In short

Bohr's three postulates for the hydrogen atom (1913):

  1. The electron orbits the proton in circular orbits, held by the Coulomb force, just as a classical particle would — but only certain orbits are allowed.
  2. The allowed orbits are those for which the electron's angular momentum is quantised: L = m_e v r = n \hbar, where n = 1, 2, 3, \ldots and \hbar = h/(2\pi).
  3. In an allowed orbit the electron does not radiate, despite being accelerated. It radiates only when jumping between orbits, emitting or absorbing a single photon of energy hf = |E_\text{final} - E_\text{initial}|.

Consequences, derived below:

Radius of the n-th orbit:

r_n = \frac{n^2 \hbar^2}{m_e k e^2} = n^2 a_0, \qquad a_0 = 0.5292\ \text{Å}.

Energy of the n-th level:

E_n = -\frac{m_e k^2 e^4}{2 \hbar^2}\cdot\frac{1}{n^2} = -\frac{13.6\ \text{eV}}{n^2}.

Photon emitted on the transition n_i \to n_f:

hf = E_i - E_f = 13.6\ \text{eV}\cdot\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right).

The ground state (n = 1) has r_1 = a_0 and E_1 = -13.6 eV. The ionisation energy of hydrogen is |E_1| = 13.6 eV — measurable in any school lab via a gas-discharge experiment, and one of the most accurately confirmed numbers in physics.

Bohr's model is wrong in detail (it ignores electron spin, it cannot handle atoms beyond hydrogen, and its picture of little classical orbits is not how electrons actually exist in atoms). But it gets the hydrogen energy levels exactly right for the purposes of JEE Advanced, and it was the first theory to explain why atoms emit discrete spectral lines. Everything since is an elaboration on this foundation.

By 1912 physics had a big problem it could not ignore. Rutherford's gold-foil experiment had just shown that the atom is a tiny, dense, positively charged nucleus surrounded by mostly empty space with electrons somewhere in it. Everyone immediately pictured this the way any reasonable classical physicist would: the electron orbits the nucleus, just as a planet orbits the sun, with the Coulomb attraction standing in for gravity. The picture was natural, neat, and — according to Maxwell's electromagnetism — completely impossible.

The reason is that any accelerating charged particle radiates electromagnetic waves. An electron in circular orbit is accelerating (changing direction). So it must radiate. The radiation carries away energy, so the orbit must shrink. Classical electrodynamics predicts — with no wiggle room — that the electron spirals into the nucleus in about 10^{-10} seconds, emitting a continuous spread of frequencies as it goes. Atoms should collapse almost instantly; the spectrum of hydrogen should be a smooth blur across all frequencies.

Neither thing is true. Hydrogen atoms have been stable for the 13.8 billion years since the Big Bang. And when you look at the light from a hydrogen discharge tube through a prism, you see a handful of sharp, isolated lines at wavelengths so precise they can be measured to eight decimal places. No continuum. No smooth blur. Just thin bright lines at particular wavelengths, always the same wavelengths, the same for every hydrogen atom in the universe.

Into this puzzle Niels Bohr, a 28-year-old postdoctoral fellow in Manchester, proposed in 1913 a fix that amounted to throwing half of classical physics out of the window. He kept Newton's mechanics and Coulomb's force, but he added a new rule: the electron's angular momentum can only take values that are integer multiples of \hbar = h/(2\pi). Everything else — the radii, the energies, the spectral lines, the number 13.6 eV — follows from that rule. It worked. And decades later, when de Broglie and Schrödinger built proper quantum mechanics, they showed that Bohr's rule was exactly right for hydrogen and not an arbitrary postulate at all — it was the condition for the electron's matter wave to form a standing wave around the orbit.

The article on Rutherford's nuclear model ends at the stability problem. This article picks up from there. You will derive every one of Bohr's famous results — the orbit radii r_n = n^2 a_0, the energy levels E_n = -13.6/n^2 eV, the emission formula — from Newton's second law, Coulomb's law, and one new rule about angular momentum.

Bohr's three postulates — stated carefully

Write them down in the order Bohr himself proposed them.

