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Emission spectrum wavelengths.

  1. Nov 21, 2013 #1
    Something that bugged me when doing this lab. Standard little glass vials of gas, toss it in a 5kV potential make pretty color, look through diffraction grating see the individual wavelengths that are the finger prints of the element.

    My question however are all wavelengths equally represented? I'm presuming that there's a rather large number of gas atoms even in a thin glass tube, but for something like hydrogen where the Balmer series has 4 colors, are there an equal number of photons represented at each color when you take the average of the "large" number of transitions occurring? Or are some colors more likely than others?

    I'm trying to get my finger on the pulse of the truth behind why some colors look brighter, my initial thought is that our eye sensitivity to certain wavelengths is the sole factor in determining brightness, but now I'm wondering if some colors simply get more photons emitted on average than others.
     
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  3. Nov 21, 2013 #2

    TumblingDice

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    Each line is created by photons emitted by electrons dropping from higher energy levels to lower levels. Contributions to specific lines will depend on the electron levels during the measurement.

    The relative brightness of the lines will depend on the distribution of the energy levels of the electrons emitting photons. In the case of your hot gas emission, the amount of energy you introduce to the cloud is what raises the electrons to higher levels, after which they drop and emit photons of discrete energy levels (frequency) based on the starting and ending energy levels. I imagine you could alter the representation of a measurement by adjusting the amount of energy you introduce or the length of time you spread it over.

    Brightness of lines also varies with the abundance of atoms. So atomic emission spectra like this can be used to not only determine the presence of a gas, but also information to help determine the density and temperature.
     
  4. Nov 22, 2013 #3
    Yes but what I'm asking for, using Balmer as an example, if you put in enough energy to raise electrons to n=5, is there anything ratio or percentage change that the electron will drop directly to n=2 or go through a n=4 or 3 before dropping to 2.

    Basically, there are 4 visible colors with hydrogen. If you have N particles in the gas, will you get N/4 photons of each color per emission? Or are some transitions more likely than others?
     
  5. Nov 22, 2013 #4

    Bill_K

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    Some are more likely than others. The line intensities derive from the "oscillator strengths". Allowed transitions are electric dipole transitions, in which going from one state |n, , m> to another state |n', ', m'>, the selection rules are m = m' and = ' ± 1.

    What remains is to calculate the off-diagonal matrix elements of r, that is <n'|r|n> and square it. Thus it's an integral over a power of r and the product of two Laguerre polynomials.

    You can get the results in closed form (it's rather complicated!) or look the values up in a table. :wink: The details can be found for example in Bethe and Salpeter, "QM of One and Two Electron Systems" p 262.
     
    Last edited: Nov 22, 2013
  6. Nov 22, 2013 #5

    DrClaude

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    That's actually ##\Delta m \equiv m- m' = 0, \pm1##.
     
  7. Nov 22, 2013 #6

    Bill_K

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    Thanks, you're right!
     
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