Bohr Model Calculator and Hydrogen Emission Spectrum Simulator

Point a prism at a glowing hydrogen discharge tube and you don't see a rainbow. You see four thin, isolated lines — red, cyan, blue-violet, violet — floating on a field of black. Nothing in between. Classical physics had no way to explain why an atom would only glow at those exact colours and nowhere else. It got worse: classical electromagnetism says an accelerating charge radiates energy continuously, so an electron circling a nucleus should lose energy, spiral inward, and crash — matter itself shouldn't hold together for more than an instant.

In 1913, Niels Bohr fixed both problems with one strange rule: electrons are only allowed on certain specific orbits, and they don't radiate while sitting on them. Eleven years later, Louis de Broglie explained why those specific orbits and no others — every electron is also a wave, and only a whole number of wavelengths can fit around a circle without cancelling itself out. This page puts both pieces in your hands: pick any transition between hydrogen's first six energy levels and watch the emitted wavelength update live, then drag a single slider and watch de Broglie's standing-wave condition build — and break — the quantization rule from scratch.

What Is the Bohr Model of the Atom?

The Bohr model describes a hydrogen atom as a single electron orbiting a proton on one of a discrete ladder of allowed circular paths, numbered by an integer $n = 1, 2, 3, \ldots$ called the principal quantum number. Three rules define the whole model:

1. Stationary states. An electron on an allowed orbit does not radiate energy, no matter how classical electromagnetism says it should.
2. Quantized angular momentum. Only orbits where $mvr = n\hbar$ are allowed — angular momentum comes in whole multiples of $\hbar = h/2\pi$.
3. Quantum jumps. An electron changes orbits by absorbing or emitting a single photon whose energy exactly matches the gap between the two levels: $E_\text{photon} = |E_\text{final} - E_\text{initial}|$.

None of this was derived from anything deeper in 1913 — Bohr essentially guessed rule 2 because it made the numbers come out right. That's the part de Broglie explained a decade later.

Why Are Electron Orbits Quantized? The de Broglie Standing-Wave Explanation

In 1924, Louis de Broglie proposed that every particle carries a wavelength, not just light: $\lambda = h/p = h/(mv)$. Take that seriously for an orbiting electron, and Bohr's quantization rule stops looking arbitrary. Picture a wave sent around a circular track and left to interfere with itself, lap after lap — for the wave to survive at all, the circumference has to hold a whole number of wavelengths:

$2\pi r = n\lambda \qquad (n = 1, 2, 3, \ldots)$

Substitute de Broglie's $\lambda = h/(mv)$ directly into that condition:

$2\pi r = n \cdot \frac{h}{mv} \quad\Longrightarrow\quad mvr = n\frac{h}{2\pi} = n\hbar$

That's Bohr's angular-momentum rule, derived rather than assumed. Quantization isn't a strange extra ingredient bolted onto the atom — it's what standing waves do on a ring, and an electron happens to be a wave.

Drag the slider slowly through an integer and watch what happens at the seam. At $n = 3.00$ the wave laps itself perfectly — three full crests, closed loop, stable pattern. Nudge it to $n = 3.20$ and the wave that "should" close instead runs into itself out of phase; the second panel shows exactly why, with the closure gap tracing a smooth curve that touches zero only at the integers. Every non-integer value is a wave destroying itself. That's a physically real reason those orbits can't exist, not a bookkeeping rule someone imposed from outside.

For the ground state ($n=1$), the numbers work out to a radius of 52.9 pm (the famous Bohr radius, $a_0$) and a de Broglie wavelength of 332.4 pm — exactly $2\pi \times 52.9 \text{ pm}$, the orbit's own circumference, confirming the identity baked into the simulator above.

What Is the Formula for Hydrogen's Energy Levels?

Balance the Coulomb attraction against the centripetal force needed to hold the electron in a circular orbit, then plug in the quantization condition $mvr = n\hbar$ and solve for the radius. Two things fall out together: the orbit radius grows as $n^2$, and the total energy (kinetic plus electrostatic potential) becomes:

$E_n = -\frac{13.6 \text{ eV}}{n^2}$

The energy is negative because the electron is bound — you have to add energy to pull it away from the proton entirely. As $n \to \infty$, $E_n \to 0$: a free electron, no longer part of the atom.

Why Is the Ground-State Energy Exactly −13.6 eV?

That number isn't arbitrary. It's built entirely from the electron mass, the electron charge, and Planck's constant — no adjustable parameters anywhere. Bohr's 1913 calculation of $13.6\text{ eV}$ from first principles matched hydrogen's already-measured ionization energy almost exactly, and that's what convinced physicists to take a semi-classical model this strange seriously in the first place. The energy needed to fully remove the electron from the ground state ($n=1 \to n=\infty$) works out to exactly $13.6 - 0 = 13.6$ eV — hydrogen's ionization energy.

How Do You Calculate the Wavelength of an Emitted Photon? (The Rydberg Formula)

When an electron falls from a higher level $n_i$ to a lower level $n_f$, the atom loses energy and a photon carries the difference away:

$\Delta E = E_{n_f} - E_{n_i}, \qquad E_\text{photon} = |\Delta E|, \qquad \lambda = \frac{hc}{E_\text{photon}}$

Rewritten in terms of wavelength directly, this is the Rydberg formula, which Swedish physicist Johannes Rydberg wrote down empirically in 1888 — a quarter-century before Bohr explained where it actually came from:

$\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right), \qquad R \approx 1.097 \times 10^7 \text{ m}^{-1}$

Pick a transition above and watch both panels move together: the term diagram shows exactly which two rungs of the ladder the electron jumped between, and the spectrum strip shows where that photon actually lands — deep in the UV for anything ending on the ground state, comfortably visible only for transitions ending on $n_f = 2$, and out in the infrared for anything ending on $n_f = 3$ or higher.

