The Bohr Model of the Atom

Niels Bohr proposed that electrons orbit the nucleus only at specific, allowed radii. This single idea explained something important. Hydrogen atoms emit light at only a few sharp colors, not a continuous rainbow. The model builds directly on an earlier idea. Light energy comes in discrete packets, from Quantized Energy and Photons. Bohr applied that quantization to the electron itself. The Bohr model is not the final picture of the atom. Real electrons are described by wave-like Electron Shells and Orbitals. But it is the first model that gets the hydrogen spectrum numerically right. It does this using nothing more than algebra and one bold postulate. Every number in this article is checkable by hand. That is exactly why the model remains the standard entry point into atomic structure. We will build the model from a force balance. Then we quantize the angular momentum. Next we derive the energy-level formula. Finally, we use it to predict the exact wavelengths of hydrogen's spectral lines.

Key takeaways

Electrons orbit the nucleus only in allowed circular orbits labeled by an integer $n = 1, 2, 3, \dots$
Angular momentum is quantized: $m_e v r = n\hbar$ , which forces the radius and energy to take discrete values.
Orbit energies follow $E_n = -\dfrac{2.18 \times 10^{-18}\ \text{J}}{n^2}$ for hydrogen.
Light is absorbed or emitted only when an electron jumps between two allowed levels, with photon energy $\Delta E = E_{\text{final}} - E_{\text{initial}}$ .
The model reproduces the Rydberg formula for hydrogen's line spectrum to four significant figures.
The Bohr model fails for multi-electron atoms and cannot explain orbital shapes — that job belongs to quantum mechanics.

Key idea: Confining an electron's angular momentum to whole-number multiples of $\hbar$ automatically restricts it to a discrete ladder of energies, and jumps between rungs produce the sharp lines seen in atomic spectra.

The core insight

An atom's electron cannot have just any energy — it sits on one rung of a fixed energy ladder, and light is the receipt for a jump between rungs.

The physical picture

Picture the hydrogen atom as a single electron circling a proton, held in orbit by the same kind of centripetal reasoning used for planets — except the attractive force here is electrical (Coulomb's law), not gravitational. Classically, an orbiting electron could have any radius and any energy, and — because it is accelerating — classical electromagnetism says it should continuously radiate energy and spiral into the nucleus within a fraction of a second. Atoms obviously do not do this. Bohr's fix was to declare, by postulate, that only certain orbits are allowed: those in which the electron's angular momentum is an integer multiple of $\hbar = h/2\pi$ . Every other radius is simply forbidden. Because the radius is quantized, the energy is quantized too, and the electron can only gain or lose energy in the specific jumps between allowed rungs. Each jump absorbs or emits a photon of exactly that energy, which is why hydrogen's emission spectrum is a set of sharp lines rather than a smear of color.

Figure

Allowed circular orbits in the Bohr model, radii growing as n²

The core relations

The allowed radii and energies of the hydrogen atom follow from three linked equations.

m_e v r_n = n\hbar, \qquad n = 1, 2, 3, \dots

where: $m_e$ is the electron mass, $v$ is its orbital speed, $r_n$ is the orbit radius, $n$ is the (positive integer) principal quantum number, and $\hbar = h/2\pi$ .

r_n = n^2 a_0, \qquad a_0 = 0.529 \times 10^{-10}\ \text{m}

where: $a_0$ is the Bohr radius, the smallest allowed orbit ( $n=1$ ) for hydrogen.

E_n = -\frac{2.18 \times 10^{-18}\ \text{J}}{n^2}

where: $E_n$ is the total energy of an electron in orbit $n$ ; the negative sign means the electron is bound to the nucleus, with $E_n \to 0$ as $n \to \infty$ (a free electron).

\Delta E = E_{\text{photon}} = h\nu = \frac{hc}{\lambda}

where: $h = 6.626 \times 10^{-34}\ \text{J}\,\cdot\,\text{s}$ is Planck's constant, $c = 3.00 \times 10^{8}\ \text{m/s}$ , $\nu$ is photon frequency, and $\lambda$ is photon wavelength.

Derivation AFrom force balance to the hydrogen energy ladder

Goal: show that quantizing angular momentum forces the electron's radius and energy onto a discrete ladder, and derive the constant $2.18 \times 10^{-18}$ J.

Step 1. Set up the force balance. The Coulomb attraction between the electron (charge $-e$ ) and proton (charge $+e$ ) supplies the centripetal force for circular motion:

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

where $k = 8.99 \times 10^9\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ is Coulomb's constant. This is pure classical mechanics — nothing quantized yet.

