Part 5 · The old quantum theory

5.4 Bohr's atom

Source: Postorino, week 4 slides, pp. 23–28

Objectives

Theorem. Bohr’s postulates, with circular orbits and $L = n ℏ$ , give $r_{n} \propto n^{2}$ , $v_{n} \propto 1/ n$ , $E_{n} = - \frac{1}{2} m (\frac{Z e ^{2}}{4 π ε _{0} ℏ})^{2} \frac{1}{n ^{2}}$ , and Rydberg’s formula with $R (\infty) = 109737$ cm⁻¹.
Recipe. Line wavelengths from $h ν = ∣ E_{a} - E_{b} ∣$ ; ground-state numbers $a_{0} = 0.529$ Å, $v_{0} = α c = c /137$ , $E_{1} = - 13.6$ eV.
Picture. The level diagram: series of lines grouped by their lower level, crowding towards a limit.
Vocabulary. Stationary orbit, quantum number, Bohr radius, fine-structure constant, reduced mass, ionisation energy, Lyman, Balmer, Paschen.

The postulates

Bohr (1913) makes ad hoc hypotheses, along the lines of Planck’s for the oscillator and Einstein’s for radiation.

Definition 5.4.1Bohr’s postulatesslides w4, p. 23

Electrons move on circular orbits, called stationary orbits, without emitting electromagnetic radiation.
The stationary orbits satisfy the quantisation of angular momentum: $L = m v r = n ℏ = n \frac{h}{2 π}, n = 1, 2, 3, \dots$
Electrons emit or absorb radiation (a photon) only when they jump from one stationary orbit to another.
The photon emitted or absorbed has energy $h ν_{a, b} = ∣ E_{a} - E_{b} ∣,$ with $E_{a}$ and $E_{b}$ the energies of the two orbits.

Postulate 1 simply declares the classical collapse (Key observation 5.1.4) not to happen. Postulates 3–4 explain why emission and absorption lines coincide (Key observation 5.1.1): both are jumps between the same pairs of levels, in opposite directions beyond the lectures.

Plan: compute the energies of the stationary orbits, and from them Rydberg’s formula.

Radius of the orbits

The Coulomb force between the nucleus (charge $Z e$ ; $Z = 1$ for hydrogen) and the electron is $F_{C} = \frac{Z e ^{2}}{4 π ε _{0} r ^{2}}$ . The orbits are circular, so it equals the centripetal force: $\frac{Z e ^{2}}{4 π ε _{0} r ^{2}} = \frac{m v ^{2}}{r} .$

Theorem 5.4.2Radius of the stationary orbitsslides w4, p. 24

$r_{n} = \frac{4 π ε _{0}}{m Z e ^{2}} (n ℏ)^{2} .$ The radius is discrete and depends on the quantum number $n$ , as $n^{2}$ .

Proof

Multiply both sides of the force balance by $m r^{2}$ . The left side becomes $\frac{m Z e ^{2}}{4 π ε _{0}}$ ; the right side becomes $\frac{m v ^{2}}{r} \cdot m r^{2} = m^{2} v^{2} r = \frac{( m v r ) ^{2}}{r} = \frac{( n ℏ ) ^{2}}{r}$ , using $n ℏ = m v r$ . So $\frac{m Z e ^{2}}{4 π ε _{0}} = \frac{( n ℏ ) ^{2}}{r} .$ Solve for $r$ .

Speed and energy

Theorem 5.4.3Speed on the stationary orbitsslides w4, p. 25

$v_{n} = \frac{Z e ^{2}}{4 π ε _{0}} \frac{1}{n ℏ} .$ The speed is discrete and depends on $n$ , as $1/ n$ .

Proof

Multiply the force balance by $r$ instead: $\frac{Z e ^{2}}{4 π ε _{0}} = m v r \cdot v = n ℏ v$ .

Theorem 5.4.4Energy of the stationary orbitsslides w4, p. 25

$T_{n} = \frac{1}{2} m [\frac{Z e ^{2}}{4 π ε _{0}} \frac{1}{n ℏ}]^{2}, V_{n} = - m [\frac{Z e ^{2}}{4 π ε _{0}} \frac{1}{n ℏ}]^{2}, E_{n} = V_{n} + T_{n} = - \frac{1}{2} m [\frac{Z e ^{2}}{4 π ε _{0} ℏ}]^{2} \frac{1}{n ^{2}} .$

Proof

$T_{n} = \frac{1}{2} m v_{n}^{2}$ with $v_{n}$ above. The potential energy, using $r_{n}$ : $V_{n} = - \frac{Z e ^{2}}{4 π ε _{0}} \frac{1}{r _{n}} = - \frac{Z e ^{2}}{4 π ε _{0}} \frac{m Z e ^{2}}{4 π ε _{0}} \frac{1}{( n ℏ ) ^{2}} .$ Add the two.

So $V_{n} = - 2 T_{n}$ , true of any circular orbit under an inverse-square force beyond the lectures. $E_{n}$ is negative, since the electron is bound, and rises towards 0 as $n \to \infty$ .

