Can the Bohr Model Accurately Predict Electron Energy in a Magnetic Field?

[Gavroy]

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?

 

[ytuab]

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?

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).

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

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

\frac{mv^2}{r} = qvB

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

\omega = 2\pi \times f = 2\pi \times \frac{v}{2\pi r} = \frac{v}{r} = \frac{qB}{m}

K = \frac{1}{2}mv^2 = \frac{n\hbar qB}{2m} = \frac{1}{2}n\hbar \omega

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

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

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

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

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

E = V + K = n\hbar \omega

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

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

where h/2q is "magnetic flux quantum"