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Natural broadening of lithium sub-levels

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data
    In the 2p state of lithium, two fine structure components are formed due to spin-orbit interaction. They produce wavelengths of 670.960nm and 670.975nm. Compare Doppler and natural broadening at 700K. The lifetime of the 2p state is 27ns and [itex]\frac{\Delta \lambda}{\lambda} ~= \frac{\Delta \nu}{\nu} [/itex] at [itex]\frac{\Delta \nu}{\nu} << 1[/itex]


    2. Relevant equations

    Given Doppler broadening: [itex]\Delta \nu = \frac{2 \nu_{0}}{c}\sqrt{\frac{2KT}{M}*ln(2)}[/itex]


    3. The attempt at a solution
    My Natural broadening derivation:
    uncertainty principle: [itex]\Delta E \tau = \hbar[/itex] (tau is lifetime)
    so [itex]\Delta \lambda = 2 \pi c \tau[/itex] (which comes out as ~50m)
    This is clearly wrong since the doppler broadening comes out as in the order of 10^-12.

    Any clues?
     
  2. jcsd
  3. Nov 11, 2012 #2

    mfb

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    How did you get ##\Delta \lambda## based on ##\Delta E##? That relation depends on the wavelength itself.
    It is easier to convert ##\Delta E## to ##\Delta \nu##.
     
  4. Nov 11, 2012 #3
    E=hc/λ
    and either way (nu or lambda) the answer is stupidly large
     
  5. Nov 11, 2012 #4

    mfb

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    E=hc/λ, but not ΔE=hc/Δλ!
    $$E+\Delta E=\frac{hc}{\lambda+\Delta \lambda} \approx \frac{hc}{\lambda}(1-\frac{\Delta \lambda}{\lambda})$$
    Simplified:
    $$\Delta E=-\Delta \lambda \frac{hc}{\lambda^2}$$

    It is easier to use relative values everywhere:
    $$\frac{\Delta E}{E}=\frac{\Delta \nu}{\nu}=-\frac{\Delta \lambda}{\lambda}$$
     
  6. Nov 11, 2012 #5
    Now it's so obvious!!
    Thanks a lot mfb!
     
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