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Question about electrons, energy levels and orbits

  1. Feb 20, 2009 #1
    To make my question clear, i'll talk about the hydrogen atom. The energy of the electron level is -13.6 eV / n^2. This means that the higher the orbit (n), the higher the energy. Still, i've always read that K-shell (n=1) electrons are strongest bound. Yet when an electron goes from a high level to a lower one, a photon is released.

    Maybe the minus sign confuses me? But an electron with n=5, as rare is it may be, would have an energy of -0.5 eV, while one in the ground state has -13.6 eV. So, you'd think the n=1 state is more tightly bound, but if that is the case, then it would seem logic for an electron to REQUIRE energy (say, a photon) to go down to lower n-state, but instead, it releases energy. How does it work?
     
  2. jcsd
  3. Feb 20, 2009 #2

    Doc Al

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    Staff: Mentor

    An electron in the ground state would require 13.6 eV to knock it out of the atom, whereas an electron with n=5 only requires 0.5 eV to be knocked out. The lower the energy of the electron, the harder it is to knock out. (Relate this to mechanical stability--generally lower energy positions are more stable.)

    The greater the n value, the greater the energy. When an electron drops to a lower n-state, it releases energy.

    It's probably the negative sign throwing you off. When the electron is "infinitely" far from the nucleus, the energy is nominally called "zero". But of course the negative electron is attracted to the positive nucleus, thus closer positions have negative energy.
     
  4. Feb 20, 2009 #3
    Thanks, doc. I still don't quite get how the following two statements are congruent:

    1. The ground state has the highest bound electrons, i.e. it requires the maximum amount of energy to free them from the atom.

    2. When an electron demotes from a higher state to a lower one, energy is RELEASED, yet it results in a higher bound state.


    It sounds like "falling" to the top of a mountain, if you get the analogy.
     
  5. Feb 20, 2009 #4

    Doc Al

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    Think of the ground state as being at the bottom of the valley. To free them from the atom, electrons in the valley need to be brought up to the top of the mountain, which requires giving them energy. (The term "highest bound" might be confusing you. More tightly bound means being lower in the valley with less energy.)

    Sure. You've dropped from the higher-energy mountain top to the lower-energy valley, thus releasing the extra energy. They are now in the lower-energy, more tightly bound state.

    The analogy is fine, but you have it backwards. Think of the side of the mountain being labeled with n-values. n = 1 in the valley, n = 2 a bit higher, all the way to the top of the mountain.

    (Since the top of the mountain is labeled zero energy, the low-energy valley must have negative energy.)
     
  6. Feb 20, 2009 #5

    jtbell

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    Or better, think of a deep pit in the ground, and define the (potential) energy to be zero at ground level. An object down inside the pit has negative potential energy. The deeper it is, the lower (more negative) the potential energy is. If an object falls into the pit, the deeper it goes, the more energy it "releases" on the way down. Conversely, if the object is already in the pit, the deeper it is, the more energy you have to supply, in order to get it out.

    For a hydrogen atom, the potential energy "pit" is infinitely deep, but there are "platforms" at fixed levels, with the "lowest" ("deepest") one being the ground state.
     
  7. Feb 20, 2009 #6
    Thanks a lot, i get it now. The valley and pit examples made it all clear.:approve:
     
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