Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is there a difference between ionization energy and ionization potential?

  1. Sep 28, 2011 #1
    Wikipedia says they are synonymous.

    Ionization energy is how much energy it takes to abstract an electron from an atom. A molecule with a low ionization energy can more easily be ionized. It seems weird to call this a low ionization potential though. You are trying to say it can easily be ionized, but you would call that a "low ionization potential"? That makes it sound like it ISN'T easy to ionize it.

    So, are they the same thing or are they opposites?

    Also, why does IE decrease down a group? The explanation given is that the electrons are further from the nucleus. But Z(effective) increases down a group. Who cares if they are further away if they are feeling more charge from the nucleus?
    Last edited: Sep 28, 2011
  2. jcsd
  3. Sep 28, 2011 #2
    Yes, but as you move down a group, the valence shell is now "shielded" in part by the electrons in the non-valence shells as well.
  4. Sep 28, 2011 #3
    I know there are core electrons that shield. But as you move down the core e- are not as effective at shielding due to increased core shell size. The effect is that Z* increases slightly down a group. My question is how can the electromagnetic force on an electron at that point be larger but it is also easier to pull that electron away?
  5. Sep 28, 2011 #4
    Ionization energy and ionization potential are used interchangeably in my experience. If you want to argue for differentiating between the two, be my guest, but chemists can be awfully slow in adopting new terminology standards (I still call ethene "ethylene," after all).

    I presume that you've seen the spherical harmonics plotted for the higher-energy shells. (They're on Wikipedia somewhere, in case you haven't - I remember seeing them not too long ago.) It's not just that they're farther from the nucleus in a general sense - the probability of finding an electron is noticeably smaller nearer the nucleus for, say, an electron in the sixth energy level, relative to the second energy level.

    There's probably also some benefit to reducing electron-electron repulsion by knocking out an electron further down the groups, although that's kind of a handwavy thing to say, and I can never remember just how well it holds for heavier elements in all cases.
  6. Sep 28, 2011 #5
    I understand this but I think the problem still stands. Trends in ionization energy should be directly related to Z*. It makes sense to do it this way, and it is how the trend across a period is explained (Z* inc. across a period, so it is harder to pull an electron off). I don't see how any sort of distance relation should matter.
  7. Sep 29, 2011 #6
    Look at it in the classical E&M sense. The force exerted on a charged particle from another charged particle is directly proportional to the product of the charges and inversely proportional to the square of the distance between those charges, so if distance increases, then the force exerted is less, and the electron is less bound to the nucleus.
  8. Sep 29, 2011 #7
    So Z* at one distance away has a different force exerted on it than the same Z* at a different distance?

    Ah ok, for some reason I was thinking that it just mattered how much charge it "felt", not how much it felt and how far away it was.
  9. Sep 30, 2011 #8
    As daveb mentioned, it's Coulomb's law all over again. The force a charge "feels" is explicitly dependent on the square of the distance. If you have a distance of 2 length units between charges, you'd divide the product of the charges by 4 (=22). If you have a distance of 4 units, you'd divide the product by 16 (=42).

    To take a simplified example (numbers from WebElements) = Z* for Be's valence electron is 1.91, while the valence shell orbital radius is 2.05 AU. For Ba, Z* is 7.6 and the radius is 4.45 AU. Let the product of the electron charge and the coefficient in Coulomb's law equal x. Do this simplified calculation given Coulomb's law, and for Be, you get ~ 0.45x, while for Ba you get ~ 0.38x - that's a 15% difference. The gap in their first ionization energies is greater than this, but chalk that up to a purely classical treatment.

    (I know I played a bit fast and loose with units in this post. No need to flagellate me over it.)
    Last edited: Sep 30, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook