Explain the discrete energy levels of atoms and how transitions between levels produce photons with $E_{\text{photon}} = E_i - E_f$, including the appearance of line emission and absorption spectra

What this dot point is asking

VCAA wants you to explain why atomic energy levels are discrete (quantised), how transitions between levels produce photons of specific energies, the resulting line emission and absorption spectra, and the connection to spectroscopic identification of elements. The dot point ties the photon picture from the photoelectric dot point to atomic structure.

Discrete atomic energy levels

In the early 20th century, atomic emission spectra (notably hydrogen's) showed that atoms emit light at very specific, discrete wavelengths rather than a continuous range. This forced an abandonment of the classical picture in which the atom's electron could have any energy. Bohr (1913) proposed that:

• Electrons in atoms occupy discrete, allowed energy levels $E_n$.
• Electrons in level $n$ do not radiate (despite their classical acceleration in circular motion).
• Electrons can transition between levels by emitting or absorbing photons.

In modern quantum mechanics, the energy quantisation arises from the wave nature of the electron (matter waves) confined in the Coulomb potential of the nucleus. The Bohr picture is a useful, although superseded, model that captures the essentials.

Photon emission and absorption

When an electron transitions from a higher energy level $E_i$ to a lower energy level $E_f$, the atom emits a photon whose energy equals the energy lost by the electron:

The photon's frequency and wavelength are determined by its energy:

Conversely, an atom in a lower level $E_f$ can absorb a photon whose energy exactly equals $E_i - E_f$, exciting the electron to level $E_i$. Photons whose energies do not match any allowed transition pass through without interaction.

Useful shortcut: $h c \approx 1240$ eV nm, so a photon of energy 1 eV has wavelength about 1240 nm; a photon of wavelength 656 nm has energy about 1.89 eV.

The hydrogen energy levels

Hydrogen is the simplest atom and has the cleanest spectrum. The Bohr-model energy levels are:

for $n = 1, 2, 3, \ldots$.

• IMATH_13 eV (ground state).
• IMATH_14 eV.
• IMATH_15 eV.
• IMATH_16 eV.
• IMATH_17 (ionised; electron free).

The negative sign indicates the electron is bound to the proton; a free electron at infinity has zero energy by convention. The energy gap from level $n$ to ionisation is $|E_n| = 13.6 / n^2$ eV.

Spectral series

Transitions ending at the same lower level form a series of spectral lines.

• Lyman series. Transitions to $n = 1$ (UV). Highest-energy transitions in hydrogen.
• Balmer series. Transitions to $n = 2$ (visible). H-alpha (3 to 2) is red at 656 nm; H-beta (4 to 2) is blue-green at 486 nm; H-gamma (5 to 2) is blue at 434 nm; H-delta (6 to 2) is violet at 410 nm.
• Paschen series. Transitions to $n = 3$ (infrared).

The Balmer series is the canonical visible spectrum of hydrogen and is observed in any high-voltage hydrogen discharge tube.

Emission spectra

An emission spectrum is produced when atoms in an excited state relax to lower states, emitting photons. The spectrum appears as bright lines on a dark background, each line a specific wavelength corresponding to one transition.

Observed in:

• A hot, low-density gas (a discharge tube, a flame test).
• A nebula illuminated by ultraviolet from nearby stars.
• The corona of the sun (at the limb, against the dark sky).

The pattern of lines is the fingerprint of the element. Helium was first identified in the solar spectrum (1868) before being found on Earth, named for the sun.

Absorption spectra

An absorption spectrum is produced when light from a continuous source passes through a cool gas. Atoms in the gas absorb photons whose energies match their transition energies, exciting electrons from lower to higher levels. The spectrum looks like a continuous (rainbow) spectrum with dark lines at the absorbed wavelengths.

