How this instrument works
In 1913 Niels Bohr proposed that the electron in a hydrogen atom cannot orbit at any distance it likes, only at the handful of radii where its angular momentum equals a whole multiple of ħ, n·ħ. Combine that quantization rule with the ordinary balance between Coulomb attraction and centripetal force, and every allowed orbit collapses to one integer, the principal quantum number n = 1, 2, 3 …, and one energy: E_n = −13.6 ⁄ n² electronvolts. The constant 13.6 eV is the Rydberg energy, built entirely from the electron's mass and charge, Planck's constant, and the permittivity of free space — nothing about it was fitted to hydrogen's spectrum after the fact.
The minus sign marks a bound state: zero energy is defined as the electron sitting at rest infinitely far from the proton, so every occupied orbit sits below that, more negative the tighter it is bound. Because orbit radius grows as n² while Coulomb potential energy falls as 1/r, the two effects compound into a 1/n² energy law — the ground state at n = 1 is bound by the full 13.6 eV, the next shell at n = 2 by only a quarter of that, and by n = 10 the electron is barely held at all. Levels do not step down evenly like rungs on a ladder; they bunch tightly near zero as n grows, which is exactly the crowding astronomers see as a spectral series approaches its limit.
The formula is exact only for one electron orbiting one point charge. Swap hydrogen for a stripped ion carrying nuclear charge Z — He⁺, Li²⁺ — and it still works once −13.6 is multiplied by Z²; add a second electron, as in neutral helium, and electron-electron repulsion breaks the clean closed form entirely. Bohr's picture also hands the ground state one unit of orbital angular momentum, which the full quantum treatment later showed is wrong — hydrogen's true ground state has zero angular momentum — yet the energies themselves survive unchanged in the Schrödinger solution, which is why a chemistry student checking a textbook problem, an astronomer reading off which hydrogen line sits where in a stellar spectrum, and anyone confirming a quantum-mechanics derivation all still reach for this exact number.
- Enter Principal quantum number — the positive integer n identifying which allowed orbit the electron occupies; 1 is the ground state.
- Read Orbital energy, eV — the instrument evaluates E_n = −13.6/n² the instant n changes; no other input is needed.
- To find a transition energy, such as an emission or absorption line, compute Orbital energy, eV at two values of n and subtract one result from the other.
- Raise Principal quantum number toward large values to watch Orbital energy, eV approach zero — the ionization limit, where the electron is no longer bound.
- For a hydrogen-like ion rather than hydrogen itself, multiply the result by the square of the ion's nuclear charge, Z², since this instrument assumes Z = 1.
Worked example — the hydrogen ground state, n = 1
Set Principal quantum number to n = 1, hydrogen's ground state. The formula gives E_1 = −13.6 ⁄ 1² = −13.6 eV exactly, so Orbital energy, eV reads −13.6. That figure is not an approximation fitted after the fact; it is the same 13.6 eV that chemists call hydrogen's first ionization energy, the minimum energy needed to strip the electron away entirely, and Bohr's 1913 model reproducing it from a simple quantization postulate was the result that made physicists take his orbits seriously.
Change Principal quantum number to n = 2 and Orbital energy, eV drops in magnitude to −3.4 eV, since −13.6 ⁄ 2² = −3.4. Subtracting the two, −3.4 minus −13.6, gives 10.2 eV — the energy of the Lyman-alpha transition, the brightest line in hydrogen's ultraviolet spectrum, observed at a wavelength of 121.6 nanometres in everything from laboratory discharge tubes to the light of distant quasars.
Questions
Why is the orbital energy negative?
Because zero energy is defined as the electron at rest infinitely far from the nucleus — a fully ionized atom. An electron actually bound to the proton has less energy than that free state, so the result comes out negative, and the more negative the value, the more tightly held the electron is. The n = 1 ground state at −13.6 eV is hydrogen's most tightly bound configuration; every higher n sits closer to zero.
Does this formula apply to atoms other than hydrogen?
Only to one-electron systems: hydrogen itself, or ions stripped down to a single electron such as He⁺ or Li²⁺, provided −13.6 is multiplied by the square of the nuclear charge, Z². Add a second electron and electron-electron repulsion breaks the closed form — neutral helium's energy levels cannot be read off this way and need a numerical or perturbative treatment instead.
Why do the energy levels crowd together at higher n?
Because the energy falls off as 1/n², not in equal steps. The jump from n = 1 to n = 2 is 10.2 eV, but the jump from n = 9 to n = 10 is under 0.03 eV. Levels pile up ever closer to zero as n grows, converging on the ionization limit where the electron is no longer bound at all — they are not evenly spaced rungs on a ladder.
What is the physical meaning of the constant 13.6?
It is the Rydberg energy for hydrogen, the ionization energy of its ground state, built from the electron's mass and charge, Planck's constant, and the permittivity of free space — hc times the Rydberg constant, expressed in electronvolts. Bohr did not fit it to match hydrogen's spectrum; it fell out of his quantization postulate and matched the measured value, which is why the model was taken seriously immediately.
How do I get the energy of a spectral line from this?
Compute Orbital energy, eV for the lower and the upper level involved, then subtract. Absorption lifts the electron from a lower n to a higher one and consumes that much energy; emission drops it back down and releases a photon carrying exactly that energy. This is the same arithmetic behind the Rydberg formula used to predict hydrogen's Lyman, Balmer, and Paschen spectral series.
What does the Bohr model get wrong?
It assigns the ground state one unit of orbital angular momentum, ħ, when the true quantum-mechanical answer is zero, and it leaves no room for the s, p, d sublevels or the fine structure that spin-orbit coupling produces. Despite that, the energy levels themselves come out numerically identical to the full Schrödinger solution for hydrogen, which is why this simple formula still gives the right number fastest.