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Calculate the electron kinetic energy ejected from cathode?

  1. Jan 9, 2016 #1
    I would like to calculate the electron kinetic energy ejected from a cathode as a function of the electric field and pressure without known of the temperature and velocity.
     
  2. jcsd
  3. Jan 9, 2016 #2

    mfb

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    Thermal emission? There is no fixed energy, and the energy spectrum depends on the temperature of the cathode. In general, the energy is small (<1 eV) and often negligible.
     
  4. Jan 9, 2016 #3

    Astronuc

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    There would be a distribution of energies based on the electric field imposed on the thermal spectrum. If the imposed potential difference between cathode and anode is much greater than one or a few eV, then the thermal distribution could be ignored. Consider 100s of ev, or keV compared to eV.

    Some theory and applications. The first reference is perhaps more appropriate for the OP.
    http://uspas.fnal.gov/materials/10MIT/Lecture2_EmissionStatisticsCathodeEmittance_text.pdf
    http://uspas.fnal.gov/materials/12UTA/Lecture1.pdf

    ZapperZ could provide some helpful information.
     
  5. Jan 10, 2016 #4

    ZapperZ

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    This is extremely vague. I often wish that people post the complete context (i.e. what exactly are you calculating, or what experimental scenario is this being used for etc.) when they post something like this.

    First of all, there is no indication of the nature of the emission. Is this a thermionic emission? A photoemission?

    Secondly, how bit is the external field and what is its geometry? If you have an external field of the order of 100s or 1000s eV, does it really matter anymore what the initial energy of the emission was?

    Thirdly, the energy spectrum of the emitted electrons can be wide, depending on which type of emission (thermionic, photo..) and also the temperature or laser energy being used. A photoemission done on on a 5 eV work function photocathode and using a laser with energy 5.5 eV will give a very different energy spectrum of the photoelectrons when compared to using a laser of 10 eV.

    In other words, the original question to this thread cannot be answered.

    Zz.
     
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