The Bohr model and the correspondence principle

The previous section left the third dead end standing: a classical atom lives under $10^{-11}$ s and ought to emit a continuous spectrum.

Start by looking at what experiment actually offered.

Spectral lines are not random — they have a formula

Nineteenth-century spectroscopy accumulated a great deal of data: each element, heated or discharged, emits only at certain wavelengths, fixed precisely enough to identify the element.

Hydrogen is the simplest, with only a handful of visible lines. In 1885 Balmer, a Swiss schoolteacher of mathematics, stared at those four numbers and found a formula. Rydberg later generalised it:

$$\frac{1}{\lambda}=R_\infty\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),\qquad n_2>n_1\tag{1.4.1}$$

with $R_\infty=1.097\times10^7\ \text{m}^{-1}$ fitted purely from experiment.

Bohr’s two postulates

In 1913 Bohr proposed two postulates in direct conflict with classical physics. He knew perfectly well they were unjustified; the plan was to accept them and see what followed.

The first postulate cuts off the collapse directly: an electron in a stationary state does not radiate, so it cannot spiral in; and $n=1$ is the lowest rung, with no step below it, so the atom is stable.

The second turns the “difference of two terms” in the Rydberg formula into “difference of two energy levels”.

Derivation: from two postulates to 13.6 eVbasic~8 minexpand

Step 1: classical mechanics gives the orbit condition. The Coulomb force supplies the centripetal force:

$$\frac{e^2}{4\pi\varepsilon_0r^2}=\frac{mv^2}{r} \quad\Longrightarrow\quad mv^2=\frac{e^2}{4\pi\varepsilon_0r}\tag{1.4.4}$$

Note this step is entirely classical — Bohr invoked quantum assumptions only where he had to, and used Newton everywhere else.

Step 2: impose the quantisation condition. From $mvr=n\hbar$ we get $v=n\hbar/mr$; substituting,

$$m\cdot\frac{n^2\hbar^2}{m^2r^2}=\frac{e^2}{4\pi\varepsilon_0 r} \quad\Longrightarrow\quad r_n=\frac{4\pi\varepsilon_0\hbar^2}{me^2}\,n^2\tag{1.4.5}$$

The orbital radii are discrete and go as $n^2$. The smallest ($n=1$) is the Bohr radius:

$$a_0=\frac{4\pi\varepsilon_0\hbar^2}{me^2}=0.529\ \text{Å}=5.29\times10^{-11}\ \text{m}\tag{1.4.6}$$

This matched the atomic size then estimated from gas viscosity and similar methods. A length assembled purely from fundamental constants landing at the right order of magnitude — this was the model’s first genuinely persuasive moment.

Step 3: compute the energy. Total energy is kinetic plus potential:

$$E=\frac12mv^2-\frac{e^2}{4\pi\varepsilon_0r} =\frac{e^2}{8\pi\varepsilon_0r}-\frac{e^2}{4\pi\varepsilon_0r} =-\frac{e^2}{8\pi\varepsilon_0r}\tag{1.4.7}$$

(using step 1’s $mv^2=e^2/4\pi\varepsilon_0r$; note in passing that the kinetic energy is exactly half the magnitude of the potential energy, a consequence of the virial theorem for a Coulomb force).

Substituting $r_n$:

$$E_n=-\frac{me^4}{32\pi^2\varepsilon_0^2\hbar^2}\cdot\frac{1}{n^2}=-\frac{13.606\ \text{eV}}{n^2}\tag{1.4.8}$$

Step 4: match the Rydberg formula. By postulate two,

$$h\nu=E_{n_2}-E_{n_1}=13.606\ \text{eV}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\tag{1.4.9}$$

In terms of wavelength,

$$\frac1\lambda=\frac{13.606\ \text{eV}}{hc}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\tag{1.4.10}$$

That leading coefficient works out to $1.097\times10^7\ \text{m}^{-1}$ — agreeing with the measured Rydberg constant to four significant figures.

Thirty years of numerical puzzle, solved.

