de Broglie’s explanation of Bohr’s postulate
A wave on a ring
Bohr’s second postulate says the angular momentum is a whole-number multiple of h over 2π. It works beautifully, but why should nature pick these orbits? Bohr gave no reason.
Ten years later, in 1923, Louis de Broglie gave one. He proposed that a moving electron behaves like a wave, of wavelength λ equals h over m v. We met this in the chapter on dual nature.
First recall a sitar string, fixed at both ends. When it vibrates, only certain waves survive: those that fit a whole number of half wavelengths between the ends. These are standing waves.
Now bend the string into a circle. There are no ends, but the wave travels round and round and meets itself. The orbit plays the role of the string.
Suppose the circumference holds 4.4 wavelengths. After one trip the wave arrives out of step with itself: a crest lands where the last trip had a trough. Trip after trip, the waves cancel out.
Now make the circumference exactly 4 wavelengths. After each trip the wave lands crest on crest. It reinforces itself and forms a steady standing wave round the orbit, with fixed nodes.
So an electron wave can survive only on an orbit whose circumference is a whole number of wavelengths: 2π r n equals n λ, with n equal to 1, 2, 3 and so on.
In the first orbit, one wavelength fits. In the second, two. In the third, three. The orbit number n counts the wavelengths round the orbit.
Bohr’s rule from waves
Start from the standing-wave condition: 2π r n equals n λ.
Put in de Broglie’s wavelength, λ equals h over m v n. Then 2π r n equals n h over m v n.
Rearrange: m v n r n equals n h over 2π. This is exactly Bohr’s quantisation of angular momentum. His mysterious postulate is just the condition for a standing electron wave.
Let us check with numbers for the first orbit. r₁ is 0.529 × 10⁻¹⁰ metre, so the circumference 2π r₁ is about 3.3 × 10⁻¹⁰ metre.
The electron’s speed there is 2.18 × 10⁶ metres per second. Its de Broglie wavelength is h over m v: 6.63 × 10⁻³⁴, divided by 9.11 × 10⁻³¹ times 2.18 × 10⁶. That is also about 3.3 × 10⁻¹⁰ metre. Exactly one wavelength fits.
For the nth orbit, r grows as n², while v falls as 1 over n, so λ grows as n. The circumference is n² times bigger and λ is n times bigger, so n wavelengths fit.
This idea, that electrons are waves confined in the atom, grew into quantum mechanics, the full theory of atoms.
The Bohr model has limits. It works for hydrogen and for ions with just one electron, like He plus and Li two plus. It cannot handle atoms with two or more electrons, and it cannot explain why some lines are brighter than others.
Still, from one whole number n, it explains the stability of the atom, the energy levels, and every line of hydrogen’s spectrum.
Try it: fit the electron wave
- The wave starts with 4.4 wavelengths round the orbit. The 2nd and 3rd trips do not line up with the 1st: the wave cancels itself.
- Slide to a whole number. The trips merge into one standing wave with fixed nodes.
- Switch to “Bohr orbits” and step n from 1 to 6. The circumference grows as n², λ as n, so exactly n wavelengths fit.
- Check the readout mvr = nh/2π: Bohr’s rule falls out of the standing-wave condition.
Wavelength in the second orbit
How does the de Broglie wavelength of the electron in the second Bohr orbit of hydrogen compare with that in the first?
- It is the same
- It is twice as long
- It is four times as long
- It is half as long
Answer
B. λ = 2πr_n/n. Since r_n ∝ n², λ ∝ n²/n = n, so λ₂ = 2λ₁. Equivalently, v_n ∝ 1/n, so λ = h/mv ∝ n. Four times is the ratio of the radii (and circumferences), not of the wavelengths.
The third orbit
The radius of the third Bohr orbit of hydrogen is 4.76 Å, so its circumference is 29.9 Å. What is the de Broglie wavelength of the electron in this orbit?
- 3.3 Å
- 10.0 Å
- 29.9 Å
- 89.7 Å
Answer
B. 2πr₃ = 3λ, so λ = 29.9/3 ≈ 10.0 Å. Three wavelengths fit round the third orbit. 3.3 Å is the wavelength in the first orbit (dividing by n² instead of n); 29.9 Å forgets that n = 3 waves fit; 89.7 Å multiplies by n.
Summary
- de Broglie wavelength:
λ = h/mv - Standing wave on an orbit:
2πr_n = nλ, n = 1, 2, 3, … - Gives Bohr’s postulate:
mv_n r_n = nh/2π
- An electron wave survives on an orbit only if a whole number of wavelengths fits round it; otherwise it cancels itself.
- Putting λ = h/mv into 2πr = nλ gives exactly Bohr’s quantum condition L = nh/2π.
- In the nth orbit λ ∝ n while the circumference ∝ n², so n wavelengths fit. In the first orbit λ ≈ 2πr₁ ≈ 3.3 × 10⁻¹⁰ m.
- The Bohr model works only for one-electron (hydrogen-like) atoms and cannot explain line intensities; quantum mechanics replaced it.