## UnkleRhaukus one year ago Oscillating electron in an atom. Light of wavelength 590 nm is emitted by an electron in an atom behaving as a lightly damped simple harmonic oscillator with a Q value of 8.0 x 10^7. From the resonance bandwidth the width of the spectral line from such an atom in units of metres is:

1. UnkleRhaukus


2. UnkleRhaukus

is this right?

3. Michele_Laino

from the theory of resonant circuit, we have the subsequent expression for the Quality factor Q: $\Large Q = \frac{{{\omega _0}}}{{2\Delta \omega }}$ where \omega_zero is the resonant frequency Now we have this: $\Large Q = \frac{{{\omega _0}}}{{2\Delta \omega }} = \frac{{2\pi {\nu _0}}}{{2 \cdot 2\pi \Delta \nu }} = \frac{{{\nu _0}}}{{\Delta \nu }}$ on the other hand we can write this: $\large \Delta \lambda = \frac{c}{{{\nu _1}}} - \frac{c}{{{\nu _2}}} = c\frac{{{\nu _2} - {\nu _1}}}{{{\nu _1} \cdot {\nu _2}}} \cong c\frac{{\Delta \nu }}{{v_0^2}} \Rightarrow \Delta \nu = \frac{{v_0^2 \cdot \Delta \lambda }}{c}$

4. Michele_Laino

oops.. I have made a typo: $\Large Q = \frac{{{\omega _0}}}{{2\Delta \omega }} = \frac{{2\pi {\nu _0}}}{{2 \cdot 2\pi \Delta \nu }} = \frac{{{\nu _0}}}{{2 \cdot \Delta \nu }}$ so substituting into the expression for Q, we get: $\Large Q = \frac{{{\nu _0}}}{{2\Delta \nu }} = \frac{{{\nu _0}}}{{2 \cdot \frac{{v_0^2 \cdot \Delta \lambda }}{c}}} = \frac{{{\lambda _0}}}{{2 \cdot \Delta \lambda }}$

5. Michele_Laino

and finally: $\Large \Delta \lambda = \frac{{{\lambda _0}}}{{2Q}}$

6. Michele_Laino

oops.. I have made a typo: please instead $\Large {v_0}$ read $\Large {\nu _0}$

7. anonymous

you smart huh @michele_Laino

8. UnkleRhaukus

Where do you get the factor of two from, (in the denominator)?

9. UnkleRhaukus

@Michele_Laino

10. Michele_Laino

I have used the definition which comes from the theory of resonant circuits