We know that Bohr model is outdated and we know that there is no such thing as an "electron orbital circumference" then how is $2\pi r=n\lambda$ still valid?

Edit :

If the electrons for higher orbitals are not moving in a circular path then how do we write $2\pi r=n\lambda$?


The Bohr model is a semi classical model, treating the electrons like satellites of the proton, in the successful hydrogen atom solution. The success relied that the Bohr assumptions reproduced the series that fitted the hydrogen emission spectra.

The solution of the Schrodinger equation for the hydrogen atom reproduces the success of the Bohr model in fitting the spectra, and gives a theoretical basis for quantum mechanics, with the interpretation of the $ψ*ψ$ of the ψsolutions as probability of finding the electron at x,y,z around the proton.

It has been shown that the most probable radius of the hydrogen ground state is the same as the Bohr radius, explaining the success of the Bohr model for this simple potential case.

Edit after edit of question:

If the electrons for higher orbitals are not moving in a circular path then how do we write 2πr=nλ?

It is a useful approximation for rule of thumb, not accurate. One would have to go through the calculations for each n. After all it is the most probable radius of the ground state that is identified with the lowest Bohr orbit radius. The expectation value of the radius ( the average) even for the ground state is 1.5 of the Bohr radius. After all it is a different mathematical model.

  • $\begingroup$ My question was that if the electrons for higher orbitals are not moving in a circular path then how do we write $2\pi r=n\lambda$? $\endgroup$ – ami_ba Feb 17 '18 at 8:30
  • $\begingroup$ @ami_ba Actually your question does not ask that. If you want it to, then please edit your question to include that. $\endgroup$ – StephenG Feb 17 '18 at 8:44
  • $\begingroup$ @anna v I have edited. $\endgroup$ – ami_ba Feb 17 '18 at 9:50

I must add some considerations.

  1. The fact that a model is outdated does not mean it must be discarded. Besides it is really useful to understand the development and the history of what we do, there are still some processes that can be explained with them. Thomson's model can still be used to derive some models of matter-radiation interation that work surprisingly well. Do not throw models to the bin, you shouldn't use quantum mechanics for the movement of a car, for example.

  2. Yes, Bohr model is outdated, but if it once was a model that's because it somehow worked. Rememebr that Bohr's model was only a model for hydrogen, not the rest.

  3. It was just a fortunate coincidence that, for a spinless particle (only orbital angular momentum) and for a Coulomb's potential (electromagnetic force only), it happens to occur that angular momenta can only be multiples of $\hbar$.

$$ 2\pi r=n\lambda \ \ \Rightarrow \ \ \frac{2\pi}{\lambda} r = n \ \Rightarrow \ kr=n \Rightarrow \hbar k=n\hbar \Rightarrow rp=n\hbar \Rightarrow L=n\hbar$$

And this is why Bohr's model worked, altough it is "incorrect" (for sure any theory we develop will always be innacurate). Of course, this is just an approximation. Spin changes it all, plus there are some more things, like the nucleus, and so on. This is a very ideal model.

  • $\begingroup$ I want to know that if electrons of higher orbitals do not move in a circle then how do we write the equation? $\endgroup$ – ami_ba Feb 17 '18 at 12:57
  • $\begingroup$ If you are still on a single electron case, Bohr model can work, improved with Sommerfield generalisation and a few approximations. But, when you have more electrons, you really need to consider more basic quantum effects, like spins... and talking about "movement" in QM is not always the correct question. $\endgroup$ – FGSUZ Feb 17 '18 at 14:09
  • $\begingroup$ My question is simple...if there is no circular orbit, from where does the circumference of the circular orbit come and we write it to be a multiple of the wavelength associated with the electron? $\endgroup$ – ami_ba Feb 17 '18 at 16:20
  • $\begingroup$ Are we not considering very non circular 3d orbitals and not 2d circular orbits? $\endgroup$ – ami_ba Feb 17 '18 at 16:25
  • $\begingroup$ Look to comment no 2(on Q) at physics.stackexchange.com/q/318621/2451 $\endgroup$ – ami_ba Feb 17 '18 at 16:43

The Bohr model of the hydrogen atom is simply wrong. It relies on an assumed quantization of the classical angular momentum of the electron orbitals $L=n\hbar$. The ground state $n=1$ is thus assumed to have the angular momentum $\hbar$. Quantum mechanics shows that it is actually zero. The correspondence with the principal hydrogen energy levels is fortuitous.


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