How Do You Use the Rydberg Formula to Find Wavelength?

1. Identify the initial level $n_i$ and final level $n_f$ of the transition.
2. Compute $E_{n_i}$ and $E_{n_f}$ from $E_n = -13.6/n^2$.
3. Find the photon energy: $E_\text{photon} = |E_{n_f} - E_{n_i}|$.
4. Convert to wavelength: $\lambda = hc / E_\text{photon}$, using $hc \approx 1240$ eV·nm as a convenient shortcut.

What Are the Lyman, Balmer, and Paschen Series?

Every transition that lands on the same final level $n_f$ belongs to the same named series, and each series lives in its own region of the spectrum:

The Balmer series is the only one your eye can see at all, and it's the one that gave hydrogen away spectroscopically in the first place. Johann Balmer noticed the pattern in 1885 just by staring at four measured wavelengths, with zero idea why any of it worked. Those four visible lines — Hα at 656.5 nm (red), Hβ at 486.3 nm (cyan), Hγ at 434.2 nm (blue-violet), and Hδ at 410.3 nm (violet) — are the transitions $3\to2$, $4\to2$, $5\to2$, and $6\to2$. You can reproduce every one of them in the simulator above by holding $n_f = 2$ and stepping $n_i$ from 3 to 6.

How Is the Bohr Model Different from the Photoelectric Effect?

It's easy to conflate these, since both involve photons and electrons — but photons and electrons interact very differently here than in the photoelectric effect. The photoelectric effect is a one-way, absorption-only event: a single photon hits a metal surface and completely knocks a free electron out into the vacuum, governed by a fixed work function $\phi$ that doesn't care about discrete levels at all — $KE_\text{max} = hf - \phi$. The Bohr model describes something that stays put: a bound electron hopping between two specific rungs inside a single atom, in either direction, governed by that atom's own discrete energy ladder rather than any threshold. One is a one-shot ejection. The other is a two-way transition between bound states.

That same discrete ladder is also the entire reason lasers work — see how lasers work for how stimulated emission uses precisely this kind of level structure to build coherent light.

What Are the Limitations of the Bohr Model?

The model gets hydrogen's energies exactly right, and then it stops generalizing. Give it a second electron and it falls apart, because electron-electron repulsion breaks the clean single-particle picture the whole derivation depends on — so much for helium. It has nothing to say about fine structure, electron spin, or why some spectral lines show up brighter than others on a real spectrometer. And the orbits themselves were never literally circular paths to begin with; full quantum mechanics, via the Schrödinger equation, replaces them with probability clouds — orbitals with no sharply defined radius at all.

What survives, remarkably, is the energy formula itself. For any single-electron ion — helium with one electron stripped away, doubly-ionized lithium, and so on — $E_n = -13.6 Z^2/n^2$ eV (with $Z$ the nuclear charge) comes out exactly right even under the full quantum treatment. None of this touches the nucleus, by the way — Bohr's levels are purely electronic. The nucleus can be separately unstable, which is the entirely different physics behind our half-life calculator: radioactive decay changes which element you have, while an electron transition only changes how much energy the atom is carrying.

Worked Examples for Physics Exams

Example 1: The Red Hydrogen Line (Balmer-α)

Find the wavelength of light emitted when an electron in a hydrogen atom falls from $n=3$ to $n=2$.

$E_3 = -1.51$ eV, $E_2 = -3.40$ eV, so $E_\text{photon} = |{-3.40} - ({-1.51})| = 1.89$ eV. Then $\lambda = 1240 / 1.89 = 656$ nm — the red line you'd see through a prism. Check it in the simulator above: set $n_i = 3$, $n_f = 2$, and read off 656.5 nm.

Example 2: Exciting a Ground-State Atom into the Lyman Series

What's the minimum-energy UV photon that can excite a ground-state hydrogen atom ($n=1$) up to $n=3$?

$E_3 = -1.51$ eV, $E_1 = -13.60$ eV, so $E_\text{photon} = |{-1.51} - ({-13.60})| = 12.09$ eV, giving $\lambda = 1240/12.09 = 102.6$ nm. Hydrogen gas is essentially transparent to visible light, but it absorbs this exact ultraviolet line strongly — the same transition, run in reverse.

Example 3: The de Broglie Wavelength of an Accelerated Electron

An electron starts from rest and is accelerated through a potential difference of 150 V. Find its de Broglie wavelength.

The kinetic energy gained is $KE = eV$, and since $KE = p^2/(2m)$, the momentum is $p = \sqrt{2m \cdot eV}$. Work through the numbers and you get $\lambda = h/p \approx 100$ pm $= 1.00$ Å — about the size of an atom. Same relation as everywhere else on this page, just applied to a free electron instead of one bound to a nucleus. In 1927, Clinton Davisson and Lester Germer accelerated electrons through 54 V and measured a de Broglie wavelength of 0.167 nm by diffracting them off a nickel crystal — the first direct experimental proof that de Broglie's hypothesis was real physics, not just a convenient assumption. Davisson won the 1937 Nobel Prize for it.