Observed in:

• Stellar spectra. The continuous spectrum from the photosphere passes through cooler chromosphere atoms, producing the dark Fraunhofer lines named for their discoverer.
• Solar transit. Sodium lines in sunlight pass through Earth's sodium-vapour streetlights; the lines are absorbed (briefly) when looking through the lamp at the sun.

The absorbed photons are re-emitted shortly after, but in random directions, so the original beam is depleted at those wavelengths.

Why the lines have the same wavelengths in emission and absorption

The set of allowed transitions $E_i - E_f$ is the same for the atom regardless of whether the atom is gaining or losing energy. Emission lines and absorption lines therefore appear at identical wavelengths.

The spectroscopic fingerprint of an element is universal: hydrogen's red emission line at 656 nm is the same wavelength as the dark line in solar absorption spectra at 656 nm.

Ionisation

When an electron absorbs enough energy to escape the atom completely (from level $n$ to $n = \infty$, energy 0), the atom is ionised. The minimum photon energy to ionise from level $n$ is $|E_n|$. For hydrogen:

• Ionisation from $n = 1$: 13.6 eV (the ionisation energy of hydrogen).
• Ionisation from $n = 2$: 3.4 eV.

Photons with energies between the ionisation thresholds produce a continuous component to the absorption spectrum (no longer discrete because the free electron can carry any kinetic energy).

Multi-electron atoms

Hydrogen has one electron; its spectrum is the simplest. Multi-electron atoms (helium, oxygen, iron) have many more allowed transitions because of multiple electrons and inner-shell effects. Their spectra contain many more lines, often grouped into series corresponding to specific shells.

The fundamental principle is the same: quantised energy levels, transitions emit or absorb photons of energy equal to the level difference. The complexity of multi-electron spectra is exploited in astrophysics and analytical chemistry to identify elements by their characteristic line patterns.

Calculating wavelengths

For any transition, the photon wavelength is:

Quick examples for hydrogen:

• 3 to 2 (H-alpha). $E = -1.51 - (-3.4) = 1.89$ eV. $\lambda = 1240 / 1.89 \approx 656$ nm. Red.
• 4 to 2 (H-beta). $E = -0.85 - (-3.4) = 2.55$ eV. $\lambda = 1240 / 2.55 \approx 486$ nm. Blue-green.
• 2 to 1 (Lyman-alpha). $E = -3.4 - (-13.6) = 10.2$ eV. $\lambda = 1240 / 10.2 \approx 122$ nm. Far UV.

Cross-link: see the Rydberg spectrum calculator for hydrogen line wavelengths from the Rydberg formula directly.

Common errors

Forgetting the sign of energy levels. $E_n$ is negative (bound state). Transitions emit photons with positive energy $E_i - E_f$ (since $E_i > E_f$ for a downward transition, both being negative, $E_i - E_f$ is the smaller magnitude minus the larger, giving positive).

Wrong direction of transition. Photon emission requires a downward transition (higher to lower level). Absorption requires an upward transition.

Trying to absorb arbitrary photons. Only photons whose energy matches a transition can be absorbed. A 5 eV photon cannot be absorbed by ground-state hydrogen (it would require a 5 eV transition; the smallest hydrogen transition from $n=1$ is 10.2 eV).

Using kinetic energy instead of photon energy. The photon energy is $E_{\text{photon}} = h f$. The energy of the electron after a transition is different (the atom keeps the difference).

Confusing levels with shells. The principal quantum number $n$ corresponds to shells in the Bohr model. In modern quantum mechanics there are sub-levels (s, p, d, f) within each shell; VCE Physics Unit 4 uses the Bohr-level picture.

In one sentence

Atomic electrons occupy discrete energy levels, and transitions between levels emit or absorb photons of energy $\Delta E = E_i - E_f$ (equivalently wavelength $\lambda = hc / \Delta E$); the result is line emission spectra (bright lines on dark) and absorption spectra (dark lines on bright continuum) whose patterns serve as fingerprints for identifying elements in laboratory samples and in stellar atmospheres.