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Collapse of a Bohr atom

  1. Dec 10, 2003 #1
    I have a question. Suppose we have an electron orbiting around a proton in a Bohr atom. It is accelerating due to centripetal motion yet travelling at v<<c so Newtownian physics applies. Since it is accelerating, it is radiating energy. Assuming we are using larmor's formula for power radiated, how would we calculate how fast it will take for the electron to crash into the proton, and this cause the Bohr atom to collapse?
  2. jcsd
  3. Dec 10, 2003 #2


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    What is the point of your question? Quantum theory says it won't happen.
  4. Dec 10, 2003 #3
    It's sounds like an excercise from modern physics class. A reasonable excercise too, unless you take everything you hear at face value, or don't value history.

    Anyway, I'll try to add something actually useful and constructive towards answering your question.

    The Larmor formula tells us (remember, we take E to be negative)
    P=-\frac{dE}{dt} = \frac{e^2a^2}{6 \pi \epsilon_0 c^3}
    As usual, take [tex]a=v^2/r[/tex]. Now, we also know that the classical radius is
    \qquad r = - \frac{e^2}{8 \pi \epsilon_0 E}
    v^2 = \frac{2E}m
    Now you should be able to put all that together and get simple separable differential equation for dE/dt that you can integrate from the starting energy (about -14 eV) to the final energy (negative infinity) that will give you the time for collapse.
  5. Dec 10, 2003 #4
    The point of the question is to show that Bohr's model is not the final model of an atom for quantum mechanics. Its just a mathematical exercise using only Newtonian physics. Thank you, Big Red Dot.
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