A community for students.
Here's the question you clicked on:
 0 viewing
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:
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:

This Question is Closed

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[ \newcommand\metre{\text m} \newcommand\U[1]{[#1]} \newcommand\E[1]{\times10^{#1}} \boxed{\ast}\\[3ex] Q %= \frac{\omega}{\Delta\omega}\\ %= \frac{\omega}{\Gamma}\\ %= \frac{\nu}{\Delta\nu}\\ = \frac{\lambda}{\Delta\lambda}\\[3ex] \Delta\lambda\ \ \ = \frac\lambda Q\\ \qquad= \frac{590\E{9}\U{\metre}}{8.0\E7}\\ % 8.428571429e18 \qquad= 8.4\E{16}\U\metre \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1from 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: \[\large \Delta \lambda = \frac{c}{{{\nu _1}}}  \frac{c}{{{\nu _2}}} = c\frac{{{\nu _2}  {\nu _1}}}{{{\nu _1} \cdot {\nu _2}}} \cong c\frac{{\Delta \nu }}{{v_0^2}} \Rightarrow \Delta \nu = \frac{{v_0^2 \cdot \Delta \lambda }}{c}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops.. 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: \[\Large Q = \frac{{{\nu _0}}}{{2\Delta \nu }} = \frac{{{\nu _0}}}{{2 \cdot \frac{{v_0^2 \cdot \Delta \lambda }}{c}}} = \frac{{{\lambda _0}}}{{2 \cdot \Delta \lambda }}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and finally: \[\Large \Delta \lambda = \frac{{{\lambda _0}}}{{2Q}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops.. I have made a typo: please instead \[\Large {v_0}\] read \[\Large {\nu _0}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you smart huh @michele_Laino

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0Where do you get the factor of two from, (in the denominator)?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I have used the definition which comes from the theory of resonant circuits
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.