Postulate 1 — Circular orbits held by Coulomb attraction. The electron moves in a circular orbit of radius r around the proton. The centripetal force needed to keep it in this orbit is supplied by the Coulomb attraction between them. Classically,

\frac{m_e v^2}{r} = \frac{k e^2}{r^2},

where k = 1/(4\pi\varepsilon_0) = 8.988 \times 10^9 N·m²/C² is Coulomb's constant, e is the electron charge, and v is the orbital speed.

Postulate 2 — Angular momentum quantisation. Not every orbit is allowed. The electron's angular momentum about the nucleus can only take the discrete values

L = m_e v r = n \hbar, \qquad n = 1, 2, 3, \ldots

The integer n is the principal quantum number. \hbar = h/(2\pi) = 1.0546\times 10^{-34} J·s is the reduced Planck constant.

Postulate 3 — No radiation in stationary orbits; a photon per transition. In any allowed orbit the electron is in a stationary state and does not radiate, even though it is accelerating. (This directly contradicts Maxwell's theory — Bohr's solution was simply to decree it so.) Radiation happens only when the electron jumps from one allowed orbit to another. A jump from level n_i to level n_f emits (or absorbs, if n_f > n_i) a single photon whose energy equals the energy gap:

hf = E_{n_i} - E_{n_f}.

Postulates 1 and 2, combined, determine the radii and speeds of the allowed orbits. Postulate 3 then determines what the atom does next.

Deriving the radius of the n-th orbit

Step 1. The Coulomb-centripetal balance (postulate 1).

\frac{m_e v^2}{r} = \frac{k e^2}{r^2}.

Multiply both sides by r:

m_e v^2 = \frac{k e^2}{r}. \tag{1}

Why: the net inward force equals m_e v^2 / r (centripetal); the Coulomb attraction provides it. This is the force-balance form of Newton's second law for a circular orbit, identical in structure to Kepler's for planets but with ke^2 in place of GMm.

Step 2. The quantisation condition (postulate 2).

m_e v r = n\hbar \quad\Rightarrow\quad v = \frac{n\hbar}{m_e r}. \tag{2}

Why: solving the angular-momentum quantisation for v lets you eliminate v from the force balance. This is the trick that turns two equations into a single equation for r.

Step 3. Substitute (2) into (1) to eliminate v.

m_e \left(\frac{n\hbar}{m_e r}\right)^2 = \frac{k e^2}{r}
\frac{n^2 \hbar^2}{m_e r^2} = \frac{k e^2}{r}.

Why: expand v^2 = n^2 \hbar^2 / (m_e r)^2. The m_e in the denominator of v^2 combines with the m_e outside to leave one factor of m_e in the denominator of the left side.

Step 4. Solve for r.

Multiply both sides by r^2 and divide by k e^2:

r = \frac{n^2 \hbar^2}{m_e k e^2}.
\boxed{\;r_n = \frac{n^2 \hbar^2}{m_e k e^2} = n^2 a_0\;}

where

a_0 \equiv \frac{\hbar^2}{m_e k e^2} = 5.292\times 10^{-11}\ \text{m} = 0.5292\ \text{Å}

is the Bohr radius — the radius of the ground-state (n=1) orbit.

Why: r grows as n^2. Doubling the quantum number quadruples the orbit size. The ground-state radius a_0 is a combination of five fundamental constants (Planck, electron mass, electron charge, vacuum permittivity, speed of light never enters because no magnetism was used), and it is the natural unit of atomic length. Every length in atomic physics is expressed in Bohr radii.

Check the numbers. \hbar^2 = 1.112\times10^{-68} J²·s². m_e k e^2 = 9.109\times10^{-31} \cdot 8.988\times10^9 \cdot (1.602\times10^{-19})^2 = 2.102\times10^{-58}. Divide: a_0 = 1.112\times10^{-68}/2.102\times10^{-58} = 5.29\times10^{-11} m. Consistent.

Deriving the energy of the n-th level

Step 1. The total energy is kinetic plus potential.

E = \tfrac12 m_e v^2 - \frac{k e^2}{r}.

Why: kinetic energy is positive, and Coulomb potential energy with a fixed zero at infinity is negative for an attractive interaction. A bound state has total energy negative — you would have to do positive work to separate the electron from the proton.

