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How does QM solve the bohr atom problem?

  1. May 19, 2005 #1
    I have only taken QM for a semester so I'm sure there are a lot of gaps in my understanding. Here is my question: The original electron orbiting around the nucleus like planets around the sun model was rejected partly because the radiation (energy loss) was not taken into consideration. I thought the new QM model was supposed to fix it. However with the hydrogen atom we learned, there are still these stationary eigenstates of energy. Even though we don't have a point electron anymore (replaced with a wave of probabibility), doesn't the electron still have angular momentum/acceleration/etc which allows it to radiate? If this is true, then how can there be any stationary states? Why don't the solutions to the hydrogen atom we learn involve a decay in energy? Thanks a lot in advance!!!!!
     
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  3. May 19, 2005 #2

    Tom Mattson

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    Hi Welkin,

    The prediction of radiation from electron orbits comes from the classical dynamics of charged particles. If you start from quantum mechanics instead of classical electrodynamics you don't even encounter the problem. So I suppose the answer to your question is that QM "fixes" the problem by making it a non-issue.
     
  4. May 19, 2005 #3
    Hmm... does that mean there is no radiation from the electron in the QM model?
    And from that shall one draw the conclusion that the effective magnetic field created by the electron does not change?
    Somewhere I must be confused... v_v
     
  5. May 19, 2005 #4

    dextercioby

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    1.Bohr-Sommerfeld's theory of the H-atom is called "old QM"."New" (1925-->) QM doesn't "solve the problems of the H atom".

    2.We DO have a pointlike electron.

    3.Measurable quantities for the H atom in a stationary state are time independent.

    Daniel.
     
  6. May 19, 2005 #5

    dextercioby

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    Unless the Hamiltonian is perturbed,no.


    Guess what ?It doesn't.But spin-orbit coupling is a relativistic effect.

    Daniel.
     
  7. May 19, 2005 #6
    isn't all magneticism relativistic effect anyway?
    so how does the magnetic field from the Bohr model differ from the one used in QM?
     
  8. May 19, 2005 #7

    dextercioby

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    Magnetism derives from angular momentum,not necessarily spin (orbital,too).So it's not a relativistic effect.Pauli's Hamiltonian can accomodate it pretty niftly.

    What magnetic field in the Bohr model ...?

    Daniel.
     
  9. May 20, 2005 #8
    Hydrogen Atom Stability

    Excellent question! QM avoids the issue by discounting classical physics and Maxwell's predictions, which is an extraordinary claim which requires convincing evidence. I suggest a new theory is needed to explain the stability of the hydrogen atom. Continue to ponder this question and you will eventually resolve this issue.
     
  10. May 20, 2005 #9
    Did I misunderstand you or do you literally mean what you're saying here?
     
  11. May 20, 2005 #10

    arivero

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    How does QM solve the photoelectric effect and the question of a cut-off in light intensity?
     
  12. May 20, 2005 #11

    dextercioby

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    All possible electron-photon interactions are handled remarkably by QED...

    Daniel.
     
  13. May 20, 2005 #12
    This is a classic one. Conform classical physics, stable atoms would not exist because the electron would radiate; lose energy and spiral inward towards the nucleus. Eventually the electron will crash onto the nucleus. What does this tell you ? Well, the fact that classical physics is NOT able to explain atomic scaled phenomena. This is the biggest thing that you need to realize. Then, mr Bohr came with his electrons moving on shells. The angular momentum is quantized as well as the energy. When an electron is on a certain shell, it does not radiate, it has one constant energy. Radiation only occurs when the electron goes from one energy level to the other (ie absorption or emission of EM-radiation)...Then, the wavefunctions came and the particle-wave duality. Electrons do not move on definite circular orbits but we can only talk about a region in space where we have a certain probability of finding an electron. These are the orbitals...

    regards
    marlon
     
  14. May 20, 2005 #13
    I am utterly confused. So I'll just ask one more time and if I can't get it, then so be it... v_v

    So... even though we only have probability when we're dealing with QM, the probability cloud still moves, and the electron would still have some expectation value for velocity/momentum/angular momentum, etc... so why wouldn't this produce a current and changing magnetic field --> radiation?
     
  15. May 20, 2005 #14

    jtbell

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    No. The probability distribution for the electron for a hydrogen atom in the ground state, for example, is

    [tex]P(r) = \frac {1}{\pi a_0^3}e^{-2r/a_0}[/tex]

    where [itex]a_0[/itex] is the radius of the ground-state orbit in the Bohr model. (Of course, there's no such thing as an "orbit" here, but it's a convenient constant for simplifying formulas.)

