F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Quantum MechanicsThe Bohr Model and Atomic Spectra
F1.1 F3.2 F2.1 F2.2About 75 minutes
Learning goals- I can describe emission and absorption line spectra and explain why classical physics could not account for them.
- I can state Bohr's postulates and use Eₙ = −13.6 eV/n² to calculate the energies and wavelengths of photons emitted or absorbed by hydrogen.
- I can explain Bohr's quantized orbits using de Broglie standing waves (2πr = nλ).
- I can evaluate the strengths and limits of the Bohr model and describe how it changed scientific thinking.
Words to know
line spectrum
A spectrum made of separate bright lines (emission) or dark lines (absorption) at specific wavelengths, characteristic of an element.
energy level
One of the allowed energies of an electron in an atom.
ground state
The lowest energy level (n = 1 for hydrogen).
excited state
Any energy level above the ground state.
principal quantum number (n)
The whole number 1, 2, 3, ... that labels an energy level in the Bohr model.
ionization energy
The energy needed to remove an electron completely from an atom in a given state; 13.6 eV for hydrogen in its ground state.
Balmer series
Hydrogen's spectral lines produced by transitions down to n = 2; four of them are visible.
Every element has a barcode
Heat a gas of one element until it glows, or run an electric current through it (as in a neon sign), and look at its light through a prism or grating. Instead of a continuous rainbow like a blackbody, you see a few sharp bright lines at particular wavelengths: an emission line spectrum. Shine white light through a cool gas of the same element and dark lines appear at exactly those same wavelengths: an absorption spectrum.
Each element's pattern is unique, like a barcode. Astronomers use these barcodes to learn what stars are made of. In fact, helium was first identified from an unexplained yellow line in the Sun's spectrum in 1868, decades before it was found on Earth.
If you are taking Chemistry (SCH4U), you met line spectra and the Bohr model in the lesson Light, Line Spectra and the Bohr Model. Here we focus on the physics: where the energy levels come from, how to calculate with them, and how matter waves explain them.
Why classical physics failed
By 1911 Rutherford's gold-foil experiment had shown that an atom is a tiny, dense, positive nucleus with electrons around it. Picture the electron orbiting like a planet. But an orbiting charge is accelerating, and Maxwell's theory says accelerating charges radiate electromagnetic waves. The electron should lose energy, spiral into the nucleus in a tiny fraction of a second, and emit a smear of all frequencies on the way.
Two predictions, both wrong: atoms are stable, and they emit sharp lines, not a smear.
Bohr's model of hydrogen (1913)
The Danish physicist Niels Bohr combined Rutherford's nucleus with Planck's and Einstein's quanta. His postulates were:
- The electron moves only in certain allowed circular orbits, called stationary states. In these it does not radiate, even though it is accelerating.
- Each allowed orbit has a definite energy. For hydrogen, Eₙ = −13.6 eV / n², with n = 1, 2, 3, ... and radii rₙ = (0.0529 nm) n².
- The atom emits or absorbs a single photon only when the electron jumps between levels, and the photon carries the energy difference: hf = E(upper) − E(lower).
The energies are negative because the electron is bound: you must add energy to pull it free (to E = 0 at n = ∞). From the ground state that takes 13.6 eV, the ionization energy of hydrogen.
Bohr's formula reproduced the known hydrogen lines to high accuracy. The visible ones (the Balmer series) come from drops to n = 2. Drops to n = 1 give the ultraviolet Lyman series; drops to n = 3 give the infrared Paschen series.
Worked exampleExample 1: The red line of hydrogen
Find the energy and wavelength of the photon emitted when a hydrogen electron drops from n = 3 to n = 2.
- E₃ = −13.6 ÷ 9 = −1.511 eV; E₂ = −13.6 ÷ 4 = −3.400 eV.
- Photon energy: E = E₃ − E₂ = −1.511 − (−3.400) = 1.889 eV.
- λ = 1240 eV·nm ÷ 1.889 eV = 656 nm.
1.89 eV, 656 nm: the red H-alpha line that gives glowing hydrogen clouds in space their pinkish-red colour.
| Transition | Photon energy (eV) | Wavelength (nm) | Region |
|---|
| 3 → 2 | 1.89 | 656 | red |
| 4 → 2 | 2.55 | 486 | blue-green |
| 5 → 2 | 2.86 | 434 | violet-blue |
| 6 → 2 | 3.02 | 410 | violet |
| 2 → 1 | 10.2 | 122 | ultraviolet |
| 4 → 3 | 0.661 | 1876 | infrared |
Worked exampleExample 2: Absorption and ionization
A hydrogen atom is in its ground state. (a) Can it absorb a 10.2 eV photon? (b) A 11.0 eV photon? (c) What is the longest-wavelength photon that can ionize it?
- (a) E₂ − E₁ = −3.40 − (−13.6) = 10.2 eV. This matches a jump from n = 1 to n = 2, so yes.
