de-Broglie's Explanation of Bohr's Second Postulate of Quantisation

What this subtopic is examined on

• The standing-wave derivation of $L=nh/2\pi$ — the standard 2- or 3-mark answer, set repeatedly.
• Describing the circular standing-wave diagram.
• Numericals on the de Broglie wavelength of an electron in a given Bohr orbit, both by $\lambda=2\pi r_n/n$ and by $\lambda=h/\sqrt{2mK}$.
• Ratios of de Broglie wavelengths in two orbits.
• The limitations of Bohr's model, which NCERT places at the end of this section.

The puzzle

• Of Bohr's three postulates, the second is the most arbitrary: why should the angular momentum be restricted to integral multiples of $h/2\pi$, and not to something else? Bohr had no reason for it beyond the fact that it worked.
• Louis de Broglie answered it in 1923, ten years after Bohr's paper, using the wave nature of matter. C. J. Davisson and L. H. Germer verified the wave nature of electrons experimentally in 1927.

The argument (write it in these steps)

• Step 1. By de Broglie's hypothesis, a material particle of momentum $p$ has a wavelength $\lambda=\dfrac{h}{p}$. The electron in its Bohr orbit must therefore be treated not as a point particle at a definite place but as a particle wave spread around the whole orbit.
• Step 2. The analogy is a string. When a string is plucked a vast number of wavelengths are excited, but only those that have nodes at the fixed ends survive as standing waves; all others interfere destructively with themselves on reflection and their amplitudes drop to zero at once.
• Step 3. Apply the same resonance condition to a circular orbit. Here the wave, after one complete circuit, returns to its starting point. It will reinforce itself — form a standing wave — only if the total distance travelled, the circumference $2\pi r_n$, contains a whole number of wavelengths:

$$2\pi r_n=n\lambda,\qquad n=1,2,3,\dots$$

• Step 4. Put in the de Broglie wavelength of the electron, $\lambda=\dfrac{h}{p}=\dfrac{h}{mv_n}$ (the speed being much less than $c$, the momentum is simply $mv_n$):

$$2\pi r_n=\dfrac{nh}{mv_n}\qquad\Longrightarrow\qquad mv_nr_n=\dfrac{nh}{2\pi}\qquad\Longrightarrow\qquad L=\dfrac{nh}{2\pi}$$

• Step 5. This is exactly Bohr's second postulate. So the quantisation of angular momentum is not an extra assumption at all: it is the resonance condition for the electron's own wave. Only resonant standing waves persist, so only certain orbits and only certain energies are possible.
• Sentence to finish with: the quantised electron orbits and energy states of Bohr's model are a consequence of the wave nature of the electron.

The diagram, described in words

• Draw a circle to represent the $n$th Bohr orbit, with the nucleus marked at the centre.
• Along the circle draw a smooth wavy curve — crests outside the circle, troughs inside — that closes exactly on itself, so that going once round the circumference you count exactly $n$ complete wavelengths. NCERT's Fig. 12.8 draws the case $n=4$: four de Broglie wavelengths fit into the circumference.
• If the wave did not close on itself, the returning wave would be out of step with the outgoing one and would cancel it — that orbit is not allowed.

Standard results and numericals

• de Broglie wavelength of the electron in the $n$th orbit:

$$\lambda_n=\dfrac{2\pi r_n}{n}=\dfrac{2\pi n^2a_0}{n}=2\pi n a_0=2\pi n(0.529\ \text{\text{Å}})=3.32\,n\ \text{\text{Å}}$$

• So $\lambda_n\propto n$, and the ratio for two orbits is simply the ratio of their quantum numbers:

$$\dfrac{\lambda_{n_a}}{\lambda_{n_b}}=\dfrac{n_a}{n_b}$$

• Example: second excited state ($n=3$) to third excited state ($n=4$) gives $\dfrac{\lambda_3}{\lambda_4}=\dfrac{2\pi\times3\times0.529\ \text{\text{Å}}}{2\pi\times4\times0.529\ \text{\text{Å}}}=\dfrac{3}{4}$.
• The other route, from the kinetic energy — use it whenever the question gives an energy:

$$\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mK}},\qquad K=-E_n=\dfrac{13.6}{n^2}\ \text{eV}$$

• $n=2$ (first excited state), $K=3.4$ eV $=3.4\times1.6\times10^{-19}$ J:

$$\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times3.4\times1.6\times10^{-19}}}\approx0.67\ \text{nm}=6.7\ \text{\text{Å}}$$

• $n=3$ (second excited state), $K=\dfrac{13.6}{9}=1.51$ eV $=2.41\times10^{-19}$ J:

$$\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times2.41\times10^{-19}}}\approx1\times10^{-9}\ \text{m}=1\ \text{nm}$$

• Both routes agree: $\lambda_n=3.32n$ Å gives $6.6$ Å for $n=2$ and $10.0$ Å for $n=3$.

• Radius from the quantisation condition, the hydrogen-like form: from $mvr=\dfrac{nh}{2\pi}$ and $\dfrac{mv^2}{r}=\dfrac{kZe^2}{r^2}$, eliminate $v$ to get

$$r=\dfrac{n^2h^2}{4\pi^2mkZe^2}$$

• The circumference of a Bohr orbit is always an integral multiple of the de Broglie wavelength — the 1-mark fill-in-the-blank.

Limitations of Bohr's model (NCERT places these here)

• It applies only to hydrogenic (single-electron) atoms — hydrogen, singly ionised helium, doubly ionised lithium. It cannot be extended even to a two-electron atom such as helium.
• The reason: the model contains only the electric force between the nucleus and one electron. In a many-electron atom each electron also interacts with all the others, and because the charges and separations are of the same order, the electron-electron force is comparable in size to the electron-nucleus force. (Contrast the solar system, where planet-planet gravity is negligible beside the sun's pull — which is why the planetary analogy fails here.)
• It predicts the frequencies of the hydrogen lines correctly but says nothing about their relative intensities: some visible lines are strong and others weak, meaning some transitions are more favoured than others, and Bohr's model cannot account for this.
• The orbital picture — a definite radius and a definite momentum at the same instant — is inconsistent with the uncertainty principle. Modern quantum mechanics replaces Bohr's orbits by regions in which the electron is likely to be found.
• It uses a single quantum number $n$, whereas a state is really characterised by four ($n$, $l$, $m$, $s$); for a pure Coulomb potential the energy happens to depend only on $n$, which is why Bohr's energies come out right.
• Why it is still taught: it rests on just three postulates yet accounts for almost all the general features of the hydrogen spectrum, and it keeps a great deal of classical physics intact.