Coaching Centre
Bohr and the hydrogen spectrum
Bohr's model of the hydrogen atom and its postulates, energy levels and transitions with E = hf, the Balmer, Lyman and Paschen series, the Rydberg equation, ionisation energy, absorption, the limitations of Bohr's model, and measuring the Balmer series with a spectroscope.
In 1885 a Swiss schoolteacher, Johann Balmer, found a formula that gave the wavelengths of hydrogen's four visible lines to within a fraction of a nanometre. He had no idea why it worked. Twenty-eight years later Niels Bohr heard about it, and said that as soon as he saw the formula "the whole thing was immediately clear to me". The numbers in it were the energy levels of the atom.
Bohr's model
Rutherford's atom should collapse and should give a continuous spectrum. In 1913 Bohr kept Rutherford's nucleus but added rules (postulates) that broke with classical physics, borrowing Planck's and Einstein's idea that energy comes in quanta:
- Stationary states. Electrons can only exist in certain allowed circular orbits, each with a fixed energy. In these orbits they do not radiate energy, despite accelerating.
- Quantised angular momentum. The allowed orbits are those where the electron's angular momentum is a whole-number multiple of $h/2\pi$: $mvr = \dfrac{nh}{2\pi}$, $n = 1, 2, 3, \ldots$
- Transitions. An electron emits or absorbs radiation only when it jumps between allowed orbits. The photon's energy equals the difference between the two energy levels:
$$E = hf = \frac{hc}{\lambda} = E_i - E_f$$
Using these rules with Coulomb's law and circular motion, Bohr calculated the allowed energies of hydrogen's electron:
$$E_n = -\frac{13.6\ \text{eV}}{n^2}$$
The energies are negative because the electron is bound: 0 eV means the electron is just free of the atom. The ground state ($n = 1$) is −13.6 eV, so 13.6 eV is hydrogen's ionisation energy. The orbit radii are $r_n = n^2 \times 0.053$ nm.
The Rydberg equation
Put Bohr's energies into $hc/\lambda = E_i - E_f$ and you get the formula Balmer and Johannes Rydberg had found by trial and error:
$$\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$
where $n_i$ is the initial (higher) level, $n_f$ the final (lower) level, and $R = 1.097 \times 10^{7}$ m−1 is the Rydberg constant. Bohr's model predicted $R$ from $e$, $m_e$, $h$ and $\varepsilon_0$, and the predicted value matched the measured one. That was the model's great success.
| Series | nf | Region | Longest wavelength (ni = nf + 1) | Series limit (ni → ∞) |
|---|---|---|---|---|
| Lyman | 1 | ultraviolet | 121.5 nm | 91.2 nm |
| Balmer | 2 | visible and near ultraviolet | 656.3 nm (Hα, red) | 364.6 nm |
| Paschen | 3 | infrared | 1875 nm | 820.4 nm |
Calculate the wavelength, frequency and photon energy for the transition from $n = 4$ to $n = 2$ in hydrogen.
$\dfrac{1}{\lambda} = 1.097 \times 10^{7}\left(\dfrac{1}{2^2} - \dfrac{1}{4^2}\right) = 1.097 \times 10^{7} \times 0.1875 = 2.057 \times 10^{6}$ m−1, so $\lambda = 4.862 \times 10^{-7}$ m $= 486.2$ nm (blue-green).
$f = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^{8}}{4.862 \times 10^{-7}} = 6.17 \times 10^{14}$ Hz
$E = hf = 6.626 \times 10^{-34} \times 6.17 \times 10^{14} = 4.09 \times 10^{-19}$ J $= 2.55$ eV. Check with the levels: $E_4 - E_2 = -0.85 - (-3.40) = 2.55$ eV. Energy is conserved.
A hydrogen atom in its ground state absorbs a photon of wavelength 97.3 nm. To which level is the electron raised? What is the longest wavelength that could ionise a ground-state atom?
$E = \dfrac{hc}{\lambda} = \dfrac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{97.3 \times 10^{-9}} = 2.04 \times 10^{-18}$ J $= 12.75$ eV
The electron's new energy is $-13.6 + 12.75 = -0.85$ eV $= -13.6/n^2$, so $n^2 = 16$ and $n = 4$.
