I was trying to understand why a stable orbit of a hydrogen atom has to satisfy that the orbit length must be a multiple of de Broglie wavelength.

I have seen some related questions like This.

But the answers seemed not very clear to me. "For the system to be stable, that is, the electron shouldn't cancell itself out". I agree with "shouldn't cancel out" But I didn't understand why only an n*π would be chosen.

I've known it should be a condition for "standing wave", but a standing wave is not necessarily n*π, is it? I'd like to know in detail about why it's stable. Moreover, what if it's an ellipse orbit? Thank you all in advance!


marked as duplicate by Qmechanic quantum-mechanics Mar 17 at 6:40

This question was marked as an exact duplicate of an existing question.


Don’t take any of this too seriously. These are vague, heuristic arguments that now have only historical interest. Electrons don’t actually orbit nuclei in circles or ellipses like planets around the Sun! Bohr’s model is wrong and is just something you learn about in high school, or in college physics without calculus, because it does manage to “explain” certain things, but in entirely the wrong way. A proper explanation of the hydrogen atom requires the Schrodinger equation or an equivalent quantum-mechanical formalism. So just accept the Bohr quantization rules and move on.

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    $\begingroup$ One could appeal to the JWKB approximation leading to this rule $\endgroup$ – lux Mar 17 at 6:34

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