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Ionization Potential and Work Function

  1. Jan 26, 2005 #1
    Do Ionization Potential values include or exclude the work function of the electron removal process? For example: The IP value for hydrogen is 13.6 eV.

    Is the true IP 13.6 eV or 13.6 - WF (ca. 4 eV) = 9.6 eV roughly?
  2. jcsd
  3. Jan 26, 2005 #2
    No they don't include the work function.

    Ionization potential is defined as the energy required to remove the outermost electron from an isolated atom in gaseous state.

    An isolated atom in gaseous state does not have any work function.

    Work function arises from other factors including the crystal structure etc. etc.
  4. Jan 27, 2005 #3
    Does this mean that all of the IPs listed for all of the elements were measured when the elements were in a gaseous state? If so, what temperature would be used?
  5. Jan 27, 2005 #4
    according to quantum theory, since the energy is quantize, the result is independent to the temperature...

    let me do a very simple calculation for H atom
    [tex] KE \approx kT = 8.617 \times 10^{-5} T (in Kelvin)eV [/tex]

    in a room temperature, T=300 degree, KE approximately equal to 10^-2 eV, only a fraction of IP (13.6 eV in your case)... so even if you use classical theory... the error in minimal.....
  6. Jan 27, 2005 #5
    Though IPs are related to atoms in the gaseous state, they ar NOT usually measured in the gaseous state. The old method is to use Hess' Law of Constant Summation, one of the great applications of which is the Born Haber Cycle. Modern methods include sophisticated equipments and spectroscopy etc. etc.
    Ionization Potential is DEFINED as a quantity independent of such things as the temperature. It's like, you consider just an atom with a nucleus and the electrons and forget about what's going on outside it. This definition does create problems because in the laboratory, your apparatus cannot be independent of such other factors. To attain a high accuracy in the measurement of IP is still a challenge.
  7. Jan 27, 2005 #6
    OK. Got it. Many thanks for the assist.

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