Step 2. Solve for the speed. Rearranging Step 1 by multiplying both sides by $r$ and dividing by $m_e$ :

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

Step 3. Apply the quantization postulate. Bohr's rule is $m_e v r = n\hbar$ , so

v = \frac{n\hbar}{m_e r}

Substitute this into Step 2:

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

Step 4. Solve for the allowed radius. Expanding the left side and cancelling one factor of $r$ and one factor of $m_e$ :

\frac{n^2\hbar^2}{m_e r^2} = \frac{k e^2}{r}

r_n = \frac{n^2\hbar^2}{m_e k e^2}

Every quantity on the right except $n$ is a fixed constant of nature, so the radius can only take the discrete values $r_n = n^2 a_0$ , where $a_0 = \hbar^2/(m_e k e^2)$ . Plugging in $\hbar = 1.055\times10^{-34}$ J·s, $m_e = 9.109\times10^{-31}$ kg, $k = 8.99\times10^9$ N·m²/C², and $e = 1.602\times10^{-19}$ C gives $a_0 = 5.29\times10^{-11}$ m, matching the accepted Bohr radius.

Step 5. Build the total energy. Total energy is kinetic plus potential:

E_n = \frac{1}{2}m_e v^2 - \frac{k e^2}{r_n}

From Step 1, $m_e v^2 = k e^2 / r_n$ , so the kinetic term is $\tfrac{1}{2}(ke^2/r_n)$ . Substituting:

E_n = \frac{k e^2}{2 r_n} - \frac{k e^2}{r_n} = -\frac{k e^2}{2 r_n}

Step 6. Insert the quantized radius. Using $r_n = n^2\hbar^2/(m_e k e^2)$ from Step 4:

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}

The entire prefactor is a fixed number — it does not depend on $n$ — so we can evaluate it once:

\frac{m_e k^2 e^4}{2\hbar^2} = \frac{(9.109\times10^{-31})(8.99\times10^{9})^2(1.602\times10^{-19})^4}{2(1.055\times10^{-34})^2} = 2.18\times10^{-18}\ \text{J}

Step 7. State the result.

E_n = -\frac{2.18\times10^{-18}\ \text{J}}{n^2}

This constant is often called the Rydberg energy, $R_H$ (as an energy). Every hydrogen energy level is this single number divided by $n^2$ .

Limiting check: as $n \to \infty$ , $E_n \to 0^-$ , meaning an electron infinitely far from the nucleus is essentially free and unbound — exactly the physical expectation for ionization.

Worked examples

Example 1 — Radius and energy of the $n=4$ orbit, and the ionization energy from it Question: Find the radius and energy of the hydrogen electron in the $n = 4$ orbit, and find the energy needed to ionize the atom starting from that orbit.

Step 1. Radius.

r_4 = n^2 a_0 = (4)^2 (0.529\times10^{-10}\ \text{m}) = 16 \times 0.529\times10^{-10}\ \text{m} = 8.46\times10^{-10}\ \text{m}

Step 2. Energy.

E_4 = -\frac{2.18\times10^{-18}\ \text{J}}{4^2} = -\frac{2.18\times10^{-18}\ \text{J}}{16} = -1.363\times10^{-19}\ \text{J}

Step 3. Ionization energy from $n=4$ . Ionization means removing the electron to $n = \infty$ , where $E_\infty = 0$ .

\Delta E_{\text{ionize}} = E_\infty - E_4 = 0 - (-1.363\times10^{-19}\ \text{J}) = 1.363\times10^{-19}\ \text{J}

This is much smaller than the ionization energy from the ground state ( $n=1$ , $2.18\times10^{-18}$ J), which makes sense: an electron already sitting in a high, loosely bound orbit needs far less energy to escape.

Example 2 — Wavelength of a specific hydrogen emission line Question: An electron in a hydrogen atom falls from $n = 5$ to $n = 2$ . Find the wavelength of the emitted photon and identify which spectral series it belongs to.

Step 1. Energies of the two levels.

E_5 = -\frac{2.18\times10^{-18}}{25}\ \text{J} = -8.72\times10^{-20}\ \text{J}

E_2 = -\frac{2.18\times10^{-18}}{4}\ \text{J} = -5.45\times10^{-19}\ \text{J}

Step 2. Energy released. The electron drops from higher energy to lower energy, so the photon carries away the difference:

\Delta E = E_2 - E_5 = (-5.45\times10^{-19}) - (-8.72\times10^{-20}) = -4.578\times10^{-19}\ \text{J}

The negative sign just signals emission; the photon energy is the magnitude, $4.578\times10^{-19}$ J.

Step 3. Convert to wavelength.

\lambda = \frac{hc}{|\Delta E|} = \frac{(6.626\times10^{-34}\ \text{J}\,\cdot\,\text{s})(3.00\times10^{8}\ \text{m/s})}{4.578\times10^{-19}\ \text{J}}

\lambda = \frac{1.988\times10^{-25}\ \text{J}\,\cdot\,\text{m}}{4.578\times10^{-19}\ \text{J}} = 4.343\times10^{-7}\ \text{m} = 434.3\ \text{nm}

Step 4. Identify the series. Because the final level is $n_f = 2$ , this line belongs to the Balmer series (visible light region); 434 nm is a violet-blue line, consistent with the well-known Balmer lines of hydrogen (410, 434, 486, 656 nm).