    Note there's no time dependence, so the "probability cloud" doesn't move. or change its shape. Nor do any of the other energy eigenstates have any time dependence in their probablity distributions.

    However, if you describe an atom that is in the process of making a transition between two energy eigenstates, by a wave function that is a linear combination (a "mixture," so to speak) of the two states, that wave function is not a stationary state. The probabilty distribution for such an "in-between" state does have a time dependence. In fact, it oscillates in shape, with a frequency equal to that of the photon that is emitted in the transition!
     
  16. May 21, 2005 #15
    The basic QM premise regarding electron energies within the atom are that the electron may possess energy [tex]E_{1}[/tex] or energy [tex]E_{2}[/tex] but no intermediate energies – all values of energy E satisfying the condition [tex]E_{1}< E < E< E_{2}[/tex] are forbidden. This caused problems : With regard to the jump of an electron from one orbit in the atom to another. Whatever the speed of the transition it has to last for some finite period of time (otherwise it would be a violation of the basic requirements of the theory of relativity ). But then it is hard to understand what the energy of the electron should be doing during this intermediate period – the electron no longer occupies the oribit corresponding to [tex]E_{1}[/tex] and has not yet arrived at the orbit corresponding to [tex]E_{2}[/tex] . This caused Schrodinger , the originator of the probability wave to make his famous remark about ”these damned quanum jumps”
    The problem was eventually solved by resorting to the Heisenberg Uncertainty relation [tex] \Delta {E}\Delta {t}\leq {h} [/tex]. This means that the possibility exists for virtual transitions to take place. According to the Uncertainty relation it is possible for the electron to go from[tex] \E_1 to \E_2[/tex], without getting any external energy , provided it immediately returns to[tex] \E_1[/tex]. Such a journey is possible if its duration \Delta{t} is such that the inequality[tex] h/\Delta {t} > \E_{1} - \E_{2} [/tex], because in this case the uncertainty in the energy of the electron is greater than the difference in the energies of the levels under consideration.
     
  17. May 26, 2005 #16
    You did not misunderstand me. The hydrogen atom's stabililty must be explained based on sound physics. This new theory should be as simple as the Bohr Atom but takes into account the wave properties of the electron.
     
  18. May 26, 2005 #17
    Because Heisenberg's uncertainty principle trumps Maxwell's equations.

    Electrons in atoms don't continuously emit radiation (even though Maxwell's equations say they should) because quantum theory (which is more accurate than classical EM theory) says they shouldn't.
     
  19. May 26, 2005 #18
    One should simply stop trying to apply classical conceps to quantum systems, because
    it doesn't work. The only definite thing that an electron has is a wavefunction, and from that one finds observables. The magnitude squared (|psi|^2)of the wavefunction gives it's probability of being found at a given position, and other operators give other things. In a stationary state, the expectation value of energy has no time dependence, so it's energy stays constant in time. Also, in a stationary state, although the wave function may have time depedence, the magnitude squared does not, and so the probability cloud doesn't move.

    This isn't true of all states which solve schrodingers equation, but as long as the hamiltonian (operator which describes energy) is independent of time, so too will the probability cloud, and so the state doesn't change with time.
     
  20. May 31, 2005 #19
    forgive me if i'm wrong (i'm still a junior in high school w/ only physics B under my belt..)

    i thought the problem was resolved through wave duality?

    because if we consider the electron to be a wave.. then it must go around the center of its orbit with a circumference = n * wavelength... where n = any integer > 0 (eventually became the principal quantum level)...

    thus the orbits are quantisized and the wave cannot travel at any other circumference (and thus radius) otherwise it will destructively interfere with itself and kill itself off...
     
  21. May 31, 2005 #20

    selfAdjoint

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    This was DeBroglie's reasoning, and you have expressed it well. It is essentially correct, and he won a Nobel Prize for it, but note that it assumes one fixed wave length at a time. What if the wave length can change? what if it is dynamic , meaning it depends on forces acting in time, what happens then? This is the problem that Schroedinger set himself. To represent dynamic quantities, physicists use differential equations, equations between rates of change. And Schroedinger derived his famous Schroedinger equation for a particle with a variable, dynamic, wave length.

    Thus the reasoning you have seen here doesn't really contradict DeBroglie, it more accurately extends it to the general case.
     
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