# Homework Help: Linewidth from quality factor

1. Mar 7, 2016

### Aidden

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
wavelength=c/f
f=w/2pi
Q=f/delta f=w/delta w

3. 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.

Last edited: Mar 7, 2016
2. Mar 7, 2016

### drvrm

so you can find the line width as you have the Q-value as well as the frequency!

3. Mar 7, 2016

### Aidden

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.

4. Mar 7, 2016

### Aidden

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 :/

5. Mar 8, 2016

### ehild

Calculate the frequencies on the resonant curve which correspond to the half power and convert those frequencies to wavelength.

6. Mar 8, 2016

### drvrm

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