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Bohr model and relativistic electron mass

  1. Sep 21, 2015 #1
    From the semi-classical Bohr model of the hydrogen atom the velocity of the electron in a certain orbit can be determined. With these velocities the electron's relativistic masses can be determined. With E=mc2 the energy levels are in agreement with those from the Bohr model. I know Bohr's model has flaws, but what does this result mean?
  2. jcsd
  3. Sep 21, 2015 #2
    Can you show the math?
  4. Sep 21, 2015 #3
    From Bohr's model of the Hydrogen atom in the ground state

    v = e2/(2εh) = 2.18x106 m/s

    Using m=m0 / √(1-v2/c2)
    and E=(m-m0)*c2 leads to E = 13.5 eV

    Using Bohr's equations E = m*e4/(8ε2h2) also gives 13.5 eV

    I tried to see what can be derived by substituting the first equation in the second by get stuck in the math.
  5. Sep 21, 2015 #4


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    The Bohr model of the atom is not so much wrong, as it is oversimplified and incomplete:

    One other note: the relativistic correction to the mass of the electron due to its zipping around the nucleus will only be like 3 parts in 100,000, so you won't notice a difference if you calculate the energy to just the first decimal place.

    What the Bohr model does:
    Predicts the energy levels of the hydrogen atom to a first-order approximation (the coulomb potential, non-relativistic kinetic energy, and nothing else):

    What the Bohr model does not include:
    Relativistic corrections to the kinetic energy
    Corrections due to the magnetic field created by the orbiting electron interacting with its own magnetic moment (Spin-orbit coupling)
    Corrections due to the nucleus having a nonzero size (Darwin term)
    Corrections due to the interactions of the electron's and proton's magnetic dipoles (spin-spin coupling).
    among others...

    Also, even without these corrections, the Bohr model doesn't predict what the electron cloud should look like for a hydrogen atom in a given state, or what the likelihood is for an atom to decay from an excited state to the ground state in a given time.

    If you want a good model of the Hydrogen atom, you're going to want to solve the Schrodinger equation describing the electron-proton interaction. You'll get the same results as the Bohr model for the first-order approximation, but it's easier to get theoretical predictions that more closely agree with experimental results by including these extra terms.
  6. Sep 22, 2015 #5
    Thanks jfizzix

    But that's what surprises me so much. The energy difference due to this relativistic effect of "only" 3 in 100,000 (to be more precise 2.66 in 100,000) is the same as the energy level of the electron from the Bohr model:

    2.66/100000 * mc2 = 13.5 eV
  7. Sep 22, 2015 #6


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    In the limit of ##v_n<<c##, upon truncating up to the first order in ##v_n/c## with ##v_n## the rms velocity, we obtain for an electron's total energy
    mc^2 \approx m_0 \left( 1+\frac{v_n^2}{2c^2} \right) c^2
    E = (m-m_0)c^2 \approx \frac{m_0 v_n^2}{2} = \frac{m_0 e^4 Z^2}{8\epsilon_0^2 h^2 n^2}
    which is, apart from the sign, indeed exactly the same as the energy levels expression both given by Bohr model and QM. I would say this is one of those mathematical coincidence. An electron bound in an attractive potential should have total energy lower than its rest mass, but your calculation showed otherwise. Moreover, by assuming that a bound electron does orbit with a well defined velocity already violates the principle of uncertainty.
  8. Sep 22, 2015 #7
    Thanks blue_leaf77. I think I understand now.
    I simply calculated the kinetic energy since v<<c. I forgot to add the potential energy, which is twice the kinetic energy but negative, hence the sign error in my result.
  9. Sep 22, 2015 #8
    It's not a coincidence. There is a general theorem (the virial theorem) which states that the magnitude of the KE is 1/2 of the magnitude of the PE (for forces following inverse square law). The total energy of the level will then be -KE.
    As was already mentioned, the relativistic formula for KE is not necessary for this case, you could use the classic one.
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