How Do You Calculate Linewidth from Quality Factor for Light Emission?

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To calculate the linewidth from the quality factor (Q) for light emission at a wavelength of 500 nm, the resonant frequency is essential. The frequency is calculated as 6E14 Hz, leading to an angular frequency of approximately 3.77E15 radians/sec. Using the relationship Q = f(resonant)/delta f, the bandwidth (delta f) is found to be 1.2E7 Hz. However, there is confusion as the resulting wavelength calculation yields 25 meters, which is inconsistent with the expected value in nanometers. It is clarified that for lightly damped oscillators, the resonant frequency closely approximates the natural frequency of the system.
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Homework Statement


Light with wavelength 500 nm is emitted from an electron in an atom behaving as a lightly damped simple harmonic oscillator with Q = 5 x 10^7. Find the linewidth (width to half-power points) in nm.

Homework Equations


wavelength=c/f
f=w/2pi
Q=f/delta f=w/delta w

The Attempt at a Solution



So i started by calculating frequency and then angular frequency

f = 3E8/500E-9 = 6E14hz
w = 6E14*2pi = 3.77E15 radians/sec

I did some research and found that Q=resonant frequency/(half-power bandwidth)
But how do i work out the resonant frequency? The lecturer said it would take us maybe an hour or two to solve this problem so it seems to simple for the frequency i have to be the one required here.
 
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Aidden said:
I did some research and found that Q=resonant frequency/(half-power bandwidth)

so you can find the line width as you have the Q-value as well as the frequency!
 
drvrm said:
so you can find the line width as you have the Q-value as well as the frequency!
Yes but as i said at the bottom of the post, is the frequency i have the resonant frequency because this seems far too simple, the lecturer said it would take an hour or two but it's only taken 5 minutes... And he asks for it in nm but when i try the frequency i have i end up with 25m as the answer so it seems wrong to me.
 
Q=f(resonant)/delta f

=> delta f = 6E14/5E7 = 1.2E7 Hz
lamda=c/f
=>lamda = 3E8/1.2E7 = 25m

It just seems far too big when he's asking for it in nm :/
 
Calculate the frequencies on the resonant curve which correspond to the half power and convert those frequencies to wavelength.
 
Aidden said:
Yes but as i said at the bottom of the post, is the frequency i have the resonant frequency because this seems far too simple, the lecturer said it would take an hour or two but it's only taken 5 minutes... And he asks for it in nm but when i try the frequency i have i end up with 25m as the answer so it seems wrong to me.
Aidden said:
lightly damped simple harmonic oscillator with Q = 5 x 10^7. Find the linewidth (width to half-power points) in nm.

When damping is small, the resonant frequency is approximately equal to the natural frequency of the system, which is a frequency of unforced vibrations.
 
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