Step 2. Use equation (1), m_e v^2 = k e^2 / r, to rewrite the kinetic energy.

\tfrac12 m_e v^2 = \frac{k e^2}{2r}.

Why: this is the virial theorem for a 1/r force. The kinetic energy is exactly half the magnitude of the potential energy. It holds for circular orbits in any inverse-square attractive force — gravity, Coulomb, or anything with the same functional form.

Step 3. Sum them up.

E = \frac{k e^2}{2r} - \frac{k e^2}{r} = -\frac{k e^2}{2r}.

Why: the kinetic energy is positive (half of ke^2/r); the potential energy is negative (whole of ke^2/r); their sum is negative (half of ke^2/r). This is the total energy of the bound electron at radius r, expressed cleanly in terms of the orbit radius alone.

Step 4. Substitute r = r_n = n^2 a_0 = n^2 \hbar^2 / (m_e k e^2).

E_n = -\frac{k e^2}{2} \cdot \frac{m_e k e^2}{n^2 \hbar^2} = -\frac{m_e k^2 e^4}{2 \hbar^2}\cdot\frac{1}{n^2}.
\boxed{\;E_n = -\frac{m_e k^2 e^4}{2 \hbar^2}\cdot \frac{1}{n^2} = -\frac{13.60\ \text{eV}}{n^2}\;}

Why: E_n < 0 always (the electron is bound), and |E_n| decreases as n^2 (higher levels are less tightly bound). The constant 13.6 eV is the Rydberg energy for hydrogen — it sets the scale of all atomic binding.

Step 5. Verify the Rydberg numerically.

E_\text{R} = \frac{m_e k^2 e^4}{2 \hbar^2} = \frac{9.109\times10^{-31}\cdot(8.988\times10^9)^2\cdot(1.602\times10^{-19})^4}{2\cdot(1.0546\times10^{-34})^2}.

Numerator: 9.109\times10^{-31} \cdot 8.079\times10^{19} \cdot 6.585\times10^{-76} = 4.846\times10^{-87}. Denominator: 2\cdot 1.112\times10^{-68} = 2.224\times10^{-68}. Ratio: 2.179\times10^{-18} J. Convert: 2.179\times10^{-18}/1.602\times10^{-19} = 13.60 eV. ✓

The energy level diagram

The allowed energies form a ladder rising from E_1 = -13.6 eV to E_\infty = 0. The gaps between successive rungs shrink rapidly as n grows.

E_1 = -13.6,\ E_2 = -3.40,\ E_3 = -1.51,\ E_4 = -0.85,\ E_5 = -0.54\ \text{eV}\ldots
Hydrogen energy level diagram with transitionsA vertical energy axis running from about minus 14 eV at the bottom to 0 eV at the top. Horizontal lines mark the levels n = 1 at minus 13.6 eV, n = 2 at minus 3.4 eV, n = 3 at minus 1.51 eV, n = 4 at minus 0.85 eV, and n = infinity at 0 eV. Arrows point downward between several pairs of levels, each labelled with the spectral series name: Lyman (to n=1), Balmer (to n=2), Paschen (to n=3).0 eV-13.6 eVn = ∞ (ionisation)n = 5, −0.54 eVn = 4, −0.85 eVn = 3, −1.51 eVn = 2, −3.40 eVn = 1, −13.6 eV (ground)Lyman (UV)Balmer (visible)Paschen (IR)
Energy levels of hydrogen. The levels crowd together as $n$ grows and accumulate at $E = 0$. Arrows show a few of the downward transitions: any transition ending at $n = 1$ is a **Lyman** line (UV); ending at $n = 2$ is a **Balmer** line (visible); ending at $n = 3$ is a **Paschen** line (IR). Every line in the hydrogen spectrum maps to one arrow on this diagram.

Emission and absorption — where the spectral lines come from

An electron in level n_i can drop to a lower level n_f < n_i by emitting a photon whose energy equals the gap:

hf = E_{n_i} - E_{n_f} = 13.6\ \text{eV}\cdot\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right).

