Why Bohr Reworked the Atom
Rutherford's nuclear model placed the electron in orbit around a tiny dense nucleus, but classical electromagnetism predicted that an orbiting (accelerating) charge must continuously radiate energy and spiral into the nucleus within a fraction of a second. That model could not explain why atoms are stable, nor why excited hydrogen emits light only at certain discrete frequencies rather than a continuous band.
When an electric discharge is passed through gaseous hydrogen, the $\ce{H2}$ molecules dissociate and the energetically excited $\ce{H}$ atoms emit electromagnetic radiation of discrete frequencies. Passed through a prism, this light separates into a set of sharp lines — the hydrogen line spectrum — rather than a rainbow. Bohr's central achievement was to make a model that produces exactly such a discrete spectrum.
Line spectrum, not continuous spectrum
A glowing solid (a bulb filament) gives a continuous spectrum; an excited gaseous atom gives a line spectrum. Examiners exploit this: hydrogen produces a line emission spectrum precisely because its energy levels are quantised.
Discrete energy levels → discrete photon energies → sharp lines.
The Four Postulates
Bohr's model for the hydrogen atom rests on four postulates. The first two introduce stationary states; the third connects energy gaps to radiation; the fourth quantises angular momentum, which is the engine that produces all the quantitative results.
| Postulate | Statement |
|---|---|
| 1. Stationary orbits | The electron moves around the nucleus in circular paths of fixed radius and energy, called orbits, stationary states or allowed energy states, arranged concentrically about the nucleus. |
| 2. Constant energy | The energy of an electron in an orbit does not change with time. The electron jumps to a higher state only when it absorbs the required energy, and emits energy when it falls to a lower state. The change is not continuous. |
| 3. Bohr frequency rule | When a transition occurs between two states differing in energy by $\Delta E$, the frequency of the radiation absorbed or emitted is $\nu = \dfrac{E_2 - E_1}{h}$. |
| 4. Quantised angular momentum | The angular momentum of the electron is an integral multiple of $h/2\pi$: $\; m_e v r = n\dfrac{h}{2\pi}$, with $n = 1, 2, 3, \dots$ |
The angular momentum used in postulate 4 follows from treating the electron of mass $m_e$ on a circular path of radius $r$ as a rotating body. With moment of inertia $I = m_e r^2$ and angular velocity $\omega = v/r$,
$$\text{angular momentum} = I \times \omega = m_e r^2 \times \frac{v}{r} = m_e v r.$$
Quantising this quantity — allowing it to take only the values $n h / 2\pi$ — is the single radical assumption. Radiation is emitted or absorbed only when the electron jumps from one quantised value of angular momentum to another, so Maxwell's classical electromagnetic theory does not apply and only certain fixed orbits are allowed. This idea grew directly out of Planck's quantisation of energy; for the experimental clues that pushed physics toward it, see developments leading to Bohr's model.
Radius of the Stationary States
Bohr's theory gives the radii of the allowed orbits of hydrogen as
$$r_n = n^2 a_0, \qquad a_0 = 52.9\ \text{pm}.$$
The first stationary state ($n = 1$), called the Bohr orbit, has radius 52.9 pm, and the hydrogen electron is normally found here. As $n$ increases the radius increases as $n^2$, so the electron sits progressively farther from the nucleus. The stationary states are numbered $n = 1, 2, 3, \dots$; these integers are the principal quantum numbers.
| Orbit n | Factor n² | Radius (H, Z = 1) |
|---|---|---|
| 1 | 1 | 52.9 pm |
| 2 | 4 | 211.6 pm |
| 3 | 9 | 476.1 pm |
| 4 | 16 | 846.4 pm |
Energy of the Stationary States
The most important property of a stationary state is its energy. For hydrogen,
$$E_n = -R_H\left(\frac{1}{n^2}\right), \qquad R_H = 2.18 \times 10^{-18}\ \text{J}.$$
Here $R_H$ is the Rydberg constant expressed as an energy. The ground state ($n = 1$) is the lowest and most stable: $E_1 = -2.18 \times 10^{-18}\,(1/1^2) = -2.18 \times 10^{-18}$ J. For $n = 2$, $E_2 = -2.18 \times 10^{-18}\,(1/2^2) = -0.545 \times 10^{-18}$ J. In electron-volts the same expression reads $E_n = -13.6/n^2$ eV, so $E_1 = -13.6$ eV.