- (b) No level difference from n = 1 equals 11.0 eV (10.2 eV to n = 2; 12.09 eV to n = 3). A photon must be absorbed whole, so no.
- (c) Ionizing from n = 1 needs 13.6 eV or more: λ = 1240 ÷ 13.6 = 91.2 nm.
(a) yes (b) no, the photon passes by (c) 91.2 nm or shorter. Any photon with 13.6 eV or more can ionize; the extra becomes the freed electron's kinetic energy.
de Broglie explains Bohr's orbits
Bohr had simply assumed which orbits were allowed. Eleven years later, de Broglie's matter waves gave a reason. If the electron is a wave travelling around the nucleus, it can only form a stable standing wave if a whole number of wavelengths fits exactly around the circle:
2πr = nλ, with λ = h/(mv)
If it doesn't fit, the wave interferes destructively with itself and cannot persist. Substituting gives mvr = nh/(2π): the electron's angular momentum is quantized, which is exactly the condition Bohr had used. So the allowed orbits are the ones where the electron's matter wave fits, much like the standing waves on a guitar string you studied in Grade 11 (only certain wavelengths fit between the fixed ends).
Worked exampleExample 3: Checking the fit for n = 1
In the n = 1 orbit (r = 0.0529 nm), Bohr's model gives the electron a speed of 2.19 × 10⁶ m/s. Find its de Broglie wavelength and compare with the orbit's circumference.
- λ = h/(mv) = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 2.19 × 10⁶) = 3.32 × 10⁻¹⁰ m = 0.332 nm.
- Circumference = 2πr = 2π(0.0529 nm) = 0.332 nm.
They are equal: exactly one wavelength fits around the n = 1 orbit, as 2πr = nλ requires for n = 1.
Strengths and limits of the Bohr model (F1.1)
Strengths: it explained hydrogen's spectrum precisely, explained why atoms are stable and have a size, and introduced quantized energy levels in atoms, which the Franck-Hertz experiment (1914) confirmed by showing that electrons lose energy to mercury atoms only in lumps of about 4.9 eV.
Limits: it works only for one-electron atoms or ions (H, He⁺, Li²⁺); it cannot explain why some lines are brighter than others, how lines split in magnetic fields, or chemical bonding. And electrons do not really move in neat circular orbits. In 1925-1926, Heisenberg and Schrödinger replaced it with full quantum mechanics, where the electron is described by a wave function that gives the probability of finding it in each place (the orbitals of chemistry; see the next lesson). The Bohr model remains a useful stepping stone: a classic case of a model that is wrong in detail but captured key truths and pointed the way forward.
Just for funHelium was discovered in the Sun's spectrum in 1868 and named after Helios, the Greek god of the Sun. It was not found on Earth until 1895. That is a long way to travel to discover a balloon gas: about 150 million kilometres.
Hands-on activityBuild a cereal-box spectroscope
You need: an empty cereal box, an old CD or DVD, scissors (use them carefully, or ask a grown-up), tape, and aluminium foil.
- Cut a slot about 1 cm wide near one end of the box top. Cover it with two pieces of foil taped so they leave a very thin slit (about 1 mm) between them.
- Cut a 3 cm square viewing window at the other end of the box, on the top, angled away from the slit.
- Tape the CD inside the box under the viewing window, shiny side up, tilted about 45° so light from the slit reflects off it to your eye.
- Point the slit at different lights and look through the window: an old-style fluorescent tube or compact fluorescent bulb (CFL), an LED bulb, a phone screen, and light from a bright cloudy sky. Never point it at the Sun.
What to look for: a fluorescent light shows several sharp bright lines (some from mercury atoms, including a strong green line near 546 nm and a violet-blue line near 436 nm) on top of the glow from the tube's coating. Daylight from clouds gives a continuous rainbow (with faint dark absorption lines you probably cannot resolve). White LEDs give a smooth band that is strongest in blue. You are seeing the difference between blackbody-like light and atomic line spectra.
Your turn
1Find the energy and wavelength of the photon emitted in the 4 → 2 transition in hydrogen.
E₄ = −0.850 eV, E₂ = −3.400 eV. ΔE = 2.55 eV; λ = 1240 ÷ 2.55 = 486 nm (blue-green).
2What is the radius of the n = 3 orbit in hydrogen?
r₃ = 0.0529 × 3² = 0.476 nm.
3How much energy is needed to ionize hydrogen from the n = 2 state? What is the longest wavelength that can do this?
From −3.40 eV to 0: 3.40 eV. λ = 1240 ÷ 3.40 = 365 nm (UV).
4A hydrogen atom in the ground state absorbs a 12.09 eV photon. To which level does the electron jump?
E = −13.6 + 12.09 = −1.51 eV, which is E₃, so n = 3.