Figure

Energy-level diagram showing the n=5 → n=2 transition

Limiting cases & common traps

Situation	What happens	Why
$n \to \infty$	$E_n \to 0$ , $r_n \to \infty$	Electron becomes unbound (free); this defines the ionization limit.
$n = 1$	Most negative energy, smallest radius	Ground state — the lowest possible energy, most tightly bound.
Absorption vs. emission	Sign of $\Delta E$ flips	Absorption: electron jumps to higher $n$ , $\Delta E > 0$ . Emission: electron falls to lower $n$ , photon carries away $\|\Delta E\|$ .
Multi-electron atoms	Formula breaks down	Electron–electron repulsion is not included; the simple $1/n^2$ law is exact only for one-electron species (H, He⁺, Li²⁺).
Series naming	Easy to mix up	Lyman: final $n_f=1$ (UV). Balmer: final $n_f=2$ (visible). Paschen: final $n_f=3$ (IR).

The chemist behind it

Niels Bohr built his model in 1913 while working in Ernest Rutherford's laboratory, shortly after Rutherford's gold-foil experiments had shown that atoms contain a tiny, dense, positively charged nucleus. The puzzle Rutherford's nuclear model left unsolved was stability: classical physics predicted that an orbiting electron should radiate energy continuously and collapse into the nucleus in an instant, yet atoms are obviously stable and, moreover, emit light only at sharp, reproducible wavelengths — the line spectra already tabulated empirically for hydrogen by Johann Balmer and generalized by Johannes Rydberg decades earlier. Bohr's insight was to import the emerging idea of quantization, from Quantized Energy and Photons, directly into mechanics: rather than letting energy take any value, he restricted the electron's angular momentum to whole-number multiples of $\hbar$ . The resulting formula for hydrogen's energy levels reproduced the empirical Rydberg formula for spectral lines essentially exactly, an early and striking triumph for quantum ideas outside of blackbody radiation and the photoelectric effect. The model could not, however, be extended successfully to atoms with more than one electron, a limitation that helped motivate the fuller quantum mechanics of Heisenberg, Schrödinger, and Pauli roughly a decade later.

Origins · 1 minNiels Henrik David Bohr and how this was discovered

Niels Henrik David Bohr

1885–1962 · Developed the quantized orbital model of the hydrogen atom in 1913 while at Rutherford's laboratory in Manchester, correctly predicting the hydrogen line spectrum from first principles; later founded the Copenhagen school of quantum mechanics and contributed foundational ideas on complementarity and atomic structure, earning the 1922 Nobel Prize in Physics.

Ask the tutor: How would the energy-level formula change for a one-electron ion like He⁺, where the nuclear charge is $Z = 2$ instead of $1$ ?

Why does the Bohr model only work well for hydrogen?

Bohr's derivation assumes exactly one electron feeling only the attraction of one nuclear charge. In atoms with more than one electron, each electron also repels every other electron, and that extra force is not included anywhere in the force-balance equation used in the derivation. The formula still works reasonably for other one-electron species like He⁺ or Li²⁺ if the nuclear charge $Z$ is inserted correctly, but it fails for neutral helium and heavier atoms.

What does the quantum number n physically represent?

In the Bohr model, $n$ labels which allowed circular orbit the electron occupies, with $n=1$ being the smallest, most tightly bound orbit. In the fuller quantum treatment used in Electron Shells and Orbitals, $n$ becomes the principal quantum number describing the size and energy of an electron shell, without implying a fixed circular path.

Why is the energy negative in E_n = −2.18×10⁻¹⁸ J / n²?

The negative sign reflects that the electron is bound to the nucleus; energy must be supplied to pull it away. Zero energy is defined as the electron completely free of the nucleus ( $n \to \infty$ ), so any bound state, which is lower in energy than being free, must be negative by this convention.

How is the Bohr model related to the photon concept from earlier study?

The model is a direct application of the quantization ideas in Quantized Energy and Photons. There, light itself was shown to come in discrete energy packets, $E = h\nu$ . Bohr extended that same discreteness to the electron's motion, and then used the photon relation again to compute the exact wavelength of light emitted or absorbed during a jump between two allowed energy levels.

Can the Bohr model explain why atoms bond together to form molecules?

Not directly. The Bohr model only assigns hydrogen-like energies to a single electron around a single nucleus; it says nothing about how electron clouds from two different atoms overlap or share electron density. Explaining bonding, shape, and reactivity requires the richer orbital picture developed in Electron Shells and Orbitals and later bonding theories.