Converting frequency to wavelength via f = c/\lambda:

\frac{1}{\lambda} = \frac{13.6\ \text{eV}}{hc}\cdot\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) = R \cdot\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right),

where R = 13.6\ \text{eV}/(hc) = 1.097\times 10^7\ \text{m}^{-1} is the Rydberg constant. This is the formula the spectroscopist Johannes Rydberg had fit to hydrogen data experimentally in 1888 — 25 years before Bohr's model. Bohr's achievement was to derive it from first principles.

The absorption process is the reverse: a hydrogen atom in the ground state can absorb a photon of exactly the right energy to kick it from n_f = 1 to some n_i > 1. Photons at other energies — between the allowed transition frequencies — pass straight through. That is why a cool cloud of hydrogen in space absorbs starlight at precisely the same wavelengths that a hot hydrogen discharge emits. Dark lines in the absorption spectrum of a star at 656 nm, 486 nm, 434 nm mean there is cool hydrogen somewhere between the star and the telescope — the kind of fingerprint that lets Indian astronomers at the Vainu Bappu Observatory in Tamil Nadu and at the Indian Astronomical Observatory in Hanle, Ladakh identify hydrogen in galaxies halfway across the observable universe.

Interactive — watch the level and emission change with n

Interactive: radius and energy of the n-th Bohr orbitA two-panel schematic. The left panel shows the hydrogen energy-level curve E equals minus 13.6 divided by n squared as a function of n, with the allowed integer values as points on the curve. A draggable red point on the n axis selects an integer n from 1 to 10. Readouts display n, the orbit radius r-n in angstrom, the energy E-n in electronvolts, and the wavelength of the transition from n to n minus 1 (or from n to 1 for the ground-state jump) in nanometres.principal quantum number $n$energy $E_n$ (eV)012468100−3.4−6.8−10.2−13.6$E_n = -13.6/n^2$drag the red point
Drag the red point to select a value of $n$. The curve shows the energy levels $E_n = -13.6/n^2$; the readouts give the orbit radius $r_n = n^2 a_0$, the energy $E_n$, and the wavelength $\lambda$ of the photon emitted when an electron falls from $n$ down to the ground state ($n = 1$). At $n = 2$ the photon is at 121.6 nm — Lyman alpha, deep UV. At $n = 3$, 102.6 nm (Lyman beta). As $n \to \infty$, $\lambda \to 91.2$ nm — the Lyman series limit.

The de Broglie picture — why quantisation of L is not arbitrary

Bohr's quantisation rule L = n\hbar looks like it was pulled out of thin air. A decade later de Broglie gave it a simple, beautiful interpretation.

If every particle has an associated wavelength \lambda = h/p, then the electron in its orbit is a wave of wavelength \lambda = h/(m_e v). For the wave to fit around the circle without destructively interfering with itself, the circumference must be an integer number of wavelengths:

2\pi r = n\lambda = \frac{n h}{m_e v}.

Rearrange:

m_e v r = \frac{nh}{2\pi} = n\hbar.

That is exactly Bohr's quantisation condition. Bohr was demanding that the electron's matter wave form a standing wave around the orbit. Orbits that do not satisfy the standing-wave condition are not actually "forbidden" by some mystery rule — they are orbits in which the electron wave would interfere destructively with itself and so not exist as a self-consistent solution. The de Broglie hypothesis article showed the hint; this is where it cashes in.

Standing matter waves around Bohr orbitsThree circular orbits drawn side by side. The first has a sinusoidal standing wave fitting exactly 2 wavelengths around the circumference, labelled n = 2. The second fits exactly 3 wavelengths, labelled n = 3. The third fits 4 wavelengths, labelled n = 4. A fourth orbit shows a mismatched wave — the ends do not meet — labelled as forbidden.n = 2n = 3n = 4an integer number of de Broglie wavelengths fits around the orbit
Bohr's quantisation rule restated: the electron's matter wave must form a closed standing wave around the orbit. The circumference of the $n$-th orbit contains exactly $n$ de Broglie wavelengths. The forbidden orbits are those where the wave would not close up on itself — it would destructively interfere.

