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Landau levels and bohr model

  1. Oct 26, 2011 #1
    hi

    i thought that if i try to derive the energy of an electron in a magnetic field, this could be done with the assumptions of the bohr model.

    L=n h/(2π)

    mv²/r=qvB => mvr=qBr²=>n h/(2π)=qBr²

    E=p²/(2m)=q²B²r²/(2m)=n h/(2π)qB/(2m)

    so i get the energy for the first level, but all transitions are wrong, as n=2,4,6 etc. should be forbidden, but i do not get this condition.

    so my question is: why does this mistake appear?
     
  2. jcsd
  3. Oct 26, 2011 #2
    Probably, you forget about the "magnetic energy", though the electron (charge= q ) is rotating under the magnetic field. I rearrange your equations here.

    According to the Bohr model. the orbital length is an integer (n) times de Broglie's wavelength (=h/mv),
    So this fact leads to your first equation of the angular momentum (L).

    [tex]2\pi r = n \times \frac{h}{mv} \quad \to \quad L = mvr = n \times \frac{h}{2\pi} = n \hbar[/tex]

    The centrifugal force is equal to Lorentz force (= qvB), as shown in your second equation.

    [tex] \frac{mv^2}{r} = qvB [/tex]

    Using these equations, the angular frequency (= w) of the rotating electron and kinetic energy (K) are

    [tex]\omega = 2\pi \times f = 2\pi \times \frac{v}{2\pi r} = \frac{v}{r} = \frac{qB}{m}[/tex]

    [tex]K = \frac{1}{2}mv^2 = \frac{n\hbar qB}{2m} = \frac{1}{2}n\hbar \omega[/tex]

    Loretz force causes the magnetic moment (= u ) (of rotating electron), which direction is opposite to the external magnetic field,

    [tex]\mu = I\pi r^2 = \frac{qv}{2\pi r}\cdot \pi r^2 =\frac{qmvr}{2m} = \frac{qn\hbar}{2m}[/tex]

    So the magnetic energy (V) is "plus", as follows,

    [tex]V = \mu\cdot B = \frac{n\hbar qB}{2m} = \frac{1}{2}n\hbar \omega[/tex]

    As a result, the energy intervals are hw, as follows,

    [tex]E = V + K = n\hbar \omega[/tex]

    By the way, using Maxwell equation and the above equations, the magnetic flux included in the circular orbit is

    [tex]\pi r^2 B = \pi \frac{m^2v^2}{q^2 B} = \frac{h}{2q} \times n[/tex]

    where h/2q is "magnetic flux quantum"
     
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