The combination of naïve 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). I've cleaned up the derivation a bit, below, to help the reader keep up, and left discussion of its most salient difficiences until later.
If the electron is in a bound state, its total kinetic plus electrostatic potential energy is some negative E. There is them some radius R for which −E.R = e.e/(4.π.ε0): this R is the radius at which this energy sum would be all electrostatic potential, with zero kinetic (hence zero momentum and infinite de Broglie wavelength). We may reasonably suppose that the wave nature of the electron has a node at this radius and that the number of wavelengths along a diameter of the sphere must be whole. To get that number we integrate dx/λ = p.dx/h with distance coordinate x along the diameter ranging from −R to +R along it. (Here, h is Planck's constant, λ is wavelength and p is momentum.)
The electrostatic potential of the electron, in so far as it's at x along that diameter, is −e.e/(4.πε0)/abs(x) = E.R/abs(x); subtracting this from (negative) E gives us its kinetic energy as −E.(R/abs(x) −1) implying a momentum, p, whose square is 2.m times this, where m is the electron mass. (Strictly, m should be the reduced mass of the electron, the inverse of the sum of inverses of masses of electron and the nucleus it's orbiting; but this makes at most one part in 1837 of difference, so isn't a big issue.) We're integrating p.dx/h over −R ≤ x ≤ +R, and it depends on x only via its abs, which is symmetric about x = 0.R, so we can take the integral over the positive half of that range and double it, eliminating the need for abs. So we we're integrating
over 0.R ≤ x ≤ R and it's natural to do a subsitution that'll make x/R vary over the interval from 0 to 1, suggesting some sinusoid. We want the inverse of that, minus one, to have a tidy square root and x = R.Cos(v).Cos(v) immediately suggests itself as 1/Cos(v)/Cos(v) −1 = Tan(v).Tan(v). To get x varying from 0 to 1 we can have v vary from −turn/4 to 0.turn, on which Tan is negative, so the (positive) square root is −Tan(v). This gives dx = −2.R.Sin(v).Cos(v).dv/radian, so we're now integrating
over turn/4 ≤ v ≤ 0.turn. At the ends of that, Sin(2.v) is Sin(turn/2) = 0 = Sin(0.turn) so there's no change in that, while v increases by turn/4. Dividing that by radian gives us π/2, so our integral is
and we're expecting N to be some whole number. We need to eliminate R, using −E.R = e.e/4/π/ε0, to solve for E. To limit mess, transiently write Q = −E.R, whence:
giving the spectrum as:
We can substitute the fine structure constant, α = e.e/(2.h.c.ε0) = 2.π.Q/h/c into this to make it
in agrement with Bohr's model. Note that −E is equal to the (non-relativistic theory's value for the) kinetic energy of an electron whose velocity is α/N times the speed of light; α is roughly 1/137, so this is indeed a suitably non-relativistic speed even for N = 1.
The most flagrant deficiency of this is that it presumes the wave-like variation within the system is radial, which is equivalent to supposing that the momentum is radial, hence there is no angular momentum. There is no getting over this, so I just delight in the fact that it works despite that.
One may also object that the radial trajectory
thus presumed involves
passing through the centre, where the energy definitely goes relativistic. That
happens when R/abs(x) −1 is of order m.c.c/(−E) =
2.(N/α)2 ≈ 37558.N.N, so when x is of order R/(1
+37558.N.N), which is at most R/37559 and respectably tiny. In particular, x
smaller than this implies Cos(v) < O(1/194), which limits v +turn/4 to
O(radian/194), a fraction of a degree of arc. Our integrand 1 −Cos(2.v)
is then O(1/972) over a very short interval, so this part of the
integral contributed little (of order radian/194/972 in comparison to
the integral's turn/4 total; a fractional contribution of
1/973/π). As a result we should not expect any huge change in the
result if, instead of an exact diameter, we take a path close to the diameter
all along its length but departing from it by just enough near the middle to let
the electron go round the nucleus, which is physically to be expected in any
case, if only by virtue of it being wavy so diffracting round the nucleus. The
relativistic approximation can thus survive, despite the nominal violation of
its precondition.
One might also object that we only need a node at each end of the diameter,
allowing it to be a whole number of half-waves long, so that 2.N is an
integer but N might not be. This would require the phase
of the wave
process (which is present even where its amplitude is zero, at a node such as
our bounding radius R) to vary over the sphere in such a way as to give our
diameter's ends opposite phases. The phase must be continuous throughout the
ball of space within radius R, implying that the phase change along any loop
within that ball – which, by virtue of being a loop, returning to whence
it came, must be a whole number of cycles – must vary continuously as we
deform the loop. (One might try to wriggle out of this by claiming that phase
can change discontinuously at any node; but then the rate of change of the
wavyness would be discontinuous, which isn't physically plausible, except at the
atom's centre, where the wavelength goes to zero.) A continuous variable over a
(connected) continuum whose value lies in a discrete set is necessarily constant
and we can continuously deform any loop within the ball down to a point, where
the phase change around it is manifestly zero, so in fact the phase change
around any loop must be zero.