Worked examples

Example 1: Wavelength of the Lyman-α line

The strongest emission line of hydrogen in the ultraviolet is Lyman-α, the photon emitted when an electron falls from n_i = 2 to n_f = 1. Compute its wavelength and verify it lies in the ultraviolet.

Step 1. Energy difference between the levels.

\Delta E = E_2 - E_1 = -\frac{13.6}{4} - \left(-\frac{13.6}{1}\right) = -3.40 + 13.60 = 10.20\ \text{eV}.

Why: E_1 = -13.6 eV, E_2 = -3.40 eV, and the emitted photon carries away the energy gap. The electron is dropping from a higher (less negative) level to a lower (more negative) level, so energy is released as radiation.

Step 2. Convert to joules (optional — but the hc = 1240 eV·nm trick lets you skip).

\lambda = \frac{hc}{\Delta E} = \frac{1240\ \text{eV·nm}}{10.20\ \text{eV}} = 121.6\ \text{nm}.

Why: the handy number hc = 1240 eV·nm turns a photon-energy-to-wavelength calculation into one division. You saw this in the photons article.

Step 3. Classify the wavelength. Visible light spans roughly 400 to 700 nm. UV spans 10 to 400 nm. At 121.6 nm the Lyman-α photon is deep in the vacuum ultraviolet — so deep that it is absorbed by ordinary air and can only be observed from above the atmosphere (for example, by UV telescopes on spacecraft).

Step 4. Cross-check with Rydberg's formula.

\frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 1.097\times10^7\cdot \frac{3}{4}\ \text{m}^{-1} = 8.228\times 10^6\ \text{m}^{-1}.
\lambda = \frac{1}{8.228\times10^6} = 1.215\times 10^{-7}\ \text{m} = 121.5\ \text{nm}. \checkmark

Why: the two methods — energy-gap-and-hc or Rydberg's formula — must give the same answer. They do, to the rounding precision of 13.6 eV.

Result. \lambda_{\text{Ly-}\alpha} = 121.6 nm.

Lyman-alpha transition on the hydrogen energy diagramTwo horizontal lines representing the n equals 1 and n equals 2 energy levels. An arrow pointing downward from n equals 2 to n equals 1 is labelled 10.2 electronvolts and lambda equals 121.6 nanometres.n = 1, E = −13.6 eVn = 2, E = −3.4 eVΔE = 10.2 eVλ = 121.6 nm (UV)
The Lyman-α transition: an electron falls from $n = 2$ to $n = 1$, releasing a single photon of 10.2 eV. This photon is what would ionise cold hydrogen in interstellar space, and it is the main channel through which young stars produce the UV background that lights up galactic nurseries.

What this shows. A pure calculation, using only Bohr's energy formula and Planck's constant, reproduces a spectral line wavelength that spectroscopists had been measuring since the 1890s. Every line in the hydrogen emission spectrum — 121.6 nm Lyman-α, 102.6 nm Lyman-β, 656.3 nm Hα (Balmer-α), 486.1 nm Hβ, 434.0 nm Hγ, 410.2 nm Hδ — can be computed this way. Bohr's prediction matched them all.

Example 2: Orbit radius and electron speed at n = 3

For the electron in the n = 3 orbit of hydrogen, compute (a) the orbit radius, (b) the orbital speed, and (c) compare the speed with the speed of light (to check the non-relativistic assumption).

Step 1. Radius from r_n = n^2 a_0.

r_3 = 3^2 \cdot 0.5292\ \text{Å} = 9 \cdot 0.5292 = 4.76\ \text{Å} = 4.76 \times 10^{-10}\ \text{m}.

Why: the n^2 scaling means excited orbits are much larger than the ground state. n = 3 is already four times the Bohr radius, and n = 10 would be a hundred times.

Step 2. Speed from the quantisation condition m_e v r = n\hbar.

v = \frac{n\hbar}{m_e r_n} = \frac{n\hbar}{m_e\cdot n^2 a_0} = \frac{\hbar}{n\, m_e a_0}.

Why: substituting r_n = n^2 a_0 lets you combine the factors of n; the speed scales as 1/n. Excited electrons move more slowly than ground-state ones.

