IB Physics HLTopic 5 — Atomic & NuclearPaper 1 & 2E = −13.6/n2 eV~18 min read
Bohr’s Model of Hydrogen
The spectra proved atoms have discrete energy levels — but why? Niels Bohr found the answer for hydrogen, the simplest atom of all. He proposed that electrons can only orbit at certain fixed radii, where their angular momentum comes in whole-number packets. Decades later, de Broglie gave a stunning reason: the electron is also a wave, and only orbits that fit a whole number of wavelengths survive as standing waves. Together these ideas produce a single clean formula for hydrogen’s energy levels — and explain every line in its spectrum.
📘 What you need to know
In the Bohr model, electrons orbit only in certain fixed orbits with fixed radii
Hydrogen line spectra fall into series (Lyman → UV, Balmer → visible, Paschen → IR, etc.)
The discrete energy levels of hydrogen are given by E = −13.6 / n2 eV
Bohr proposed that angular momentum is quantised: L = nh / 2π
De Broglie: an electron has a wavelengthλ = h/p
A stable orbit must fit a whole number of wavelengths: nλ = 2πr (a standing wave)
Combining these gives the Bohr condition: nh/2π = mvr
Only orbits where the wave joins up in phase (constructive) are allowed
Energy levels of hydrogen
Bohr’s model gives hydrogen’s allowed electron energies as a simple formula in the quantum number n = 1, 2, 3, …
Hydrogen energy levelsE = −13.6 / n2 eVn = 1 is the ground state; larger n = higher (less negative) energy
Level n
Energy (eV)
Notes
1
−13.6
Ground state (most tightly bound)
2
−3.40
First excited state
3
−1.51
Second excited state
4
−0.85
Third excited state
∞
0
Ionised (electron free)
Notice the levels bunch up as n grows — the gap from n=1 to n=2 is huge (10.2 eV), but from n=3 to n=4 it’s tiny. That’s the 1/n2 in action. It’s also why the spectral lines within a series crowd together toward the series limit. And the energies are negative because a bound electron has less energy than a free one at 0 eV.
WE 1
An electron in a hydrogen atom makes a transition from the n = 4 level to the n = 2 level. Determine the frequency of the emitted photon. (h = 6.63 × 10−34 J s, 1 eV = 1.60 × 10−19 J)
Step 1 — energies of the two levelsE₄ = −13.6/4² = −0.85 eVE₂ = −13.6/2² = −3.40 eVStep 2 — energy of the photonΔE = E₄ − E₂ = (−0.85) − (−3.40) = 2.55 eVin joules: ΔE = 2.55 × (1.60 × 10⁻¹⁹) = 4.08 × 10⁻¹⁹ JStep 3 — frequency from E = hff = ΔE/h = (4.08 × 10⁻¹⁹) / (6.63 × 10⁻³⁴)f = 6.2 × 10¹⁴ HzThis is the blue-green Balmer line of hydrogen. Watch the double negative in Step 2: (−0.85) − (−3.40) = +2.55 eV. Convert to joules before dividing by h.
Quantised angular momentum
Bohr’s key postulate was that an electron’s angular momentum can’t take just any value — it’s restricted to whole-number multiples of h/2π.
Bohr’s quantisation of angular momentumL = nh / 2πn = 1, 2, 3, … • h = Planck’s constant • angular momentum L = mvr
So the angular momentum comes in fixed “steps” — L can be 1, 2, or 3 units of h/2π, but never 1.5. This quantisation is what forces the electron into fixed orbits.
The de Broglie standing wave
Why should angular momentum be quantised? De Broglie’s insight: the electron is also a wave, with a wavelength set by its momentum.
De Broglie wavelengthλ = h / p = h / mv
For the electron-wave to survive in a circular orbit, it must join up smoothly with itself — a standing wave. That only works if the circumference fits a whole number of wavelengths. Otherwise the wave overlaps out of phase and destructively cancels itself out.
Left: three whole wavelengths fit the circumference, so the electron-wave joins up smoothly — an allowed standing wave. Right: a non-whole number means the wave meets itself out of phase and cancels — a forbidden orbit.
