Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electromagnetic radiations in bohr model

  1. Feb 9, 2012 #1
    Bohr model had a flaw that electrons would emit EM radiations and would spiral into the nucleus - but he just said that it wouldnt happen providing no reasons

    Do the modern models of atom explain - why electrons wont emit EM radiations?
     
  2. jcsd
  3. Feb 9, 2012 #2

    jedishrfu

    Staff: Mentor

    Bohr put a stake in the ground and derived his theory from that. It got others thinking about the idea of treating the electron as a wave and it's orbit about the nucleus as a standing wave brought forth Quantum Mechanics.

    That was Bohr's genius.
     
  4. Feb 9, 2012 #3
    That was Bohr's nonsense, not genius!!
     
  5. Feb 9, 2012 #4

    jedishrfu

    Staff: Mentor

    Please explain your reasoning. By your thinking you'd say that Faraday's lines of force were nonsense too yet it inspired Maxwell to develop EM theory.
     
  6. Feb 9, 2012 #5
    Hey my knowledge does go that far in quantum mechanics so plz if any1 can help - why dont electrons emit EM radiations - is it just one of those things whose answer is - "It just doesnt happen" with no reason or does it have a reason
     
  7. Feb 9, 2012 #6

    jedishrfu

    Staff: Mentor

    When the electron is in orbit about the nucleus it doesn't mean it orbiting like a planet orbits the sun. It's in a state where it's acting like a wave and not a particle like a cloud about the nucleus so it's not a moving electric charge and hence doesn't emit EM radiation.
     
  8. Feb 9, 2012 #7
    ok assume we have some electrons emitted from an atom -so now they are not around the nucleus anymore - we accelerate them - so according to you it will have some wave nature and so it will not emit EM radiations - but it does right?
     
  9. Feb 9, 2012 #8

    jedishrfu

    Staff: Mentor

    I didn't say that. A freely moving charge will emit EM but the electron isn't moving around the nucleus. It's not a planet orbiting the nucleus. If a photon of the right energy hits the electron it will be knocked free. You can't extend classical physics and EM theory into this realm. That's the whole point that Bohr was trying to make. The electron cloud about the nucleus is a probability function saying the electron is somewhere nearby but we can't know where it is exactly.
     
  10. Feb 10, 2012 #9
    ok thanx
     
  11. Feb 10, 2012 #10

    Jano L.

    User Avatar
    Gold Member

    Hello everybody,
    If I may do so, I would like to revive the discussion. It is a very interesting question.

    In non-relativistic quantum theory, the hydrogen atom does not radiate electromagnetic waves for the same reason as in Bohr's model: electromagnetic radiation is neglected and forgotten right from the start. The Hamiltonian operator is derived from the Hamilton's function of a particle in a static force field. It is then no surprise that there is no radiation, no spontaneous emission of radiation and no problem with the electron falling into the nucleus.

    Experimentally, it is I think hard to observe whether the atomic hydrogen is stable. But the gas of hydrogen molecules is stable in certain temperature interval and can be in thermal equilibrium, which always goes with presence of the thermal radiation. It follows that in fact hydrogen does radiate even when it is in a macroscopically stable equilibrium state. It does not lose energy though, because the surroundings radiate too and on the time average, the molecules lose as much energy as they absorb.

    An interesting question is, what would happen in a more general quantum theory that would take into account Maxwell's equations of the electromagnetic field? Would the hydrogen atom still be stable in such a theory? Is there any such calculation, say, in quantum electrodynamics?
     
  12. Feb 10, 2012 #11

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    The Hamiltonian is simply H=p^2/2m-e^2/r. What's the problem with this Hamiltonian? It hasn't neglected anything, it simply just assumes a universe with 1 proton (at the origin) and 1 much lighter electron.

    What do you mean by "electromagnetic radiation is neglected and forgotten right from the start"? Which terms have I thrown away?
     
  13. Feb 10, 2012 #12

    Jano L.

    User Avatar
    Gold Member

    Dear Matterwave,

    the problem with Hamilton's function p^2/2m + Kq^2/r is that it describes a particle in a static electric field of a proton.