What the negative sign means
A free electron at rest, infinitely far from the nucleus ($n = \infty$), is assigned zero energy. A bound electron has lower energy than this, hence the negative sign. As $n$ decreases, $E_n$ grows in magnitude and becomes more negative — the electron is more tightly bound.
$E_\infty = 0$; most negative energy at $n = 1$ → ground state is the most stable.
Hydrogen-like Ions
Bohr's theory works for any species that has just one electron — the so-called hydrogen-like ions such as $\ce{He+}$, $\ce{Li^2+}$ and $\ce{Be^3+}$. The energy and radius then depend on the atomic number $Z$:
$$E_n = -2.18 \times 10^{-18}\left(\frac{Z^2}{n^2}\right)\ \text{J}, \qquad r_n = 52.9\left(\frac{n^2}{Z}\right)\ \text{pm}.$$
As $Z$ increases, the energy becomes more negative and the radius becomes smaller, so the electron is bound more tightly to the more highly charged nucleus. The velocity of the orbiting electron, by contrast, increases with $Z$ and decreases with $n$.
| Species | Z | Ground-state energy E₁ | First-orbit radius r₁ |
|---|---|---|---|
| $\ce{H}$ | 1 | −2.18 × 10⁻¹⁸ J | 52.9 pm |
| $\ce{He+}$ | 2 | −8.72 × 10⁻¹⁸ J | 26.45 pm |
| $\ce{Li^2+}$ | 3 | −19.62 × 10⁻¹⁸ J | 17.63 pm |
Bohr's fixed orbits are eventually replaced by probability clouds — see towards the quantum mechanical model.
Explaining the Line Spectrum
The triumph of the model is that it explains the hydrogen line spectrum quantitatively. Energy is absorbed when the electron jumps from a lower orbit to a higher one, and emitted when it falls from a higher orbit to a lower one. The gap between two orbits is
$$\Delta E = E_f - E_i = R_H\left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right) = 2.18 \times 10^{-18}\left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)\ \text{J}.$$
By the Bohr frequency rule the photon's frequency is $\nu = \Delta E / h$, which works out as
$$\nu = 3.29 \times 10^{15}\left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)\ \text{Hz}.$$
Dividing by the speed of light gives the wavenumber form, which is exactly the empirical formula Johannes Rydberg had fitted to the data years earlier:
$$\bar{\nu} = 1.09677 \times 10^{7}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\ \text{m}^{-1}.$$
The value $109677\ \text{cm}^{-1}$ is the Rydberg constant for hydrogen. Because the right-hand side can take only specific values (the $n$'s are integers), only specific photon energies are possible — and those are exactly the sharp lines that are observed. For absorption $n_f > n_i$ and the bracket is positive; for emission $n_i > n_f$ and $\Delta E$ is negative, meaning energy is released.
The Spectral Series
The lines group themselves into series, each defined by the orbit $n_f$ at which the electron lands. Of all the elements, hydrogen has the simplest line spectrum; the Balmer series (landing on $n = 2$) is the only one in the visible region.
| Series | Final orbit n_f | Starting orbits n_i | Spectral region |
|---|---|---|---|
| Lyman | 1 | 2, 3, 4, … | Ultraviolet |
| Balmer | 2 | 3, 4, 5, … | Visible |
| Paschen | 3 | 4, 5, 6, … | Infrared |
| Brackett | 4 | 5, 6, 7, … | Infrared |
| Pfund | 5 | 6, 7, 8, … | Infrared |
Which series is visible?
Only the Balmer series (transitions ending at $n = 2$) lies in the visible region. Lyman is ultraviolet; Paschen, Brackett and Pfund are all infrared. A common error is to call Lyman "visible" because it has the largest energy gaps — large energy means UV, not visible.
Order of energy gap: Lyman > Balmer > Paschen, i.e. UV > visible > IR.
Worked Examples
Find the energy of the electron in the $n = 2$ state of the hydrogen atom, and the energy released when the electron falls from $n = 5$ to $n = 2$.
