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Principal quantum number n

  1. Aug 3, 2006 #1
    The larger n, the longer the orbitals, the higher the energy level, and the higher the speed of electrons?
     
    Last edited: Aug 4, 2006
  2. jcsd
  3. Aug 4, 2006 #2
    Everything sounds right, except the electrons move more slowly for larger n. Its easy to see this when you consider our solar system as a crude model for the atom. The further away an object is from the parent body (corresponding to larger n), the more slowly it moves.
     
  4. Aug 5, 2006 #3
    I don't understand. Here it says, "as you add energy to the electron, it will go faster." So is there a distinction between higher energy and higher energy level?
     
  5. Aug 5, 2006 #4

    ZapperZ

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    You are trying to use the same explanation for two DIFFERENT situations. You first asked about atomic orbitals. Now you are using particle accelerators. These two are not the same conditions and different rules apply.

    Electrons in atomic orbitals are strictly governed by quantum mechanical rules. When such thing kicks in, then our concept of "orbitals" and "speed", etc. needs to be redefined, especially in how we describe the motion of electrons in an atom.

    On the other hand, the JLab page you referred to is for FREE charges in which classical laws apply. The electrons are not bound in a potential that it can see (i.e. it doesn't see a potential boundary unlike the atomic case), and there aren't any wavefunction overlap between neighboring electrons. Classical physics kicks in, which is what most people use in describing charged particles in accelerators.

    Zz.
     
  6. Aug 5, 2006 #5

    jtbell

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    In a particle accelerator, all the energy that you add to an electron goes into its kinetic energy, and therefore its speed increases. In an atom, the energy that you add to a bound electron goes into increasing its potential energy, not its kinetic energy.
     
  7. Aug 5, 2006 #6
    what is this potential energy like?
     
  8. Aug 5, 2006 #7

    jtbell

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    It's the potential energy associated with the electrical attraction force between the electron and the nucleus:

    [tex]F = \frac{k q_e q_{nuc}}{r^2}[/tex]

    [tex]PE = - \frac{k q_e q_{nuc}}{r}[/tex]

    As r increases, PE increases.
     
  9. Aug 6, 2006 #8
    how does PE increase? the negative sign shows that increasing r relates to an increasingly negative PE... :yuck:
     
  10. Aug 6, 2006 #9
    It depends on what you define as zero potential energy. When the electron is located at a point infinitely far away it's potential energy is defined to be zero and the electron is free and not bound by the nucleus. As the electron moves closer to the nucleus it loses potential energy (the potential energy becomes more and more negative), accounting for the negative sign.
     
  11. Aug 6, 2006 #10

    jtbell

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    Look at it more closely. As r increases, 1/r decreases, and -1/r increases (towards zero as a limit).

    Or if you're having problems with the words, calculate a simple example. Let r increase from 10 to 20. Calculate -1/10 and -1/20, and plot them on the vertical axis of a graph. Which one is higher?
     
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