Now consider a great circle around our ball's surface, along which the phase
must vary continuously, so the phase difference between ends of a diameter must
also vary continuously; but is constrained to always be a whole number of half
turns. As before, continuous variation over a connected continuum but with a
discrete set of values must be constant, so where one end of the diameter
changes one way, as we vary which diameter round the circle to look at, the
other end must change the same way, at the same rate, so the total change round
half a circle, as the diameter rotates to become its reverse, must be half the
total change round the whole circle, which we know to be zero. Thus, in fact,
the ends of any diameter must have the same phase – not just a whole
number of turns apart. That, in turn, is a problem for my reasoning, as it
seems to claim N = 0, but the singularity at the origin provides an escape
route: the electron needs to not be there
, as it's occupied by the
nucleus, making a node there entirely reasonable, making it reasonable to
suppose it's also a node, and the wavelength goes to zero there, making a
discontinuity in phase less implausible. Thus my reasoning turns into the
radius, rather than the diameter, being a whole number of half-waves, which
works just fine.
Lastly, I should say a few words about the fact that an accelerating charge
should be radiating energy, as a result of Maxwell's equations. This is an
issue likewise for Bohr's model; and, indeed, one could argue for it
with Schrödinger's (in so far as the electron is
at any position in the electrostatic field of the nucleus, it is accellerating,
after all). This could be averted if the radiation to be produced would
destructively inferfere with itself. That isn't possible in the classical atom,
but the quantum one allows the electron to be in a superposition of distinct
states. Although it may take some juggling to arrange for even such a
superposition's various states' radiative output to destructively interfere, it
is at least a conceivable pretext for accepting the quantum atom's failure to
radiate. This is, of course, a hand-wavy
argument that should not be
taken too seriously – a proper understanding requires quantum
electrodynamics. It is at best a slogan
to give the intuition at least
some grip on the situation: taking it seriously would meander into the realms
of quantum woo
and magical thinking.
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. So the important part of either Bohr's model or the one above is that it determins the right constant to divide by those squares of whole numbers.
Along with the spectrum, we obtain a length-scale for our atom, in its N state, as R = Q/(−E) = (N.h/π)2/Q/m/2 = N2.h/m/c/π/α, in which h/m/c/π/α ≈ 0.1 nm is twice the Bohr radius. We can compare this to the Compton wavelengths of the electron, c. 2.4 pm, and proton, c. 1.32 fm, a characteristic length-scale of the particles themselves: the Bohr radius is about 22 times that of the electron and about 40000 times that of the proton. Thus the neighbourhood of the nucleus in which the electron's speed gets relativistic is only slightly larger than we can expect the nucleus itself to seem.
When we come to look at the spectrum in terms of the frequencies of light the Hydrogen atom can absorb or emit, we see transitions with energy differences ΔE = (m/2).(α.c)2.(1/n/n −1/m/m) for positive whole numbers n, m. The frequency of the absorbed or emitted light is just this divided by h and the wavelength is c divided by the frequency,
I shall refer to the (2.h/m/c)/α2 ≈ 91 nm base
length, that this is dividing by a difference of inverses of squares, as
Rydberg's wavelength; its inverse is known as Rydberg's constant (albeit the
name Rydberg also attaches to scalings of it by c and h.c.c to deliver the
frequency and energy-per-quantum of the light). Since this wavelength scale is
considerably larger than the Bohr radius, there is a tendency for the light the
atom can absorb to simply diffract round it, instead of hitting
it and
triggering a transition. The shortest-wavelength photon that the atom can
absorb, in energy level N, are the ones from there to escape, with wavelengths
up to (2.h/m/c).(N/α)2, which is just N.N times the Rydberg
wavelength. However, since the radius of energy level N grows with N also as
its square, we are left with a wavelength/radius ratio of
at the threshold between light (of longer wavelengths) that can cause a
transition between bound states and light (of shorter wavelengths) that will
ionise the atom. It is only when λ/R is small, however, that the photon
has a high probability of hitting
the atom; and these photons have
energies O(800) times greater than the ionisation energy, so eject the electron
with significant energy, hence speed away from the ion left behind. It is thus
easier to ionise the atom than to merely bump it up to another energy level,
from which it is apt to fall back down.
It is striking that two such naïve models (mine and Bohr's) 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, agreeing not only with the now-orthodox solution based on Schrödinger's equation but, most importantly of all, with experiment.
Written by Eddy.