Plug in: \hbar = 1.0546\times10^{-34} J·s, m_e = 9.109\times10^{-31} kg, a_0 = 5.292\times10^{-11} m, n = 3.

v = \frac{1.0546\times10^{-34}}{3 \cdot 9.109\times 10^{-31}\cdot 5.292\times10^{-11}} = \frac{1.0546\times10^{-34}}{1.446\times10^{-40}} = 7.29\times 10^5\ \text{m/s}.

Step 3. Ratio v/c.

\frac{v}{c} = \frac{7.29\times10^5}{3.0\times10^8} = 2.43 \times 10^{-3}.

Why: the ground-state electron (n = 1) has v_1/c = \alpha = 1/137 = 7.3\times10^{-3}, where \alpha is the fine-structure constant. At n = 3 the speed is a third of that — 0.24% of the speed of light. Non-relativistic mechanics is accurate to about (v/c)^2/2 \sim 3\times 10^{-6}, which is well below the resolution of this calculation.

Result. r_3 = 4.76 Å, v_3 = 7.29\times10^5 m/s, v_3/c = 2.4\times 10^{-3}.

First three Bohr orbits of hydrogen to scaleThree concentric dashed circles representing the n equals 1, n equals 2, and n equals 3 orbits around a central nucleus. Radii are in ratio 1 to 4 to 9. Each orbit is labelled with its radius in angstrom.pn=1: 0.53 Ån=2: 2.12 Ån=3: 4.76 Åradii in ratio 1 : 4 : 9 (n²)
The first three Bohr orbits, drawn to scale (with the nucleus enlarged for visibility). Radii go as $n^2$: the $n = 3$ orbit is nine times the ground-state radius, still only about half a nanometre — an enormous size for an electron cloud but still smaller than a single DNA base pair.

What this shows. The Bohr atom is tiny but not point-like. The ground-state orbit is half an ångström across; an excited orbit at n = 10 would be 50 Å — big enough that two such atoms could bump into each other and disturb each other's levels. Highly excited hydrogen ("Rydberg atoms") with n \sim 100 have orbit sizes of a few hundred nanometres — big enough that you can see them with an ordinary optical microscope. Indian atomic-physics groups at the Raman Research Institute have trapped Rydberg atoms to study this directly.

Common confusions

Where this fits in the arc of physics history

Bohr's 1913 paper was written in Manchester during a postdoctoral visit with Rutherford. It came out only a year after Rutherford had announced the nuclear model. The two results together opened the door for the quantum revolution. For the next decade Bohr's model was the only quantum theory of the atom — extended, patched, and debated. When Schrödinger's equation arrived in 1926, it explained Bohr's results as a special case and went far beyond them. By 1930 the old-quantum-theory Bohr atom was obsolete as a research tool, but it remained (and remains) the canonical first encounter with atomic structure for every physics student everywhere. JEE and NEET both test it — in fact they test essentially no atomic physics beyond it. The standing-wave picture from de Broglie arrived in 1924 and retrofitted Bohr's quantisation rule with a physical reason.

One piece of Indian science history sits right in the middle of this decade. In 1924, Satyendra Nath Bose at Dhaka University sent Einstein a derivation of Planck's blackbody law based on treating photons as indistinguishable quantum particles. Einstein extended the argument to a gas of atoms and predicted what became known as Bose-Einstein condensation. Bose-Einstein statistics underlies how atomic levels are populated in thermal equilibrium — the distribution of electrons across Bohr's levels when a hydrogen gas is at temperature T. Bohr supplies the level positions; Bose supplies the statistics. Together they give you the actual observed spectrum of a hot hydrogen cloud, and that is why astrophysicists can read hydrogen column densities off a stellar absorption-line profile to three significant figures.

Going deeper

If you came for the JEE-relevant story — postulates, radius, energy, emission formula, Lyman-α — you have it. What follows is the rigorous layer: hydrogenic ions, reduced mass, fine structure, why Bohr breaks down for helium, and a sketch of the Schrödinger correction.

Hydrogen-like ions

A single-electron ion with nuclear charge Ze (helium-plus He^+, lithium-doubly-plus Li^{2+}, etc.) is handled by the same derivation with e^2 \to Z e^2 in the Coulomb attraction. The results scale as

r_n = \frac{n^2 a_0}{Z}, \qquad E_n = -\frac{Z^2\cdot 13.6\ \text{eV}}{n^2}.

