature of the molecule or atom, and the presence of any external influences such as temperature or pressure.

3. Can the lifetime of an excited state be accurately measured?

Yes, with the use of advanced technologies and precise measurement techniques, the lifetime of an excited state can be accurately measured.

4. How is the lifetime of an excited state related to its energy level?

The lifetime of an excited state is inversely proportional to its energy level. This means that the higher the energy level of the excited state, the shorter its lifetime will be.

5. Why is estimating the lifetime of excited states important?

Estimating the lifetime of excited states is important because it helps us understand the behavior and properties of atoms and molecules. It also has practical applications in fields such as materials science, chemistry, and astrophysics.

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