Electron orbit circumferences have to be in multiples of its de Broglie wavelength, but what do those 2 have in common?

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    $\begingroup$ I always thought that if it wasn't a multiple of the wavelength then it would destructively interfere with itself. $\endgroup$ Commented Mar 15, 2017 at 13:21
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    $\begingroup$ Please note that the Bohr model is outdated and we now know there is no such thing as an "electron orbital circumference". Electrons do not have orbits, they occupy orbitals. $\endgroup$
    – ACuriousMind
    Commented Mar 18, 2017 at 13:44
  • $\begingroup$ There's an explanation in this Wikipedia page on Bohr. $\endgroup$ Commented May 27, 2017 at 12:04

5 Answers 5


I will assume you are familiar with the properties of waves such as interference and diffraction.

Consider an electron orbiting the nucleus. By de Broglie's hypothesis, we would consider it to be a wave orbiting around the nucleus. Now, once the electron wave orbits once, the second time it would interfere with the first wave. For the system to be stable, the electron shouldn't cancel itself out, so the wave must constructively interfere. That means, the circumference must be an integral multiple of the electron's wavelength.

Here's a picture that shows how the electron wave is expected to exist in the atom according to the Bohr's theory.

source: www.astronomyclub.xyz

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    $\begingroup$ @ Pritt Balagopal Does that mean that the electron instantaneously cancels itself out? If it it travelling in a given time to return to its starting position, how can it cancel itself out? $\endgroup$
    – radial9174
    Commented May 7, 2017 at 19:15
  • $\begingroup$ Electron shouldn't cancel itself out. Thats why circumference is a multiple of wavelength. $\endgroup$ Commented May 8, 2017 at 3:00
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    $\begingroup$ @PrittBalagopal : would be better to call the electron a standing wave, which is the original de-Broglie hypothesis. $\endgroup$
    – vs_292
    Commented Sep 9, 2017 at 14:18
  • $\begingroup$ That's right, the electron is in a standing wave. $\endgroup$ Commented Sep 9, 2017 at 15:11
  • $\begingroup$ Why does it constructively interfere when the circumference is an integral multiple of the wavelength? $\endgroup$
    – Kaushik
    Commented Sep 3, 2019 at 19:12

Nothing actually. It was quite a wild guess by Bohr and supplied him with the spectrum of hydrogen. Pretty good guess indeed.

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    $\begingroup$ Most of the advancements in quantum science has been done by equation guessing and then trying to figure out the "whudunnit" of it later. $\endgroup$
    – Stian
    Commented Mar 15, 2017 at 9:35

De Broglie suggested the existence of matter waves and gave a relation between their wavelength and momentum.

$\lambda=\frac{h}{p}$ ,

He said that this relation is completely general. It can be applied to matter particles and even photons.

enter image description here

Bohr, in his atomic model, considered an electron to be in form of a standing electron wave and if this wave is to be continuous over the circumference of the stationary orbit that the electron lie in, the circumference must be a integral multiple of its wavelength $(n\lambda)$.

$2\pi r=n\lambda$

$2\pi r=\frac{nh}{p}$


And finally,


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    $\begingroup$ your answer has really nice start, clarifying that de Broglie's idea was applicable in general and not just sub atomic particles. However your answer and especially the equations do not answer the question asked. It looks like what you write in your school's exam - rote memorization. Please see the accepted answer and thrive to understand physics $\endgroup$
    – RinkyPinku
    Commented Apr 25, 2017 at 6:25

This comes from the Bohr-Sommerfeld quantization formula, which can be derived from the semiclassical WKB approximation of quantum mechanics, cf. e.g. this Phys.SE post. The quantization condition follows from requiring single-valuedness of the wavefunction.


Quite new here. Anyways, it was not a guess by Bohr in the first place. And it is so as to save energy else once it'll be out of phase it will start losing energy and Niels quantization will not be valid. That's the logical explanation though still incomplete considering that there's a lot going on and the theory we are referring to is outdated to a LOT of extent. Happy learning!


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