5From n = 4, how many different photon energies can be emitted as the electron makes its way down to n = 1 (by any route)?
Possible jumps: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1: 6 different photon energies.
6Why are the Lyman lines all in the ultraviolet?
Every Lyman transition ends at n = 1, so its energy is at least E₂ − E₁ = 10.2 eV, giving λ ≤ 122 nm, which is ultraviolet.
7Why do absorption lines appear at the same wavelengths as emission lines for the same element?
Both come from jumps between the same pairs of energy levels. Absorbing a photon of energy E(upper) − E(lower) moves the electron up; emitting the same energy brings it down.
More practice: warm-up
1What is the energy of hydrogen's ground state?
3To which level do Balmer transitions end?
4Who proposed the model of the atom with quantized orbits in 1913?
5What is the condition for a de Broglie standing wave on a circular orbit?
More practice: on your own
1Calculate the wavelength of the 5 → 3 transition and name the region of the spectrum.
ΔE = −0.544 − (−1.511) = 0.967 eV; λ = 1240 ÷ 0.967 = 1280 nm, infrared (Paschen series).
2A hydrogen atom in the ground state is hit by a photon of 15.0 eV. What happens? Find the kinetic energy of the freed electron.
15.0 eV > 13.6 eV, so the atom is ionized. E(k) = 15.0 − 13.6 = 1.4 eV.
3Explain why classical physics predicted that a Rutherford atom would collapse.
An orbiting electron is constantly accelerating (changing direction). Maxwell's theory says accelerating charges radiate energy, so the electron would lose energy and spiral into the nucleus almost instantly, giving off a continuous spread of frequencies.
4Which visible Balmer line has the shortest wavelength among 3 → 2, 4 → 2, 5 → 2 and 6 → 2? Why?
6 → 2 (410 nm). It has the largest energy drop of the four, so the highest photon energy and shortest wavelength.
5The ion He⁺ has one electron and a nuclear charge twice that of hydrogen; its levels are Eₙ = −54.4 eV/n². Find the wavelength of its 2 → 1 photon.
E₂ = −13.6 eV, E₁ = −54.4 eV; ΔE = 40.8 eV; λ = 1240 ÷ 40.8 = 30.4 nm (extreme UV).
Challenge
1Show that the series limit (shortest wavelength) of the Balmer series is 365 nm, and explain what "series limit" means.
The series limit is the transition from n = ∞ (E = 0) to n = 2 (E = −3.40 eV): ΔE = 3.40 eV, λ = 1240 ÷ 3.40 = 365 nm. As n increases, the lines get closer together and approach this wavelength but never pass it.
2Use 2πr = nλ and λ = h/(mv) to derive mvr = nh/(2π). Then find the electron's speed in the n = 2 orbit (r = 0.2116 nm).
2πr = nh/(mv) → mvr = nh/(2π). ✓ For n = 2: v = nh/(2πmr) = 2(6.63 × 10⁻³⁴) ÷ (2π × 9.11 × 10⁻³¹ × 2.116 × 10⁻¹⁰) = 1.09 × 10⁶ m/s (half the n = 1 speed).
3A gas of hydrogen atoms, all in the ground state, is exposed to white light containing all wavelengths from 90 nm to 700 nm. Which wavelengths will be missing (absorbed) from the light that passes through? Give the first three.
Only Lyman transitions start from n = 1: 122 nm (1 → 2), 103 nm (1 → 3), 97.3 nm (1 → 4), and more crowding toward 91.2 nm, plus wavelengths shorter than 91.2 nm (ionization). No visible lines are absorbed, because Balmer absorption would require atoms already in n = 2.
Quiz
1. An electron in hydrogen drops from n = 3 to n = 2. The emitted photon has energy
2. Why are atomic energy levels given negative values?
3. A ground-state hydrogen atom is struck by an 11.0 eV photon. The atom
4. According to de Broglie, the allowed Bohr orbits are those for which
5. Which series of hydrogen lines lies in the visible region?
6. Which is a limitation of the Bohr model?
Big idea (F1.1, F3.2): atoms have quantized energy levels, so they emit and absorb only photons whose energies match level differences. Bohr's 1913 model captured this for hydrogen; de Broglie's matter waves explained why only certain orbits are allowed.
How to help: build the cereal-box spectroscope together and compare the lights in your home. Ask your teen to explain why a fluorescent tube shows lines but daylight shows a smooth rainbow.
Common difficulties:
- Sign errors with negative energies. Photon energy is always the positive difference E(upper) − E(lower).
- Thinking an atom can absorb any photon above 10.2 eV. Below the ionization energy, only exact level differences are absorbed.
- Forgetting to square n in Eₙ = −13.6/n² or rₙ = 0.0529n².
- Treating Bohr orbits as the real picture. Modern quantum mechanics uses probability clouds (orbitals).
Safety: the spectroscope must never be pointed at the Sun. Supervise scissor use.