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 The Bohr picture is a useful, although superseded, model that captures the essentials.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 $n = 1, 2, 3, \ldots$.

• $E_1 = -13.6$ eV (ground state).
• $E_2 = -3.4$ eV.
• $E_3 = -1.51$ eV.
• $E_4 = -0.85$ eV.
• $E_\infty = 0$ (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.

Examples in context

Example 1. Mt Stromlo spectroscopy of nearby stars. Mt Stromlo's $1.88$ m telescope uses a spectrograph to identify chemical composition of stars from emission and absorption lines. The hydrogen Balmer-alpha line at $656.3$ nm arises from a transition between $n=3$ and $n=2$ levels. Photon energy is $E = hf = hc/\lambda = 6.63 \times 10^{-34} \times 3 \times 10^8 / 656.3 \times 10^{-9} = 3.03 \times 10^{-19}$ J or $1.89$ eV. This matches the predicted Bohr-Rydberg difference $E_3 - E_2 = -1.51 - (-3.4) = 1.89$ eV, confirming hydrogen in the stellar atmosphere. Doppler shifts of this line at Mt Stromlo measure stellar radial velocities to within $\pm 1$ km s$^{-1}$.

Example 2. Sodium-vapour street lighting in Carlton. Sodium-vapour street lamps emit a characteristic yellow doublet at $589.0$ and $589.6$ nm from sodium's $3p \to 3s$ transition. Photon energy is $E = hc/\lambda = 6.63 \times 10^{-34} \times 3 \times 10^8/(589 \times 10^{-9}) = 3.38 \times 10^{-19}$ J or $2.11$ eV. A $400$ W low-pressure sodium lamp emits about $1.5 \times 10^{20}$ photons per second of yellow light. The discrete-line emission means almost all electrical energy converts into the two yellow wavelengths, giving high efficacy ($\sim 200$ lumens per watt) but poor colour rendering, which is why amber sodium lighting is being replaced by white LEDs across Carlton.

Try this

Q1. Define energy-level transition and state the relationship between transition energy and emitted photon wavelength. [2 marks]

• Cue. Electron drops between discrete atomic energy levels; $E_i - E_f = hf = hc/\lambda$.

Q2. An electron in a hydrogen atom drops from $n = 4$ ($E = -0.85$ eV) to $n = 2$ ($E = -3.4$ eV). Calculate the wavelength of the emitted photon. [3 marks]

• Cue. $\Delta E = 2.55$ eV $= 4.08 \times 10^{-19}$ J. $\lambda = hc/\Delta E = 1.99 \times 10^{-25}/4.08 \times 10^{-19} = 487$ nm.

Q3. Refer to sodium street lamps. (a) Calculate the energy of a $589$ nm photon in eV. (b) Determine the number of photons per second from a $400$ W lamp assuming $100\%$ efficiency. (c) Explain why discrete-line emission gives high efficacy but poor colour rendering. [2+3+2 marks]

• Cue. (a) $2.11$ eV. (b) $400/(3.38 \times 10^{-19}) = 1.18 \times 10^{21}$ s$^{-1}$. (c) Single colour means few wasted wavelengths but cannot reproduce true colours.