The Bohr model describes the atom as a tiny positive nucleus orbited by electrons confined to fixed, quantised energy levels labelled by a whole number n. An electron emits or absorbs light only when it jumps between two levels. For hydrogen, each level’s energy equals minus 13.6 electronvolts divided by n squared.
Hold a cheap diffraction grating up to a hydrogen discharge tube and you will not see a smooth rainbow. You see four sharp lines — red, cyan, blue, violet — floating on black, and nothing in between. Every hydrogen atom in the universe emits exactly those colours.
That stubborn fact broke classical physics. In 1913 a 27-year-old Dane fixed it with one audacious rule: electrons are simply not allowed to have any energy they like. The Bohr model is what came out — a model we now know is wrong in its picture, yet still right in its numbers.
What Is the Bohr Model?
The Bohr model is a picture of the atom in which electrons orbit the nucleus only at certain fixed radii, each with a fixed energy, and light is emitted or absorbed only when an electron jumps between them. Niels Bohr proposed it in 1913, and it won him the 1922 Nobel Prize in Physics.
Think of a staircase rather than a ramp. You can stand on step one or step two, but never at 1.6 — and every time you drop a step, you release exactly the energy difference between them, no more and no less.
Bohr grafted this quantum staircase onto Rutherford’s nuclear atom. Rutherford’s gold-foil experiment had already shown the atom is mostly empty space with a dense positive core, but that model had a fatal flaw, which the next section unpacks.

Bohr orbits drawn to scale. The radius grows as n2, so n = 3 is nine times the ground-state orbit.
The problem Bohr was solving
Classical electromagnetism says an accelerating charge radiates energy. An electron circling a nucleus is accelerating constantly, so it should spiral inwards and hit the nucleus in about a hundred-trillionth of a second.
Matter, obviously, does not do this. Bohr’s escape was to declare certain orbits stationary — allowed states in which, by fiat, the electron does not radiate at all. It was not derived from anything. It was postulated because it worked.
The Bohr Model Formula
The central Bohr model formula gives the energy of a hydrogen electron in the level numbered n:
- En — total energy of the electron in level n, in electronvolts (eV). 1 eV = 1.602 × 10-19 J.
- n — the principal quantum number: a positive whole number, n = 1, 2, 3, and so on. It is dimensionless.
- 13.6 eV — the ionisation energy of ground-state hydrogen, sometimes written E0 or the Rydberg energy.
The orbit radius follows the same quantum number, and grows far faster:
- rn — radius of the nth Bohr orbit, in metres (m).
- a0 — the Bohr radius, 5.29 × 10-11 m, which is the n = 1 orbit of hydrogen.
And the rule that links the atom to light — a jump from level ni down to level nf emits one photon:
- ΔE — energy released in the jump, in joules (J) or eV.
- h — the Planck constant, 6.626 × 10-34 J s (4.136 × 10-15 eV s).
- f — frequency of the emitted photon, in hertz (Hz).
- λ — wavelength of the emitted photon, in metres (m).
- c — speed of light, 299,792,458 m/s exactly.
In practice you will almost always use the shortcut hc = 1240 eV·nm. Divide 1240 by the photon energy in eV and you get the wavelength straight in nanometres — the single most time-saving trick in atomic physics. It rests on the same photon energy formula E = hf that Einstein used to explain the photoelectric effect.
Why the energy is negative
Zero energy is defined as a free electron, infinitely far from the nucleus and at rest. Any electron actually stuck to an atom has less energy than that, so its total energy is negative.
The number tells you the escape fee. At n = 1 the electron sits 13.6 eV below freedom, so you must supply 13.6 eV to ionise the atom.
How Does the Bohr Model Work?
The Bohr model works by combining three postulates: electrons move in circular orbits held by electrostatic attraction, only orbits with angular momentum equal to a whole multiple of h/2π are allowed, and a photon is emitted or absorbed exactly when the electron jumps between two allowed orbits.
Take them one at a time.
Postulate 1 — a classical orbit
The electron circles the proton, and the electrostatic pull supplies the centripetal force. Set Coulomb’s law equal to the centripetal force requirement and you get one equation linking speed to radius:
This half is pure Newtonian mechanics. Nothing quantum has happened yet.
Postulate 2 — angular momentum is quantised
Here is the leap. Bohr asserted that the electron’s angular momentum can only take discrete values:
Combine this with postulate 1 and the two unknowns, v and r, are pinned down completely. Out drops rn = n2a0, and from the total energy (kinetic plus potential) out drops En = -13.6/n2 eV. Nothing was fitted to the spectrum; the numbers came from e, me, h and ε0 alone.
That is the moment the model earns its reputation. A single arbitrary rule about angular momentum reproduced a spectrum that had been measured for fifty years and explained by nobody. OpenStax’s University Physics derivation walks through every algebraic step if you want to see it in full.
Postulate 3 — quantum jumps
An electron in an allowed orbit radiates nothing. It only emits light in the instant it drops to a lower orbit, and the photon carries away precisely the energy gap.