Rydberg’s formula derived

Theorem 5.4.5Rydberg’s formula from Bohrslides w4, p. 26

$\tilde{ν}_{a, b} = \frac{1}{h c} ∣ E_{a} - E_{b} ∣ = \frac{m}{2 h c} [\frac{Z e ^{2}}{4 π ε _{0} ℏ}]^{2} [\frac{1}{n _{a}^{2}} - \frac{1}{n _{b}^{2}}], n_{a} < n_{b} .$ This is exactly Rydberg’s empirical formula (Theorem 5.1.2) for hydrogen, $Z = 1$ , provided that $R (\infty) = \frac{m}{2 h c} [\frac{e ^{2}}{4 π ε _{0} ℏ}]^{2} = 109737 cm^{- 1} .$

Proof

Postulate 4 with $E_{n}$ above: $h ν_{a, b} = ∣ E_{a} - E_{b} ∣ = \frac{1}{2} m [\frac{Z e ^{2}}{4 π ε _{0} ℏ}]^{2} [\frac{1}{n _{a}^{2}} - \frac{1}{n _{b}^{2}}]$ . Divide by $h c$ .

It is written $R (\infty)$ , or $R_{H} (\infty)$ , as a reminder that the proton was treated as infinitely heavy: it sits still at the centre. Compare the measured $R_{H} = 109677.58$ cm⁻¹.

Proposition 5.4.6Reduced massslides w4, p. 26

The proton’s mass $M$ is finite: electron and proton both orbit their common centre of mass. Taking this into account replaces $m$ with the reduced mass $μ = \frac{m M}{m + M}, R (M) = \frac{μ}{m} R (\infty) .$ For hydrogen, $M / m = 1836$ , so $μ / m = 1/ (1 + 1/1836) = 0.999456$ and $R (M) = 109677.6$ cm⁻¹: the measured value to better than 1 part in $1 0^{5}$ .

WarningSlide slip

The slide gives $R (M) = 109681$ cm⁻¹ for hydrogen. $109737 \times 0.999456 = 109677.6$ cm⁻¹, equal to the measured $R_{H}$ to all the digits shown; 109681 would be off by $3 \times 1 0^{- 5}$ , worse than the slide’s own “better than $1/1 0^{5}$ ”. The handwritten orange “H” over the equals sign marks the value as hydrogen’s.

RemarkWhy the reduced massbeyond the lectures

The two-body problem separates into the motion of the centre of mass and the relative motion, and the relative coordinate behaves like a single particle of mass $μ$ . Since $M ≫ m$ , $μ$ is very close to $m$ . The correction is small but measurable: deuterium, with twice the nuclear mass, has its lines shifted enough to be resolved, and that is how it was discovered.

The energy levels of hydrogen

The energies of the stationary states of hydrogen ( $Z = 1$ ): $E_{n} = - \frac{R}{n ^{2}}, R \approx 109700 cm^{- 1} \approx 13.6 eV$ ( $R$ in energy units, $h c R_{H}$ ; the two values are the lecturer’s handwritten annotations). Fowles’ level diagram: 0 at $n = \infty$ , then −0.85 eV ( $n = 4$ ), −1.51 ( $n = 3$ ), −3.40 ( $n = 2$ ), −13.6 ( $n = 1$ ).

$n = 1$ is the ground state (stato fondamentale).
The gap from $n = 1$ to $n = \infty$ is the ionisation energy, 13.6 eV: the energy needed to pull the electron off.

Above $n = \infty$ ( $E > 0$ ) the electron is free beyond the lectures. Each downward arrow on the diagram is an emission line, labelled with its wavelength in ångström. The lines group into series by their lower level:

Series	Lower level $n_{a}$	Lines shown	Region
Lyman	1	Ly $_{α}$ 1216 Å, Ly $_{β}$ 1026, Ly $_{γ}$ 972.5, Ly $_{δ}$ 949.7	ultraviolet
Balmer	2	H $_{α}$ 6563 Å, H $_{β}$ 4861, H $_{γ}$ 4340	visible
Paschen	3	18 751 Å, 12 818, …	infrared
Brackett	4		infrared

(The regions are not on the slide.)

The Balmer lines are the visible lines of The hydrogen spectrum. The level spacing shrinks as $n$ grows, so each series crowds towards a limit, the jump from $n = \infty$ .

Loading the demo…

Figure 5.4.1 Hydrogen’s levels

E_{n} = - 13.6/ n^{2}

eV. Drag the

upper

and

lower

levels of a jump: the readout gives the photon’s energy, wavenumber and wavelength, and the series. Left, the orbits drawn to scale,

r_{n} = n^{2} a_{0}

. Try

3 \to 2

, the red H

_{α}

, and

\infty \to 1

, the ionisation energy.

