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Ionization energy and speed of electron

  1. Aug 27, 2006 #1
    A neon sign is a gas discharge tube in which electrons travelling from the cathode to anode collide with neon atoms in dicharge tube and knocks electrons off of them. As electrons return to the neon ions and drop to lower energy levels, light is given off. How fast would an electron have to be moving to eject an electron from an atom of neon,which has a first ionization energy of 2080KJ per mol.

    OK, I got 2 set of solutions, buI don't know which one of them is the correct one, maybe neither is correct, too.

    For solution number 1, I calculate the ionization energy for 1 atom of neon, by dividing 2080 by Avogadro number. Then, I get E. I obtain the mass of 1 atom of neon, by dividing its molar mass by Avogadro number, too. Using E= (1/2)(mv^2), i get the value of v=1.436x10^4 m/s.

    For solution number 2, i do the same thing, except for the value of m, I substitute in the m for an electron. Finally, I get v=2.754x10^6 m/s.

    Which one is the correct solution? Or neither? Thanks. :smile:
     
  2. jcsd
  3. Aug 28, 2006 #2

    Hootenanny

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    One solution is correct, the other is incorrect. Which do you think is which and why?
     
    Last edited: Aug 28, 2006
  4. Aug 28, 2006 #3
    I think the solution which I substitute the value of m equals the mass of an electron seems correct. My understanding of the situation of this problem is as such: the kinetic energy of the high-speed moving electron has been "changed" to the energy that is required to knock out an electron from a neon atom, which is later become te light energy.

    May I know where the electron that are being accelerated comes from? Is it from the neon atom itself? Or it is "already there" in the discharge tube? Thanks.
     
  5. Aug 29, 2006 #4

    Hootenanny

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    Yes, the latter answer is correct.
    A brief introduction from howstuffworks.com;
     
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