In the future, I will need to teach students at school the proportionality $E_n\propto-\frac{1}{n^2}$ of the energy levels of a hydrogen atom. I could present this as an experimental fact, but I would like to also justify the proportionality. However, the tools we have are very basic. For instance, we can't solve the Schrödinger equation because we don't have any techniques for solving ODEs other than the classical harmonic oscillator, let alone PDEs with complex coefficients. I can do it using the Bohr model, but I would like to see if it can be justified in a more wave centric way, where electrons are standing waves, not particles on a fixed orbit.

How can we justify $E_n\propto-\frac1{n^2}$ under the following conditions?

  • No Schrödinger equation
  • Electrons are assumed to behave as waves (no Bohr model)
  • Basic quantum mechanics, like the de Broglie wavelength, are allowed
  • Classical formulae for kinetic and potential energies are allowed
  • The Coulomb potential should remain unchanged (no approximation by box potentials or similar, if possible)
  • 6
    $\begingroup$ A note of caution: if you choose to do this, please make very clear - repeatedly emphasize - that this is just a heuristic explanation, and that when they study QM more formally they will see a more rigorous derivation. Lots of beginning students assume that any mathematical derivation must be rigorous, and then they get confused about exactly what the starting postulates are. Please convey to them clearly that this is not actually a derivation or proof of anything. $\endgroup$
    – tparker
    Jun 25, 2023 at 13:07

2 Answers 2


I am not sure this will respect all of the rules, however I'll give it a try.

You can try to imagine the electron as an oscillating string wrapped in a circle. The wavelength of the string is an integer multiple of the "natural" De Broglie wavelength $\lambda = h/(mv)$ ($h$ is the Planck's constant, $m$ the electron mass and $v$ the electron velocity). To satisfy the boundary condition (the two edges of this string should always coincide), the condition is $$ n\lambda = 2\pi r$$ where $n$ is a positive integer and $r$ the radius of the circle. You can also impose Newton's law to rewrite $r$ as a function of velocity: $$ m \frac{v^2}{r} = \frac{kZe^2}{r^2} \;\;\;\;\;\; \to \;\;\;\;\;\; r = \frac{kZe^2}{mv^2}, $$ where $k=1/(4\pi\varepsilon_0)$, $Z$ the atomic number and $e$ the electron charge. Substituting $r$ and $\lambda$ we get $$ n\lambda = 2\pi r \;\;\;\;\;\; \to \;\;\;\;\;\; v = \frac{kZe^2}{\hbar} \frac{1}{n}. $$ Now you can use all of these to compute the total energy of the electron: $$ E = \frac{1}{2}mv^2 - \frac{kZe^2}{r} = - \frac{1}{2}mv^2 = - \frac{\text{const}}{n^2} $$

I think this is an idea from Arnold Sommerfeld.

  • $\begingroup$ Isn't it just a rephrased version of the Bohr model? $\endgroup$
    – Ruslan
    Jun 26, 2023 at 11:04
  • $\begingroup$ Yeah, essentially this is the Bohr model combined with De Broglie's idea of matter waves. This is a little bit more fundamental than Bohr's model because the assumption of quantized orbits is more justified by the boundary condition on the circular wave. This is the first and simplest semiclassical idea that comes to mind when combining Bohr's theory with the postulate of particle-wave duality. $\endgroup$
    – Matteo
    Jun 26, 2023 at 12:08
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    $\begingroup$ Oh interesting, I was sure Bohr had based his model on the de Broglie's hypothesis, but now that I re-read the history, he appears to have just made an ad-hoc assumption 11 years before de Broglie's proposed his idea of matter waves. $\endgroup$
    – Ruslan
    Jun 26, 2023 at 13:15

I've seen the following 'argument'. It is better viewed, in my opinion, as a memory-aid than as giving a description of an atom that is properly consistent with quantum mechanics.

Imagine an electron as a stationary de Broglie wave 'fitting around' a circle of radius $r$ centred on the nucleus. In order for the wave to be stationary there mustn't be discontinuities, so a whole number of wavelengths ($\lambda = \tfrac h{mv}$) fit round the circumference. In other words... $$2\pi r=\frac {hn}{mv}\ \ \ \ \ \ \ (n=1, 2, 3 ...).$$ If you combine this with the classical equations for a particle doing circular orbits in a Coulombic field you will reach the Bohr formula for energies of orbits.


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