Bohr's atomic model

Bohr proposed three postulates for the hydrogen atom, each fixing a specific flaw in Rutherford's picture.

Postulate 1 — stationary states. The electron moves in certain fixed circular orbits, called stationary states, without radiating energy. This resolves the stability problem: classically, an accelerating (orbiting) charge should continuously radiate energy and spiral into the nucleus within a fraction of a second, yet atoms are stable.

Postulate 2 — quantised angular momentum. Only those orbits are allowed for which the electron's angular momentum is an integral multiple of $h/2\pi$:

where: $m$ is the electron's mass, $v$ its speed, $r$ the orbit radius, $h$ is Planck's constant, and $n$ is the principal quantum number labelling the orbit. This single quantisation rule, combined with Coulomb's law providing the centripetal force, is what forces the orbit radii and energies into a discrete, calculable set rather than a continuum.

Postulate 3 — quantised transitions. Energy is exchanged only when the electron jumps between stationary states. Jumping from a higher-energy orbit $E_2$ to a lower one $E_1$ emits a photon; absorbing a photon of exactly that energy does the reverse:

where: $h\nu$ is the photon energy, $\nu$ its frequency. This is why hydrogen's spectrum is a set of sharp lines rather than a continuous glow — only photons matching an exact energy gap are emitted or absorbed. Working through the postulates gives two key results for hydrogen. The orbit radius grows as $n^2$:

and the orbit energy is:

where: $n$ is the principal quantum number of the orbit. The negative sign means the electron is bound (energy is measured relative to a free, motionless electron at infinity, taken as zero); as $n$ increases, $E_n$ rises toward zero, meaning the electron is more loosely held, until at $n \to \infty$ the electron is free — this limit is ionisation, and $13.6\ \text{eV}$ is exactly the ionisation energy of hydrogen from its ground state ($n=1$).

Achievements and limitations. Bohr's model explains atomic stability, correctly predicts hydrogen's spectral line wavelengths (Lyman, Balmer, Paschen and further series), and gives the correct ionisation energy of $13.6\ \text{eV}$. But it works quantitatively only for one-electron species (H, $\text{He}^+$, $\text{Li}^{2+}$); it fails for multi-electron atoms; it cannot explain the fine splitting of lines in a magnetic field (Zeeman effect) or electric field (Stark effect); and its picture of sharply defined circular orbits contradicts the wave nature of matter and Heisenberg's uncertainty principle, which is why it was later replaced by the quantum mechanical model.

Question. Calculate the wavelength of light emitted when an electron in a hydrogen atom falls from $n=3$ to $n=2$, and identify the spectral series and region of the spectrum.

Step 1. Apply the Rydberg equation with $n_1 = 2$ (the lower, final level) and $n_2 = 3$ (the higher, initial level): $\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right) = 1.097\times10^7\left(\dfrac{1}{4} - \dfrac{1}{9}\right)$.

Step 2. Simplify the bracket: $\dfrac{1}{4} - \dfrac{1}{9} = \dfrac{9-4}{36} = \dfrac{5}{36} = 0.1389$.

Step 3. Multiply out: $\dfrac{1}{\lambda} = 1.097\times10^7 \times 0.1389 = 1.524\times10^6\ \text{m}^{-1}$.

Step 4. Invert to get the wavelength: $\lambda = \dfrac{1}{1.524\times10^6} = 6.56\times10^{-7}\ \text{m} = 656\ \text{nm}$. Since $n_1 = 2$, this is the Balmer series, and $656\ \text{nm}$ falls in the visible region (red light) — the famous $H_\alpha$ line. Answer: $\lambda = 656\ \text{nm}$, Balmer series, visible red.

Competitive stretch

The same postulates give the excitation energy directly from $E_n = -13.6/n^2\ \text{eV}$: for an electron jumping from the ground state ($n=1$, $E_1 = -13.6\ \text{eV}$) to the second excited state ($n=3$, $E_3 = -13.6/9 = -1.51\ \text{eV}$), the energy needed is $\Delta E = E_3 - E_1 = -1.51-(-13.6) = 12.09\ \text{eV}$, absorbed as a single photon; the reverse transition emits the same $12.09\ \text{eV}$ photon in the Lyman series.

• Mixing up $n_1$ and $n_2$ in the Rydberg equation — $n_1$ is always the lower (smaller) level, and the formula gives a positive $1/\lambda$ only when $n_2 > n_1$.
• Dropping the negative sign in $E_n = -13.6/n^2\ \text{eV}$, causing sign errors when computing $\Delta E$ for absorption or emission.
• Applying $E_n = -13.6/n^2$ to multi-electron atoms like helium — the formula is quantitatively exact only for hydrogen and other one-electron species ($\text{He}^+$, $\text{Li}^{2+}$).
• Treating Bohr's circular orbits as the final, literal picture of where the electron is — the model was superseded because electrons do not follow sharply defined orbits (wave nature, uncertainty principle).

Ask the tutor: Want to practise identifying spectral series for other transitions, or see how the Rydberg-equation derivation connects to the quantum mechanical model that replaced Bohr's orbits?