Ionisation needs 13.6 eV $= 2.18 \times 10^{-18}$ J, so $\lambda_{max} = \dfrac{hc}{E} = 91.2$ nm, the Lyman series limit. Any shorter wavelength also ionises the atom, with the excess becoming the electron's kinetic energy.
- Start with the upper level 3 and lower level 2 and press Fire photon. Check the readouts: 1.89 eV and 656.3 nm, the red Hα line.
- Press All Balmer lines. Read the wavelengths of the lines from n = 4, 5 and 6 to n = 2 and check one of them with the Rydberg equation.
- Set the lower level to 1. Which series is this, and which region of the spectrum are the lines in? Find the 2 → 1 wavelength (about 121.5 nm).
- Set 4 → 3. Why can't you see this line with your eyes?
- Switch to Absorption with levels 1 and 2. Explain why the photon must have exactly 10.2 eV, and what would happen to a 10.0 eV photon.
What Bohr's model explained
- Why atoms are stable: electrons in stationary states don't radiate, and can't go below the ground state.
- Why hydrogen gives line spectra, and the exact wavelengths, including the value of the Rydberg constant from fundamental constants.
- Why the emission and absorption lines of a gas are at the same wavelengths.
- The ionisation energy of hydrogen (13.6 eV), and the spectra of other one-electron ions like He+.
Limitations of Bohr's model
- Only works for one-electron atoms. It can't predict the spectra of helium or any heavier atom, because it ignores the interactions between electrons.
- Can't explain relative intensities. Some lines are much brighter than others; the model says nothing about how likely each transition is.
- Can't explain fine structure or the Zeeman effect. At high resolution many lines are close pairs or groups, and they split further in a magnetic field.
- Its postulates have no justification. It mixes classical physics (circular orbits, Coulomb force) with quantum rules imposed by hand. Why shouldn't an accelerating electron radiate? Why is angular momentum quantised?
- It treats the electron as a particle on a definite path, which the Heisenberg uncertainty principle later showed is impossible (page 8I).
De Broglie's matter waves and Schrödinger's wave mechanics answered the last two points and replaced orbits with orbitals.
Test yourself
Which transition in hydrogen emits the photon with the longest wavelength?
4 → 3 has the smallest energy gap (0.66 eV), so the longest wavelength (1875 nm).
The Balmer series consists of transitions
Balmer lines end on n = 2.
A photon of energy 12.09 eV is absorbed by a ground-state hydrogen atom. The electron is raised to
−13.6 + 12.09 = −1.51 eV = −13.6/9, so n = 3.
Which of the following could Bohr's model NOT explain?
The model gave no way to calculate transition probabilities, so intensities were unexplained.
An emission line of hydrogen has a wavelength of 434.1 nm. The photon energy is
$E = hc/\lambda = 4.58 \times 10^{-19}$ J ÷ $1.602 \times 10^{-19}$ = 2.86 eV (the 5 → 2 line).
Label the hydrogen energy-level diagram.
Sort these observations.
In Bohr's model, electrons in allowed orbits do not energy. The energy of a photon emitted in a transition equals the between the two energy levels. The ground-state energy of hydrogen is eV. Transitions ending on n = 1 form the series, in the ultraviolet.
Relate the quantised energy levels of the hydrogen atom and the law of conservation of energy to the line emission spectrum of hydrogen. Include a calculation of the wavelength of the n = 3 to n = 2 transition.
The electron in a hydrogen atom can only have certain energies, $E_n = -13.6/n^2$ eV. In a discharge tube, collisions excite electrons to higher levels. An excited electron drops to a lower level, and by conservation of energy the energy lost by the atom is carried off by a single photon: $hf = hc/\lambda = E_i - E_f$. Because only certain energy levels exist, only certain energy differences, and so only certain wavelengths, are possible. The spectrum is therefore a set of discrete bright lines rather than a continuous band. Transitions ending on n = 2 give the visible Balmer series.
For n = 3 to n = 2: $\dfrac{1}{\lambda} = 1.097 \times 10^{7}\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 1.524 \times 10^{6}$ m⁻¹, so $\lambda = 6.56 \times 10^{-7}$ m = 656 nm (red). Check: $E_3 - E_2 = -1.51 - (-3.40) = 1.89$ eV $= 3.03 \times 10^{-19}$ J, and $hc/E = 656$ nm.