The correspondence principle: how Bohr found that condition

In the derivation above, $L=n\hbar$ was written down out of nowhere. Bohr did not guess it — he had a methodological principle.

Check: the two frequencies really do agree at large nadvanced~6 minexpand

Quantum side. The gap between adjacent levels:

$$\Delta E=E_{n+1}-E_n=13.6\ \text{eV}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{1.4.13}$$

Expanding for large $n$: $\frac{1}{n^2}-\frac{1}{(n+1)^2}\approx\frac{2}{n^3}$, so

$$\nu_\text{quantum}=\frac{\Delta E}{h}\approx\frac{2\times13.6\ \text{eV}}{hn^3}\tag{1.4.14}$$

Classical side. The orbital frequency is $f=v/2\pi r$. With $v=n\hbar/mr$ and $r=a_0n^2$:

$$f_\text{orb}=\frac{n\hbar}{2\pi mr^2}=\frac{n\hbar}{2\pi ma_0^2n^4}=\frac{\hbar}{2\pi ma_0^2}\cdot\frac{1}{n^3}\tag{1.4.15}$$

Both go as $1/n^3$, and putting the constants in shows the coefficients match too.

For a feel for the numbers: at $n=2\to1$ the two differ by more than a factor of two; at $n=100\to99$ by under 2%; at $n=1000$ they are indistinguishable.

Quantum effects do not “vanish” at large $n$; they simply become impossible to tell apart from the classical answer. The same idea reappears in another guise in section 2.7.

Where it goes wrong

The Bohr model is absurdly accurate for hydrogen, but it is a semi-classical patchwork: half Newtonian orbit, half quantum rule, and the two are fundamentally incompatible.

∎Key formulas

• Angular-momentum quantisation

  $$L=mvr=n\hbar,\qquad n=1,2,3,\dots$$

  Bohr's first postulate, reverse-engineered from the correspondence principle

• Bohr radius

  $$a_0=\frac{4\pi\varepsilon_0\hbar^2}{me^2}=0.529\ \text{Å}$$

  An atomic scale assembled purely from fundamental constants

• Hydrogen energy levels

  $$E_n=-\frac{13.606\ \text{eV}}{n^2}=-\frac{\alpha^2mc^2}{2n^2}$$

  The levels themselves are right; quantum mechanics re-derives them

• Rydberg formula

  $$\frac{1}{\lambda}=R_\infty\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

  R∞ = 1.097×10⁷ m⁻¹; theory and experiment agree to four figures

?Self-check4 questions

1. 1.

   How does the Bohr model deal with "an atom collapses within 10⁻¹¹ s"?

   • It postulates that electrons in stationary states do not radiate, and that there is no level below n = 1 to fall to
   • It postulates that the Coulomb force becomes repulsive at short range
   • It postulates that the electron mass varies with orbital radius
   • It postulates that the nucleus reabsorbs the radiated energy

   Submit
2. 2.

   What role does the correspondence principle play in Bohr's work?

   • It is the basis for deriving the quantisation condition: at large n the quantum transition frequency must equal the classical orbital frequency
   • It proves classical physics is wrong
   • It is an empirical rule fitted to spectroscopic data
   • It fixes the photon energy at hν

   Submit
3. 3.

   Which statements about the limitations of the Bohr model are correct? (Select all that apply.)

   Select all that apply

   • It cannot handle two-electron atoms such as helium
   • It gives no relative line intensities and no selection rules
   • Its energy-level formula for hydrogen is wrong
   • Its assumed circular electron orbits do not correspond to real quantum states

   Submit
4. 4.

   What is the photon energy, in eV, for the hydrogen transition n = 3 → n = 2? (This is the Balmer Hα line, and it is red.)

   eV10% relative tolerance

   Submit

What comes next

Bohr’s quantisation condition $L=n\hbar$ is still a rule with no reason behind it. Why integer multiples of $\hbar$ in particular?

Ten years later a doctoral student gave an unexpected answer: because the electron is a wave, and the orbit has to hold a whole number of wavelengths.