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CU-101 / L07 · lecture

Atomic Line Spectra

~5 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be familiar with the quantisation of energy levels in the hydrogen atom and the concept of photon emission/absorption from the previous lecture on the photoelectric effect.

Why this matters

Atomic line spectra provide the experimental fingerprint that confirms the quantum nature of the atom.

Unlike the continuous rainbow produced by a hot solid, a gas of isolated atoms emits or absorbs light only at specific, discrete wavelengths. This phenomenon, known as atomic line spectra, was the critical experimental evidence that forced physicists to abandon classical models of the atom. By analysing these lines, we can determine the electronic structure of atoms, identify elements in distant stars, and understand the fundamental interactions between matter and electromagnetic radiation. This lecture bridges the gap between the abstract energy levels and the observable light we detect.

The Fingerprint of Atoms

Imagine shining white light through a cold gas of hydrogen. Most of the light passes through, but specific colours are missing, appearing as dark lines in the spectrum. Conversely, if you energise the hydrogen gas, it glows with specific bright lines. These lines are not random; they follow a precise mathematical pattern. Each element has a unique set of lines, much like a fingerprint. This uniqueness arises because the energy levels of electrons in different atoms are different. When an electron jumps between these levels, it releases or absorbs a photon with energy exactly equal to the difference between the levels. Since the levels are fixed, the photon energies—and thus the wavelengths—are fixed. This is why we see lines rather than a continuous band of colours.

From Energy Levels to Wavelengths

The energy of a photon is given by \(E = h\nu = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. In the hydrogen atom, the energy of an electron in level \(n\) is proportional to \(-\frac{1}{n^2}\). When an electron transitions from a higher level \(n_2\) to a lower level \(n_1\), the energy difference is emitted as a photon. The Rydberg formula captures this relationship directly in terms of wavelength. It states that the reciprocal of the wavelength is proportional to the difference of the reciprocals of the squares of the principal quantum numbers.

\[ \frac{1}{\lambda} = Z^2R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]

Here, \(R_H\) is the Rydberg constant for hydrogen, approximately \(1.097\times10^7\,\text{m}^{-1}\). For hydrogen, \(Z=1\). For hydrogen-like ions (ions with only one electron, such as \(\mathrm{He^+}\) or \(\mathrm{Li^{2+}}\)), the nuclear charge \(Z\) scales the energy levels, and thus the wavenumber, by a factor of \(Z^2\). This scaling occurs because the electrostatic attraction between the nucleus and the single electron is stronger for higher \(Z\), making the energy levels more deeply bound.

Spectral Series

Depending on the final energy level \(n_1\) of the transition, the emitted lines group into distinct series. The most famous is the Lyman series, where \(n_1=1\). These transitions involve the ground state and emit ultraviolet light. The Balmer series, with \(n_1=2\), produces visible light and was the first series identified experimentally. The Paschen series (\(n_1=3\)) and Brackett series (\(n_1=4\)) emit infrared radiation. As \(n_2\) approaches infinity, the wavelength of the emitted photon approaches a limit known as the series limit. This limit corresponds to the ionisation energy of the atom from the level \(n_1\). Understanding these series allows us to map the entire electronic structure of the hydrogen atom from a single set of spectral data.

Selection Rules and Intensity

Not all transitions between energy levels are equally probable. Quantum mechanics imposes selection rules that dictate which transitions are "allowed." For electric dipole transitions, the most common type, the change in the principal quantum number \(\Delta n\) can be any integer, but the change in the orbital angular momentum quantum number must be \(\Delta l = \pm 1\). Transitions that violate these rules are "forbidden" and have much lower probabilities, resulting in very weak spectral lines. The intensity of a spectral line depends on the population of the initial state and the transition probability. In thermal equilibrium, lower energy states are more populated, so transitions starting from lower \(n\) levels are generally more intense, provided the transition is allowed. This explains why the \(n=3 \to n=2\) line (H-alpha) is often the brightest in the Balmer series.

The result

◆ RESULT · T-002 · full treatment on its own page

The Rydberg formula

For a hydrogen atom, the wavelength \(\lambda\) of light emitted or absorbed as the electron transitions between two allowed energy levels \(n_1\) and \(n_2\) (integers, \(n_2>n_1\ge1\)) satisfies \(\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\), where \(R_H\approx1.097\times10^7\,\text{m}^{-1}\) is the Rydberg constant for hydrogen. For a hydrogen-like ion of nuclear charge \(Z\), the same formula holds with \(R_H\) replaced by \(Z^2R_H\).

\[ \frac{1}{\lambda} = Z^2R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \qquad R_H\approx1.097\times10^7\,\text{m}^{-1} \]
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Tier 3 · watch out

A common misconception is that the Rydberg constant \(R_H\) is a universal constant independent of the atom. In reality, \(R_H\) is specific to hydrogen because it accounts for the reduced mass of the electron-proton system. For heavier hydrogen-like ions, the reduced mass is closer to the electron mass, leading to a slightly different Rydberg constant. However, for most introductory calculations, using \(R_H\) with the \(Z^2\) scaling is sufficient. C

Worked example

Problem. Calculate the wavelength of the photon emitted when an electron in a hydrogen atom transitions from the \(n=4\) level to the \(n=2\) level. Use \(R_H = 1.097\times10^7\,\text{m}^{-1}\).

  1. Identify the transition: \(n_1 = 2\) and \(n_2 = 4\).
  2. Substitute these values into the Rydberg formula for hydrogen (\(Z=1\)): \(\frac{1}{\lambda} = R_H\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\).
  3. Calculate the term in the brackets: \(\frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16} = 0.1875\).
  4. Multiply by \(R_H\): \(\frac{1}{\lambda} = 1.097\times10^7 \times 0.1875 \approx 2.0569\times10^6\,\text{m}^{-1}\).
  5. Invert to find \(\lambda\): \(\lambda = \frac{1}{2.0569\times10^6} \approx 4.861\times10^{-7}\,\text{m}\).
  6. Convert to nanometres: \(4.861\times10^{-7}\,\text{m} = 486.1\,\text{nm}\).
\[ \lambda \approx 486.1\ \mathrm{nm} \]

Units check. The Rydberg constant has units of \(\text{m}^{-1}\), and the term in brackets is dimensionless, so \(1/\lambda\) has units of \(\text{m}^{-1}\), making \(\lambda\) have units of metres, which is consistent.

CHECK YOURSELF

Three quick tests

  1. What is the value of \(n_1\) for the Lyman series?
    Answer1
  2. If \(Z=2\), by what factor does the wavenumber \(1/\lambda\) increase compared to hydrogen for the same transition?
    Answer4 (since the factor is \(Z^2\))
  3. Which series of hydrogen transitions falls in the visible region?
    AnswerBalmer series
UNIT

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NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Results cited: T-002 · Threads: quantum, structure