The combination of elementary electrostatics and de Broglie's relationship between the wavelength and momentum of a quantum particle is sufficient to build a crude model of the hydrogen atom, which predicts energy levels which match those actually observed, despite ignoring important physical phenomena (e.g. that accelerating charges emit electromagnetic radiation) in the model. Historically, this problem was solved by Niels Bohr: he considered electrons in classical circular orbits around the nucleus subject to the constraint that the circumference of the orbit must be a whole multiple of the electron's wavelength (determined by its momentum) when in that orbit. These turn out to have the right energies.

The last notes in my Physics file from my school-days (when I wasn't quite eighteen years old yet) include my teacher's remarks, on how one could go about building such a model, and my response to this – an almost useful model built without taking account of angular momentum (it effectively assumed zero angular momentum). My teacher was probably hoping for something more like Bohr's model: what I actually came up with was peculiarly different (and may even have been original).

In so far as the electron, in a state with energy E, is at radius r from
the nucleus, it has kinetic energy E+Q/r where Q =
e.e/4/π/ε_{0}, in which ε_{0} is Gauss'
permittivity of free space and e is the charge on the nucleus, equal to minus
that on the electron. With electron mass m tiny compared to that of the
proton, this gives the electron momentum p with p.p/2/m = E+Q/r so p =
√(2.m.(E+Q/r)) tells us the magnitude of K.h, where K is the de
Broglie wave co-vector associated with
the electron. (Strictly, m should be the reduced mass of the electron, the
inverse of the sum of inverses of masses of electron and the proton it's
orbiting; but this makes only one part in 2000 of difference, so isn't a big
issue.) For a stable solution, E is negative; introduce a speed q for which
q.q = −E/2/m, giving

- K = √(−E.(E+Q/r))/h/q

The electron is constrained to r < −Q/E. I reasoned that the electron should exhibit a whole number of half-waves along any diameter – so that the wave can manage to be zero at both extremes. Furthermore, I expected to get a whole number of waves, since it seemed intuitively that an odd number of half-waves would create problems for the sphere at radius −Q/E. Integrating K(x).dx (ignoring K's direction where, clearly, we should really integrate K(x)·dx) along a diameter, construed as the x-axis, gives twice the result of integrating along a radius,

- 2.integral(: dr.√(−E.(E+Q/r)) ←r; 0<r<−Q/E :)/h/q
substitute r = −Q.Cos(v).Cos(v)/E with v varying from −turn/4 to zero; the √ is then of −E.(E+Q/r) = E.E.(1/Cos(v)/Cos(v) −1) = E.E.Tan(v).Tan(v), so we get (as both E and, in the given interval, Tan(v) are negative, so their product is positive)

- = 2.Q.integral(: 2.Cos(v).Sin(v).dv.Tan(v) ←v; −turn/4<v<0.turn :)/h/q/radian
in which 2.Cos(v).Sin(v).Tan(v) = 2.Sin(v).Sin(v) = 1 −Cos(2.v), which is the derivative of v −Sin(2.v).radian/2, which is zero at the end of our interval and −turn/4 at its start, so we obtain the increase

- = 2.Q.turn/4/h/q/radian
- = π.Q/h/q
- = π.Q/h/√(−E/2/m)

which I require to be some whole number of half turns, i.e. some integer multiple of π radians. This makes 2.π.Q/h/√(−E/2/m)/turn an integer, N, giving the spectrum as:

- E = −2.m.(2.Q.π/N/h/turn)
^{2}= −2.m.(e.e/2/N/(h.turn)/ε_{0})^{2}

and I was expecting N to have to be even. We can substitute the fine
structure constant, α = e.e/(2.h.turn.c.ε_{0}) into this
to make it

- E = −2.m.(α.c/N)
^{2}

where Bohr's model has E = −(m/2).(α.c/n)^{2},
without constraining n to be even. This model agrees exactly with Bohr once
we take into account my intuition that N had to be even; substituting N=2.n is
all it takes. Note that −E is equal to the (non-relativistic theory's
value for the) kinetic energy of an electron whose velocity is 2.α/N
times the speed of light; α is roughly 1/137, so the non-relativistic
approximation is reasonably accurate.

It is striking that two such naïve models based on de Broglie produce
the same answer, despite one of them totally ignoring angular momentum while
the other assumes circular orbits *characterized by* their angular
momenta. What's even more surprising is that the answer they both get is
actually correct, given that each completely ignores the electromagnetic
radiation the electron should be radiating away due to being an accelerating
charge. The spectrum of Hydrogen was (prior to de Broglie) already known to
consist of various lines whose frequencies were, aside from a constant
scaling, the differences of the inverses of squares of whole numbers; from
this (and Planck's law), one could reasonably guess that the Hydrogen atom's
possible states had energies equal to some constant scalar divided by the
squares of whole numbers, yielding the observed spectral lines when it made
transitions between two such states. This is, indeed, the result predicted by
the now-orthodox solution based on Schrödinger's
equation.