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I Helium Neon Laser

  1. Jul 25, 2018 #1
    I need help to understand the energy transfer mechanism in Helium Neon Gas laser. When helium atom is excited and staying in its metastable state, it is said to have collision with neon atom and thus transfers its 20.61eV energy plus some of its kinetic energy 0.05eV to enable neon atom to reach its metastabel state of 20.66eV. I can't understand this mechanism. To my understanding , as the helium atom decays it must release a photon of 20.61eV. This photon can't be absorbed by neon atom as it does not have such an energy level. So how is this energy transferred? Pl help.
  2. jcsd
  3. Jul 25, 2018 #2


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    When you really want to describe this process as a photon transfer process, you have to take in mind that the photon exchanged is not a real photon but a virtual photon. As this photon exists only for a very short time, due to time energy uncertainty relation, it's energy needs not to be exactly 20.61 eV. Another point is that even if a helium atom would emit a real photon of 20.61 eV, if the neon atom moves at a relative speed to the helium atom, it will see a Doppler shifted frequency corresponding to a different energy. This energy may well be 20.66 eV.
  4. Jul 26, 2018 #3
    Thanks for the reply> This explanation is still beyond my understanding. Pl try some simplistic view.
  5. Aug 10, 2018 at 4:21 PM #4
    The helium atom can't decay to ground state by way of photon emission because the transition is forbidden quantum mechanically. Helium's 1s2s excited state has an energy level that is very similar to neon's 1s22p55s state, and so it can transfer this energy through collisions with neon and return to ground state. The interaction has to be electron-electron and so it is described by quantum electrodynamics where electrons exchange energy in the form of virtual photons.

    What DrDu was describing is how the relative energies between the atoms vary due to their relative velocities. In a HeNe gas laser (and any system of gas in general), there is a large distribution of velocities for the gas particles. Think of the difference between a small rock hitting the windshield of your car when you're stationary versus when you're driving at highway speeds. There is a corresponding energy of the impact based on how fast you're moving and how fast the rock is moving. In quantum mechanics the energy to make the kind of transition mentioned above has to be exactly that energy, so you'll some rocks hitting too soft, some too hard, and hopefully most hitting just right to get more neon in the excited state than the ground state. I hope this makes sense for you, I tried including some keywords so that you can search further on your own if necessary.

  6. Aug 10, 2018 at 10:34 PM #5
    Thank you very much DrDu and Marisa5 for your kind reply. It has solved my problem.
    Kind regards. Best wishes for PF for being so helpful and supportive.
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