Deriving the Bohr condition
Now combine the two ideas. The standing-wave requirement says the circumference is a whole number of wavelengths:
Standing-wave condition in a circular orbitnλ = 2πr
Substitute the de Broglie wavelength λ = h/mv:
Substitute λ = h/mv, then rearrangen(h/mv) = 2πr → nh/2π = mvrthis is the Bohr condition — and mvr is exactly the angular momentum L
Electron is a wave
standing wave nλ = 2πr
Whole-number orbits only
gives
nh/2π = mvr
This is the deep payoff of the whole topic: Bohr’s “angular momentum is quantised” isn’t an arbitrary rule — it follows from the electron being a wave. Only orbits that fit a whole number of de Broglie wavelengths make a stable standing wave, and those are exactly the orbits with L = nh/2π. Quantisation drops out of wave interference. That’s why examiners love to ask you to derive the Bohr condition from nλ = 2πr.
WE 2
Determine the speed of the electron in the first Bohr orbit (n = 1) of hydrogen. (m = 9.1 × 10−31 kg, r = 0.529 × 10−10 m, h = 6.63 × 10−34 J s)
Step 1 — start from the Bohr conditionnh/2π = mvr → v = nh / (2πmr)Step 2 — substitute n = 1v = (1)(6.63 × 10⁻³⁴) / [2π(9.1 × 10⁻³¹)(0.529 × 10⁻¹⁰)]v = 2.2 × 10⁶ m s⁻¹About 2.2 million m/s — fast, but still under 1% of the speed of light, so we don’t need relativity. Rearrange the Bohr condition for v, then plug in n=1. Keep all the powers of ten lined up to avoid slips.
WE 3
(a) Calculate the angular momentum of an electron in the n = 2 orbit of hydrogen. (b) An electron orbit fits exactly 3 de Broglie wavelengths around its circumference. State the value of n and explain what this means physically. (h = 6.63 × 10−34 J s)
(a) angular momentum at n = 2L = nh/2π = (2)(6.63 × 10⁻³⁴) / (2π)L = 2.1 × 10⁻³⁴ kg m² s⁻¹(b) 3 wavelengths around the orbitnλ = 2πr with 3 wavelengths → n = 3
This is a stable standing wave that joins up in phase.
n = 3; the electron-wave interferes constructively, so this orbit is allowedThe number of whole wavelengths that fit the orbit is the quantum number n. Three wavelengths → n = 3, a permitted orbit. Any non-whole number would cancel destructively and isn’t allowed.
Photon from a jump? Convert ΔE to J, then f = ΔE/h or λ = hc/ΔE.
Angular momentum?L = nh/2π.
Speed or radius? Use the Bohr condition nh/2π = mvr.
Standing wave?nλ = 2πr; the number of whole wavelengths = n.
Derive it? Put λ = h/mv into nλ = 2πr.
💡 Top tips
Hydrogen energy levels: E = −13.6/n2 eV. Ground state is n = 1.
Watch the double negative when finding a transition energy.
Angular momentum is quantised: L = nh/2π.
Derive the Bohr condition by substituting λ = h/mv into nλ = 2πr.
The number of de Broglie wavelengths around the orbit equals n.
Only whole-wavelength (in-phase, constructive) orbits are allowed.
⚠ Common mistakes
Sign errors in E = −13.6/n2 — the energies are negative
Botching the transition subtraction (double negatives)
Forgetting to convert eV to J before using E = hf
Writing L = nh instead of L = nh/2π
Thinking any orbit is allowed — only whole-wavelength standing waves are
Confusing the de Broglie λ = h/p with the photon E = hc/λ
Quick recap: Bohr’s model puts electrons in fixed orbits with quantised angular momentum L = nh/2π, giving hydrogen’s energy levels E = −13.6/n2 eV. De Broglie explained why: the electron is a wave (λ = h/p), and only orbits fitting a whole number of wavelengths (nλ = 2πr) form stable standing waves. Substituting one into the other gives the Bohr conditionnh/2π = mvr — quantisation emerging naturally from wave interference.
And that completes the Structure of the Atom chapter — from Rutherford’s gold foil all the way to Bohr’s quantised orbits and the electron’s wave nature. You can now describe any nucleus, explain atomic spectra, calculate photon energies and nuclear sizes, and derive the Bohr condition from first principles. Every idea here — energy levels, photons, the strong force — feeds directly into nuclear physics and quantum topics ahead. Superb work getting through the whole set.
Bohr model bending your brain?
Book a free meeting and we’ll work through E = −13.6/n2, quantised angular momentum, and the de Broglie standing-wave derivation.