Because the gaps are fixed, the colours are fixed. A line spectrum is nothing more than a photograph of an atom’s energy ladder.
Hydrogen’s Energy Levels and Spectral Series
The energy levels crowd together as n rises, because dividing 13.6 by n2 shrinks fast. Between n = 1 and n = 2 there is a 10.2 eV chasm; between n = 5 and n = 6 there is a 0.17 eV crack.
Every downward jump that ends on the same level forms a named series. Jumps ending on n = 1 are the Lyman series, on n = 2 the Balmer series, on n = 3 the Paschen series.
Only the Balmer series lands in visible light. That is why hydrogen shows four visible lines while most of its spectrum hides in the ultraviolet and infrared parts of the electromagnetic spectrum.

Hydrogen energy levels drawn to scale from E = -13.6/n2 eV. Only Balmer transitions land in visible light.
| Transition | Series | ΔE (eV) | Bohr λ (nm) | Measured λ (nm) | Region |
|---|---|---|---|---|---|
| n = 2 to 1 | Lyman α | 10.20 | 121.6 | 121.57 | Ultraviolet |
| n = 3 to 2 | Balmer α | 1.889 | 656.5 | 656.28 | Red |
| n = 4 to 2 | Balmer β | 2.550 | 486.3 | 486.13 | Cyan |
| n = 5 to 2 | Balmer γ | 2.856 | 434.2 | 434.05 | Blue |
| n = 6 to 2 | Balmer δ | 3.022 | 410.3 | 410.17 | Violet |
| n = 4 to 3 | Paschen α | 0.661 | 1875 | 1875 | Infrared |
Look at how close those columns are — agreement to four significant figures from three postulates and no fitted parameters.
The tiny residual gap in the visible rows is not a failure of the model. Bohr predicts vacuum wavelengths, while the measured Balmer figures listed in NIST’s Atomic Spectra Database are air wavelengths, which run about 0.2 nm shorter because light slows slightly in air.
Real-World Examples of the Bohr Model
Quantised energy levels are not a laboratory curiosity. You are almost certainly looking at one right now.
- Neon signs and discharge lamps. Electricity kicks electrons up; they fall back and emit fixed colours. Neon’s orange-red glow is its energy ladder made visible.
- Sodium street lights. That flat yellow comes from one dominant transition at about 589 nm, which is why colours look washed out beneath them.
- Stellar spectroscopy. Astronomers identify hydrogen in a star billions of light-years away by finding the Balmer pattern in its light — then measure the redshift from how far those lines have slid.
- Fluorescent tubes. Mercury atoms emit ultraviolet photons from a specific jump; the phosphor coating converts them into visible light.
- Fireworks and flame tests. Lithium red, copper green, potassium lilac — each colour is a fingerprint of one element’s energy gaps.
Spectroscopy works precisely because no two elements share an energy ladder. It is the closest thing chemistry has to a barcode scanner.
Common Misconceptions About the Bohr Model
“Electrons orbit the nucleus like planets”
This is the single most persistent atomic-physics error, and the Bohr model is largely to blame for it. Electrons do not follow tracks. Modern quantum mechanics replaces the orbit with an orbital: a three-dimensional cloud giving the probability of finding the electron in each region of space.
The planetary analogy also gets the force wrong. Planets are held by gravity; electrons are held by the far stronger electrostatic attraction.
“Negative energy means something unphysical”
Negative energy simply means bound. The zero point is a free, motionless electron, so anything trapped in an atom sits below zero.
A common student slip is to conclude that n = 4 at -0.85 eV has less energy than n = 1 at -13.6 eV. It has more: -0.85 is the larger number, and the n = 4 electron is closer to escaping.
“Energy levels are evenly spaced”
Textbook diagrams drawn as a neat ladder cause this. The real spacing collapses as n rises, since E depends on 1/n2.
A useful magnitude check: the n = 1 to n = 2 gap is 10.2 eV, but n = 5 to n = 6 is only 0.17 eV — sixty times smaller. That crowding is why every spectral series piles up towards a series limit.
“The Bohr model is obsolete, so it’s useless”
Obsolete as a picture, indispensable as a tool. It still delivers hydrogen’s ionisation energy, the Bohr radius and every hydrogen wavelength to several significant figures, using arithmetic a school student can do.
Chemists still draw Bohr shells for a reason — for counting valence electrons and predicting reactivity, the full quantum treatment is overkill.
Where the Bohr Model Fails
The Bohr model fails for every atom with more than one electron, because it ignores electron–electron repulsion and has no way to handle it. Try it on neutral helium and the predicted spectrum is simply wrong.
The other cracks are just as telling:
- Fine structure. High-resolution spectroscopy splits single Bohr lines into close pairs. The model has no mechanism for this — it needs electron spin and relativity.
- Line intensities. Bohr tells you which wavelengths appear, never how bright each one is.
- The uncertainty principle. A definite orbit means a definite position and momentum at once, which quantum mechanics forbids outright.