    However, proton can move, so its electromagnetic field hardly can be static.

    If the simultaneity was absolute and the changes in electric force propagated instantaneously, this would not invalidate the Hamiltonian, because it would be possible to introduce reduced mass and introduce coordinates of electron _relative_ to the proton.

    This is indeed done in more advanced discussions of the hydrogen atom in both classical and quantum theory. But it is only a non-relativistic approximation. In this approximation, there is no electromagnetic radiation.

    However, if we want to be more precise, relativity requires that changes in proton's electromagnetic field take some time to reach the electron, which means the force on the electron should be given by a more complicated electromagnetic field of the proton (say, the retarded field of proton).

    Due to the finite velocity of propagation, the force is a time dependent function of the motion of the proton. It is not given as a gradient of a function of coordinates and for this reason the whole approach based on Hamilton's function breaks down.

    Perhaps it would be possible to maintain Hamilton's picture by paying the price that the coordinates and momenta of the field were included, but I do not know how to introduce them in a mathematically correct way. I wonder whether there is such a thing as exact Hamiltonian in QED. As far as I know, there is none in the classical theory.
     
  14. Feb 10, 2012 #13

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    You're in the rest frame of the proton. As long as you don't consider external forces, this frame will be inertial. You get EM radiation when you consider time-dependent perturbations to this system. You get results such as the Fermi's golden rule in TDPT, I don't see what's wrong with this.

    One can consider relativistic effects, all this does is lead to the fine structure and hyperfine structure of hydrogen.

    If you are talking about quantizing the E&M field itself, then I'm not entirely sure on the specifics of that. I hear that QFT is not trivial to apply to bound states.
     
  15. Feb 10, 2012 #14

    Jano L.

    User Avatar
    Gold Member

    I would like to comment on your claims one by one:

    "You're in the rest frame of the proton. As long as you don't consider external forces, this frame will be inertial. "

    I do not think the rest frame of the proton can be considered strictly inertial, because the proton is not an isolated particle. From the law of conservation of momentum, it is necessary that the electron acts on the proton.

    "You get EM radiation when you consider time-dependent perturbations to this system. You get results such as the Fermi's golden rule in TDPT, I don't see what's wrong with this."

    I understand you probably mean that the effect of _external_ electromagnetic radiation on the system can be described by a time-dependent addition to the Hamiltonian. I agree. I see nothing wrong with this.

    But the original question was concerned with the behaviour and the radiation of the hydrogen atom itself, not under with the action of external radiation on the atom. In the Hamiltonian model of an isolated hydrogen atom, there is nothing that would correspond to electromagnetic radiation.

    "One can consider relativistic effects, all this does is lead to the fine structure and hyperfine structure of hydrogen."

    This is only partially true. Standard textbook calculations consider relativistic effects only in a very limited way. They use a slightly modified Hamiltonian with terms containing the speed of light c (Darwin, Breit Hamiltonian). They can describe basic magnetic effects, but as far as I know, none of these models claims to be exact and if I understand them well, none of them takes into account the retardation.

    See also
    http://en.wikipedia.org/wiki/Darwin_Lagrangian

    the author claims that if higher powers of v/c (third and higher) were to be included, the retardation has to be dealt with too.

    The retardation is necessary if the radiation due to the electron and the proton is to be described. Hence these standard models are not sufficiently accurate to address the question of the radiation of the constituting particles or stability of the ground state.

    In short, the problem is that:

    Description of radiation due to the proton and the electron requires that we take into account retardation. But the standard Hamiltonian description (even its relativistic enhancements) is not able to do that by default. It is about conservative systems right from the start.

    "If you are talking about quantizing the E&M field itself, then I'm not entirely sure on the specifics of that. I hear that QFT is not trivial to apply to bound states."

    I hear exactly the same thing. Perhaps the question of stability of hydrogen atom is still open in the relativistic quantum theory?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electromagnetic radiations in bohr model
  1. Bohr model (Replies: 1)

  2. Bohr Model (Replies: 11)

  3. Bohr's model (Replies: 3)

  4. Bohr model (Replies: 3)

Loading...