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Classically predicted ionisation energy of H

  1. Nov 25, 2011 #1
    So, as a matter of interest, I tried predicting the ionisation energy of a Hydrogen atom according to classical electrostatics. I started from the point of escape velocity, with
    Ve=√(2kq1/r)

    The following bit I think is wrong, but fetches an interesting result.
    As E=1/2mv2, the analogous equation for electrostatics (with m=q) is E=1/2qv2, so
    Ee=kq1q2/r

    The charge on the electron and proton is e, (1.6x10-19).
    Ee=ke2/r
    To convert this from joules to electronvolts
    Ee=(ke2/r)/e
    so
    Ee=ke/r
    So using k=8.99x109, e=1.6x10-19, and the Bohr radius of r= 5.3x10-11
    Ie=27.14eVe-1
    Ie=27.14eVe-1

    The experimental value of the ionisation of H is Ie = 1312kJ.mol-1,
    Ie = 1.312x107J.mol-1
    So convert to J.e-1 (per electron, not the inverse of the charge of an electron)
    Ie = 1.312x107J.mol-1*mol
    Ie = 1.312x107J.mol-1/6.022x1023
    Ie = 2.18x10-17J.e-1

    So to convert from J to eV
    Ie = 2.18x10-17J.e-1/e
    Ie = 2.18x10-17J.e-1/1.6x10-19
    Ie = 136.17eVe-1

    This experimental value is ridiculous as an ionisation value, so I imagine I have forgotten something. Interestingly though, the value I have found from various sources is 13.6eVe-1, exactly 1 decimal place shifted. I will assume then that the experimental value is 13.6, and I have messed up somewhere.

    Interestingly, the correct experimental of 13.6 is nearly EXACTLY half of 27.14. Is this a coincidence, or does it imply that Ie=ke/(2r)?
    And as F=kq1q2/r2, does this then mean that
    Ie=(1/2)Fr?
    Perhaps this is an equation I do not know of?

    Please feel free to pick apart my logic and findings - it seems a bit simple. ^^
     
  2. jcsd
  3. Nov 25, 2011 #2
    Found my mistake with the experimental value - Ie = 1.312x106J.mol-1, not Ie = 1.312x107J.mol-1. Ie = 13.6eVe-1
     
  4. Nov 25, 2011 #3
    If you're using the Bohr radius, then your calculation is not classical electromagnetism.

    "Classically", of course, the accelerating electron emits radiation continuously and does a death spiral into the nucleus pretty quickly....

    BBB
     
  5. Nov 25, 2011 #4
    Ahhh! How silly of me.

    I'm aware classical physics couldn't explain it, I was merely curious as to the difference predicted in this way.

    Are my calculations incorrect then?
     
  6. Nov 27, 2011 #5

    edguy99

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    Your calculations are correct, at the bohr radius of 53 picometers, the coulomb force amounts to 13.6 evolts. I also find it more than a coincidence that the binding energy of a hydrogen proton and electron never gets past this amount.

    It means that proton is acting like a large shell (think of a chemistry molecule, hydrogen is usally drawn using the bohr radius). Once an electron is inside this shell, it no longer feels a force from the proton and becomes free floating somewhere inside the bohr radius.

    This kind of model also addresses the concern of bbbeard, related to the death spiral. Once inside the shell, the electron no longer feels a force from the proton, hence does not accelerate and emit radiation continuously. The falling electron only emits once when it passes the bohr radius at a specific wavelength, or is bent enough to emit a Bremsstrahlung (braking radiation) photon at various wavelengths on the way by.
     
  7. Nov 27, 2011 #6
    The coulomb (or other) force is not measured in eV. The 13.6 eV is the binding energy and not the force. So the next sentence does not make much sense.
    And the following ones even less.
     
  8. Nov 27, 2011 #7
    As far as I can see, Nasu, that is his only slip up. The following sentence DOES refer to binging energy, and make perfect sense...
     
  9. Nov 28, 2011 #8
    I think most of the quantum physics developed since Bohr has demonstrated the inadequacy of these kind of models.

    In non-relativistic quantum mechanics (the kind governed by the Schrodinger equation), the potential to which the electron is subject is the Coulomb potential ([itex]\propto -1/r[/itex]). There is no "shell" inside of which the potential is constant (i.e. with zero force). The Bohr radius is the length scale in the radial direction, but nothing magic happens to the potential at the Bohr radius.
     