- Magnetic and electric field effects. The Zeeman and Stark splittings need quantum numbers Bohr never introduced.
- Chemical bonding. Molecular shape and bond angles come from orbital geometry, which circular orbits cannot express.
Schrödinger’s 1926 wave equation replaced the orbits with wavefunctions and cured all of it. Reassuringly, solving it for hydrogen returns exactly Bohr’s energy levels — the right answer arrived at for the right reasons.
| Feature | Bohr model (1913) | Quantum model (1926) |
|---|---|---|
| Electron path | Definite circular orbit | Probability cloud (orbital) |
| Quantum numbers | n only | n, l, ml, ms |
| Works for | Hydrogen and one-electron ions | All atoms and molecules |
| Fine structure | Not predicted | Predicted (spin and relativity) |
| Hydrogen energies | -13.6/n2 eV | -13.6/n2 eV (same result) |
How the Bohr Model Relates to Photons, Matter Waves and Ionisation
Bohr’s third postulate is really Einstein’s photon hypothesis applied inside an atom. Light arrives and departs in indivisible packets of energy hf, which is exactly what the photoelectric effect had already forced physicists to accept in 1905.
The angular-momentum rule looked arbitrary for a decade — until 1924. Louis de Broglie proposed that electrons have a wavelength λ = h/mv, and suddenly Bohr’s condition meant something physical.
Fit a whole number of de Broglie wavelengths around the circumference — set 2πr = nλ — and you recover mvr = nh/2π exactly. The allowed orbits are the ones where the electron’s own wave closes on itself instead of cancelling out.
Ionisation ties the model to chemistry. Push n to infinity and E goes to zero, so the 13.6 eV needed to free a ground-state hydrogen electron is precisely its ionisation energy — a value you can look up in any periodic table and check against the formula.
Worked Problems
Show Solution
Solution:
Step 1: Use the Bohr energy formula, En = -13.6/n2 eV.
Step 2: Substitute n = 3. E3 = -13.6 / 32 = -13.6 / 9 eV.
Step 3: Divide. E3 = -1.5111 eV.
Answer: E3 = -1.51 eV (3 s.f.)
Show Solution
Solution:
Step 1: Ionisation means raising the electron to E = 0, so the energy needed is |E2|.
Step 2: E2 = -13.6 / 22 = -13.6 / 4 = -3.40 eV.
Step 3: Energy required = 0 – (-3.40 eV) = 3.40 eV.
Answer: 3.40 eV — exactly one quarter of the 13.6 eV needed from the ground state.
Show Solution
Solution:
Step 1: ΔE = E4 – E2 = (-13.6/16) – (-13.6/4) eV.
Step 2: ΔE = -0.850 + 3.400 = 2.550 eV.
Step 3: Convert to wavelength using λ = hc/ΔE with hc = 1240 eV·nm.
Step 4: λ = 1240 / 2.550 = 486.3 nm.
Answer: ΔE = 2.55 eV, λ = 486 nm — the cyan Balmer β line.
Show Solution
Solution:
Step 1: Use rn = n2 a0.
Step 2: Substitute n = 3. r3 = 32 × 5.29 × 10-11 m = 9 × 5.29 × 10-11 m.
Step 3: Multiply. r3 = 4.761 × 10-10 m.
Answer: r3 = 4.76 × 10-10 m = 0.476 nm — nine times the ground-state radius.
Show Solution
Solution:
Step 1: Find the photon energy. ΔE = 1240 / 97.3 = 12.74 eV.
Step 2: Absorption from n = 1, so ΔE = 13.6 (1/12 – 1/n2) eV.
Step 3: Rearrange. 1/n2 = 1 – (12.74 / 13.6) = 1 – 0.9370 = 0.0630.
Step 4: n2 = 1 / 0.0630 = 15.9, so n = 3.98.
Answer: n = 4. The small shortfall from a whole number is rounding in hc = 1240 eV·nm, not physics.
Show Solution
Solution:
Step 1: The shortest wavelength means the largest ΔE, which comes from a very high level falling to nf = 2. As n gets very large, 1/n2 tends to zero.
Step 2: ΔE = 13.6 (1/22 – 0) = 13.6 / 4 = 3.40 eV.
Step 3: λ = 1240 / 3.40 = 364.7 nm.
Answer: 365 nm (3 s.f.) — just past violet, in the near ultraviolet, which is why the visible Balmer lines bunch up towards the blue end.
Show Solution
Solution:
Step 1: Ground state, n = 1, Z = 2. E1 = -13.6 × 22 / 12 = -13.6 × 4 = -54.4 eV.
Step 2: E2 = -13.6 × 4 / 22 = -13.6 eV.
Step 3: ΔE = E2 – E1 = -13.6 – (-54.4) = 40.8 eV.
Step 4: λ = 1240 / 40.8 = 30.4 nm.
Answer: E1 = -54.4 eV and λ = 30.4 nm — extreme ultraviolet. Bohr still works here because He+ has only one electron.