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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:

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  1. UnkleRhaukus
    • one year ago
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    \[ \newcommand\metre{\text m} \newcommand\U[1]{[#1]} \newcommand\E[1]{\times10^{#1}} \boxed{\ast}\\[3ex] Q %= \frac{\omega}{\Delta\omega}\\ %= \frac{\omega}{\Gamma}\\ %= \frac{\nu}{\Delta\nu}\\ = \frac{\lambda}{\Delta\lambda}\\[3ex] \Delta\lambda\ \ \ = \frac\lambda Q\\ \qquad= \frac{590\E{-9}\U{\metre}}{8.0\E7}\\ % 8.428571429e-18 \qquad= 8.4\E{-16}\U\metre \]

  2. UnkleRhaukus
    • one year ago
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    is this right?

  3. Michele_Laino
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    and finally: \[\Large \Delta \lambda = \frac{{{\lambda _0}}}{{2Q}}\]

  6. Michele_Laino
    • one year ago
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    oops.. I have made a typo: please instead \[\Large {v_0}\] read \[\Large {\nu _0}\]

  7. anonymous
    • one year ago
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    you smart huh @michele_Laino

  8. UnkleRhaukus
    • one year ago
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    Where do you get the factor of two from, (in the denominator)?

  9. UnkleRhaukus
    • one year ago
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    @Michele_Laino

  10. Michele_Laino
    • one year ago
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    I have used the definition which comes from the theory of resonant circuits

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