  10. Nov 28, 2011 #9
    You mean this:
    "I also find it more than a coincidence that the binding energy of a hydrogen proton and electron never gets past this amount"?
    Once we agree that the 13.6 eV is the binding energy (and not a force), the above quoted sentence states that is more than a coincidence that the binding energy never gets past the value of the binding energy. Would you call this a "coincidence"?

    The rest about shells and so on is a mixture of confusion and incorrect models, as pointed already by bbbeard.
     
  11. Nov 28, 2011 #10

    edguy99

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    The above words (amounts to) were used to indicate the OP was correct in assuming the classically predicted calculations used were indeed correct and lead to the answer of 13.6 evolts.

    The classical model fails at the bohr radius if you let the binding energy (and speed and acceleration..) increase to infinity as the electron gets closer and closer to the proton. If you assume the electron no longer feels a force within the Bohr radius, the failure quoted by bbbeard no longer fails in the classical model.

    A 3d game along this line (it assumes normal coulomb forces up to the bohr radius and no force once inside the bohr radius) can be found here.

    Assume you are a bug with antenna 100 picometers apart looking at a classical model of hydrogen. You shoot tiny red dots (electrons going at 5 picometers/attosecond) at a large shell (53 picometers - proton) and try and dislodge the electron already inside the shell.
     
  12. Nov 28, 2011 #11
    The OP "derivation" is not valid, even in the classical frame.
    Playing with formulas may gave some numbers. The binding energy is not equal to the electrostatic energy at the Bohr radius.

    But this is not the point I was talking about. It was about the "coincidence" mentioned in this paragraph:

    "Your calculations are correct, at the bohr radius of 53 picometers, the coulomb force amounts to 13.6 evolts. I also find it more than a coincidence that the binding energy of a hydrogen proton and electron never gets past this amount."
     
  13. Nov 28, 2011 #12

    edguy99

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    I will try and use the words "calculations involving the coulomb force" rather then "coulomb force" in isolation.

    A couple of questions to help clarify:

    Are you suggesting the binding energy will exceed 13.6 evolts at times? Do you agree that the calculations involving the coulomb force at 53 picometers works out to 13.6 evolts?
     
  14. Nov 28, 2011 #13
    No, I am not suggesting it. And I don't think it makes sense. What do you mean by "at times"? Do you consider the binding energy as a dynamic variable, which is function of time?

    The semi-classical calculations based on Coulomb force predict these values for ionization energy and radius of the classical orbit (in ground state).
     
  15. Nov 28, 2011 #14
    You have several misunderstandings here. First, what you are doing is analogous to what Bohr did when he proposed his explanation of the spectral lines of hydrogenic atoms. That is, his "old quantum theory" was a semi-classical model that included one key assumption: that the orbital angular momentum of the electron was an integral multiple of Planck's constant over 2π, i.e.

    [itex]mvr=n\hbar[/itex]

    By starting in the middle of Bohr's calculation, using the value of the Bohr radius derived from the quantization condition (by equating the centripetal force with the Coulomb force [itex]mv^2/r=e^2/r^2[/itex]), what you are finding is not the "classically predicted ionisation energy" at all -- you are finding the ionization energy predicted by the Bohr model.

    Second, you are not calculating the ionization energy correctly. I have a hard time following your reasoning, but it appears that you forgot that an orbiting electron has kinetic energy. The energy of the electron when it is in orbital equilibrium at radius r is

    [itex]E=\frac{1}{2}mv^2-\frac{e^2}{r}=-\frac{e^2}{2r}[/itex]

    which is where your factor of 1/2 came from.

    BBB
     
  16. Nov 28, 2011 #15

    edguy99

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    I agree with everything you have said here. One of the things that caught my eye on the original post is that he started with escape velocity. I have assumed (although I never actually checked) that if an electron falls from very far from a proton, that when it passes the bohr radius of the proton, it will have 13.6 evolts worth of energy and that conversely, to shoot an electron from the bohr radius of a proton out to infinity you would need to start with 13.6 evolts worth of energy.

    In other words, he is not so much doing the calculation starting from a circular orbit of an electron, but rather from a stationary electron sitting 53 picometers from the proton.
     
  17. Nov 29, 2011 #16
    Thanks for the criticism guys (no sarcasm).
    I am aware that the model is incorrect. I don't know quantum physics, and I was more just curious as to what the results would fetch. As it was, it was a very interesting result, and you guys have helped me learn through it as well.

    I have only completed high school physics at this point, thus my lack of knowledge of quantum physics. I assume you that, by the end of my first year of uni (next year), I should have a more firm grasp of the topic.

    Cheers all, Kael.
     
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