Ground-state numbers

Using $h = 6.626 06876 \dots \times 1 0^{- 34}$ J·s and $ℏ = 1.054 571817 \dots \times 1 0^{- 34}$ J·s:

Proposition 5.4.7Bohr radiusslides w4, p. 28

$r_{1} = a_{0} = \frac{4 π ε _{0} ℏ ^{2}}{m e ^{2}} = 5.29 \times 1 0^{- 11} m = 0.529 \overset{˚}{A} .$

The size of an atom (Sizes and masses). In general $r_{n} = n^{2} a_{0} / Z$ beyond the lectures.

Proposition 5.4.8Speed in the ground stateslides w4, p. 28

$v_{1} = v_{0} = \frac{e ^{2}}{4 π ε _{0} ℏ} = α c, α = \frac{e ^{2}}{4 π ε _{0} c ℏ} = \frac{1}{137},$ the fine-structure constant. $v_{0} = c /137 \approx 2.2 \times 1 0^{6}$ m/s: the electron in the hydrogen atom is not relativistic.

$α$ is dimensionless beyond the lectures: it compares the strength of the electromagnetic interaction, $e^{2} /4 π ε_{0}$ , with the quantum-relativistic scale $ℏ c$ .

For a hydrogen-like atom (one electron, nuclear charge $Z$ ): $v_{0} = Z α c$ . For $n$ not too large the electron stays non-relativistic.
Inner electrons of heavy atoms ( $Z \sim 80$ ) reach about $0.6 c$ , and relativity does matter there: it is why gold is yellow beyond the lectures.

Ground-state energy (from the formulas above, not on this page): $E_{1} = - \frac{1}{2} m (α c)^{2} = - 13.6$ eV, half the electron’s rest energy $m c^{2} = 511$ keV times $α^{2}$ .

Problems of the Bohr model

The slide lists five; the explanations are added.

Ad hoc hypotheses. Quantisation is imposed, not derived. Why should $L$ be quantised, and why don’t stationary orbits radiate?
Circular orbits. Elliptical orbits are allowed classically, and the model has no rule for them. (Quantum mechanics later shows the ground state has $L = 0$ , not $ℏ$ .)
Atoms with more electrons. It fails already for helium: it can’t handle the electron–electron interaction.
Descriptive, not predictive (transition rates). It gives the line positions, not their intensities: how often each jump happens.
States of the continuum. It says nothing about unbound electrons, $E > 0$ .

Quantum mechanics, with the Schrödinger equation, fixes all of these.

Exercises

Exercise 5.4.1beyond the lectures

He⁺ is hydrogen-like with $Z = 2$ . Find its ground-state energy, radius and electron speed.

Solution

$E_{n} \propto Z^{2}$ : $E_{1} = - 4 \times 13.6 = - 54.4$ eV. $r_{1} = a_{0} / Z = 0.265$ Å. $v_{1} = Z α c = 2 c /137 = 4.4 \times 1 0^{6}$ m/s.

Exercise 5.4.2beyond the lectures

Compute $a_{0}$ and $R (\infty)$ from the constants.

Solution

$a_{0} = \frac{4 π ε _{0} ℏ ^{2}}{m e ^{2}} = \frac{1.113 \times 1 0 ^{- 10} \cdot 1.112 \times 1 0 ^{- 68}}{9.109 \times 1 0 ^{- 31} \cdot 2.566 \times 1 0 ^{- 38}} = 5.29 \times 1 0^{- 11}$ m. $R (\infty) = \frac{m}{2 h c} (\frac{e ^{2}}{4 π ε _{0} ℏ})^{2} = \frac{m ( α c ) ^{2}}{2 h c} = \frac{13.606 eV}{h c} = \frac{13.606}{1.2398 \times 1 0 ^{- 4} eV \cdot cm} = 109740$ cm⁻¹.

Exercise 5.4.3beyond the lectures

A hydrogen atom in $n = 4$ can decay to lower levels by one or more jumps. How many different lines can it emit, and which are visible?

Solution

Jumps among 4 levels: $(2 4) = 6$ lines: $4 \to 3$ (1875 nm, IR), $4 \to 2$ (486 nm, visible), $4 \to 1$ (97 nm, UV), $3 \to 2$ (656 nm, visible), $3 \to 1$ (103 nm, UV), $2 \to 1$ (122 nm, UV). Two are visible.

Exercise 5.4.4beyond the lectures

Show that the orbital period of Bohr orbit $n$ goes as $n^{3}$ , and compute it for $n = 1$ .

Solution

$T_{n} = 2 π r_{n} / v_{n} \propto n^{2} / (1/ n) = n^{3}$ . $T_{1} = 2 π a_{0} / (α c) = 2 π \cdot 5.29 \times 1 0^{- 11} /2.19 \times 1 0^{6} = 1.52 \times 1 0^{- 16}$ s.

Exercise 5.4.5beyond the lectures

By how much is the deuterium H $_{α}$ line shifted from hydrogen’s?

Solution

$λ \propto 1/ R (M) \propto 1/ μ$ . $μ_{D} / μ_{H} = (1 + 1/1836) / (1 + 1/3670) = 1.000272$ , so $λ_{D} = 656.5/1.000272 = 656.3$ nm: 0.18 nm shorter, easily resolved by a good spectrometer.