For He^+ (Z = 2): the ground-state radius is a_0/2 = 0.26 Å, and the ionisation energy is 4 \times 13.6 = 54.4 eV. This was confirmed spectroscopically well before any deeper theory existed.

Reduced mass

The derivation above assumed the proton is infinitely massive and fixed at the centre. In reality both proton and electron orbit their common centre of mass. The correct replacement is

m_e \to \mu = \frac{m_e m_p}{m_e + m_p} = m_e\cdot\frac{1}{1 + m_e/m_p}.

Since m_e/m_p \approx 1/1836, the correction to E_n is about 0.054% — small but measurable. The hydrogen-deuterium isotope shift (deuterium has a different reduced mass because its nucleus has twice the mass) was one of the ways physicists first detected deuterium in a sample.

Fine structure

The Bohr model predicts each level n is a single sharp line. High-resolution spectroscopy shows that each line is actually split into a few closely spaced components — fine structure. The splittings come from two effects that Bohr left out:

  1. Relativistic kinetic energy. The electron's true kinetic energy is E_k = \sqrt{(pc)^2 + (m_ec^2)^2} - m_e c^2, not p^2/2m_e. The correction is of order \alpha^2 \approx 5\times 10^{-5} times E_n.
  2. Spin-orbit coupling. The electron's intrinsic spin interacts with the magnetic field it experiences in its own rest frame as the proton moves around it. This further splits levels with the same n but different total-angular-momentum quantum number j.

Adding these corrections to Bohr's formula gives the Dirac fine-structure formula:

E_{nj} = -\frac{13.6\ \text{eV}}{n^2}\left[1 + \frac{\alpha^2}{n^2}\left(\frac{n}{j + 1/2} - \frac{3}{4}\right)\right].

You can still read the dominant -13.6/n^2 term; fine structure is a small correction on top of it.

Why Bohr fails for helium

A helium atom has two electrons. The would-be Bohr energy would include electron-nucleus attractions (2\cdot ke^2/r_i each) and an electron-electron repulsion (ke^2/r_{12}). The last term is not central — it does not permit the separation of variables that made the hydrogen problem tractable. No amount of postulating quantised orbits for individual electrons captures the correlations between the two electrons. The full quantum theory (Schrödinger's equation with the two-electron wavefunction) handles this via approximation methods; the simple Bohr picture does not.

From Bohr to Schrödinger

In the Schrödinger wave picture, the electron is described by a wave function \psi(\vec r) that satisfies

-\frac{\hbar^2}{2 m_e}\nabla^2 \psi - \frac{ke^2}{r}\psi = E \psi.

Solving this three-dimensional differential equation (separable in spherical coordinates) gives the energy eigenvalues

E_n = -\frac{13.6\ \text{eV}}{n^2} \qquad (n = 1, 2, 3, \ldots)

— exactly Bohr's formula, now as a mathematical theorem rather than a postulate. The eigenfunctions \psi_{n\ell m} are not circular orbits; they are three-dimensional probability clouds with three quantum numbers (n, \ell, m). The \ell = 0 state is spherically symmetric (the famous "s-orbital"); higher \ell states have angular lobes. None of them are circles. The price of getting the right physical picture is giving up Bohr's picture. The pay-off is that the same method generalises to helium, molecules, solids — everything.

Correspondence principle

Bohr himself formulated the correspondence principle: in the limit of large quantum numbers, quantum mechanics reproduces classical mechanics. At high n, the spacing between adjacent levels becomes small compared to the level energy, and the spectrum of emitted frequencies approaches the classical orbital frequency of the electron. You can check this: the frequency of the transition n+1 \to n at large n is approximately

f_{n\to n-1} \approx \frac{2\cdot 13.6\ \text{eV}}{h \cdot n^3},

while the classical orbital frequency at the n-th orbit works out to exactly the same expression. Quantum mechanics and classical mechanics agree at large n — as they must, since classical mechanics is what you see in everyday macroscopic objects.

Where this leads next