Using $E_n = -2.18 \times 10^{-18}/n^2$ J:
$$E_2 = -\frac{2.18 \times 10^{-18}}{2^2} = -0.545 \times 10^{-18}\ \text{J} = -5.45 \times 10^{-19}\ \text{J}.$$
For the $5 \to 2$ emission (a Balmer line):
$$\Delta E = 2.18 \times 10^{-18}\left(\frac{1}{5^2} - \frac{1}{2^2}\right) = 2.18 \times 10^{-18}(0.04 - 0.25) = -4.58 \times 10^{-19}\ \text{J}.$$
The magnitude $4.58 \times 10^{-19}$ J is emitted. The corresponding frequency is $\nu = \Delta E/h = (4.58 \times 10^{-19})/(6.626 \times 10^{-34}) \approx 6.91 \times 10^{14}$ Hz, a line in the visible Balmer series.
Calculate the wavelength of the $\ce{H}$ emission line for the $n = 3 \to n = 2$ transition (the first Balmer line, H-alpha).
Use the wavenumber form with $n_f = 2$, $n_i = 3$:
$$\bar{\nu} = 1.09677 \times 10^{7}\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 1.09677 \times 10^{7}\left(\frac{1}{4} - \frac{1}{9}\right)\ \text{m}^{-1}.$$
$$\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} = 0.13889.$$
$$\bar{\nu} = 1.09677 \times 10^{7} \times 0.13889 = 1.5233 \times 10^{6}\ \text{m}^{-1}.$$
Therefore $\lambda = 1/\bar{\nu} = 6.565 \times 10^{-7}\ \text{m} = 656.5$ nm — the familiar red H-alpha line.
Calculate the energy and the radius of the first orbit of $\ce{He+}$.
For $\ce{He+}$, $Z = 2$ and $n = 1$. Using $E_n = -2.18 \times 10^{-18}(Z^2/n^2)$ J:
$$E_1 = -2.18 \times 10^{-18} \times \frac{2^2}{1^2} = -8.72 \times 10^{-18}\ \text{J}.$$
Using $r_n = 52.9\,(n^2/Z)$ pm:
$$r_1 = 52.9 \times \frac{1^2}{2} = 26.45\ \text{pm} = 0.02645\ \text{nm}.$$
Limitations of the Model
Bohr's model was a major advance over Rutherford's, accounting for the stability and the line spectrum of hydrogen and hydrogen-like ions. Yet it is too simple to survive closer scrutiny.
| Shortcoming | What it fails to explain |
|---|---|
| Fine structure | The closely spaced doublet lines seen with high-resolution spectroscopy. |
| Multi-electron atoms | The spectrum of any atom with more than one electron — even neutral helium ($\ce{He}$), which has two electrons. |
| Zeeman effect | The splitting of spectral lines in a magnetic field. |
| Stark effect | The splitting of spectral lines in an electric field. |
| Chemical bonding | The ability of atoms to combine into molecules through chemical bonds. |
These failures stem from a deeper problem: the model treats the electron as a particle on a definite circular path, which violates the wave nature of matter and the Heisenberg uncertainty principle. Recognising this drove physicists toward a fully quantum mechanical description of the atom.
Bohr's Model in One Glance
- Four postulates: stationary orbits, constant energy, frequency rule $\nu = \Delta E/h$, and quantised angular momentum $m_e v r = nh/2\pi$.
- Radius: $r_n = 52.9\,(n^2/Z)$ pm; first Bohr orbit of hydrogen = 52.9 pm.
- Energy: $E_n = -2.18 \times 10^{-18}\,(Z^2/n^2)$ J $= -13.6\,(Z^2/n^2)$ eV; negative because the bound electron lies below the free-electron zero.
- Transition: $\Delta E = R_H\left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)$, reproducing the Rydberg formula $\bar{\nu} = 1.09677 \times 10^{7}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ m⁻¹.
- Series by final orbit: Lyman (n = 1, UV), Balmer (n = 2, visible), Paschen/Brackett/Pfund (n = 3,4,5, IR).
- Valid only for one-electron species; fails on fine structure, multi-electron spectra, Zeeman/Stark effects and bonding.