# My de Broglie model of the Hydrogen atom

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).

## My model

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/ε0, in which ε0 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

= 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)/ε0)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.ε0) 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.

## Conclusion

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.

Written by Eddy.