Emission and Absorption Spectra: Study Guide (Physics 2 15.3)

The atom as a system

In Topic 15.2 you met the idea that an electron in an atom can only have certain energies. This topic asks what happens when light meets those energy states.

Choose the system carefully: the atom is the electron plus the nucleus. The energy of the atom is the interaction (electric potential) energy between them, together with the electron’s kinetic energy. When we say “the electron moves to a higher level”, what really changes is the energy of the electron–nucleus system.

The energy states are fixed by the atom. Three words describe them:

• Ground state. The lowest energy state (n = 1). Atoms at room temperature are almost all in this state.
• Excited state. Any higher state (n = 2, 3, …). An atom in an excited state does not stay there long.
• Ionized. The electron has left the atom completely. The zero of energy is set here, for an electron at rest far from the nucleus. So every bound state has a negative energy.

Photons in, photons out

Energy is conserved when an atom meets light. The atom changes from one state to another, and a single photon carries the difference.

• Absorption. An atom in a lower state takes in a photon and jumps to a higher state. This can only happen if the photon’s energy exactly equals the gap between the two states.
• Emission. An atom in an excited state drops to a lower state on its own (spontaneously) and gives out one photon. The photon’s energy equals the gap.

In both cases:

E_photon = |E_upper − E_lower| = hf = hc/λ

Because the gap between two particular states is fixed, each transition gives a photon of one frequency and therefore one wavelength. That is why atoms produce sharp lines rather than a smear of colour.

Units: the electronvolt

Atomic energies are tiny in joules, so we use the electronvolt: 1 eV = 1.60 × 10⁻¹⁹ J. With h = 4.14 × 10⁻¹⁵ eV·s and c = 3.00 × 10⁸ m/s, the product hc is about 1.24 × 10⁻⁶ eV·m, or 1.24 × 10³ eV·nm. This gives a fast route:

λ (in nm) ≈ 1240 ÷ E (in eV)

A 2.0 eV photon has λ ≈ 620 nm (visible, orange-red). A 10 eV photon has λ ≈ 124 nm (ultraviolet). Bigger energy gaps give shorter wavelengths. If you work in joules with h = 6.63 × 10⁻³⁴ J·s instead, your answer may differ in the last figure. Both routes are fine.

Energy level diagrams

An energy level diagram shows each allowed state as a horizontal line. Higher lines are higher energies. The vertical distance between two lines is the energy of the photon for that transition. An arrow pointing down is emission; an arrow pointing up is absorption. The horizontal position means nothing.

Figure 1 shows an invented single-electron atom that we will use for this guide. It is a model, not a real element. In this course you only need diagrams for atoms with one electron.

How many different lines can this atom emit? Any pair of levels gives one line, so four levels give 4 × 3 ÷ 2 = 6 lines. An atom excited to n = 4 can drop straight to n = 1, or go down in steps (4 → 2 → 1, for example). Each step is a separate photon.

Binding energy and ionization

The binding energy of a state is the energy needed to take the electron from that state to E = 0, so that it is free. For the atom in Figure 1:

• from the ground state: 0 − (−12.0 eV) = 12.0 eV;
• from n = 2: 4.0 eV; from n = 4: only 1.0 eV.

The ground state is the most tightly bound, so it needs the most energy to ionize. A photon with at least the binding energy can ionize the atom. Above that threshold the energy does not have to match anything, because a free electron can have any kinetic energy: the extra energy becomes the electron’s kinetic energy.

Line spectra and what they tell you

A spectrum is light spread out by wavelength, for example with a diffraction grating (Topic 14.8) or a prism.

• Continuous spectrum. All wavelengths are present with no gaps. A hot, dense object such as a lamp filament gives this (Topic 15.4).
• Emission (bright-line) spectrum. A hot, low-density gas, for example in a discharge tube, gives bright lines on a dark background. Collisions keep lifting atoms into excited states, and each drop down emits a photon at one of the atom’s line wavelengths.
• Absorption (dark-line) spectrum. Continuous light passes through a cooler gas. Atoms in the gas absorb photons that match their gaps. Those wavelengths are missing from the beam, so dark lines appear in the continuous spectrum. The excited atoms soon re-emit, but in all directions, so very little of that energy continues along the original beam.

Figure 2 shows an important detail. A cool gas has nearly all its atoms in the ground state, so its absorption lines are only the transitions that start at n = 1. The emission spectrum includes every downward transition, so it usually has more lines. Every absorption line still matches an emission line of the same element.

Spectra as fingerprints. Each element has its own set of energy levels, so it has its own set of line wavelengths. That makes a spectrum a reliable fingerprint:

• An emission spectrum tells you which elements are present in a light source, such as a gas lamp or a hot gas cloud.
• An absorption spectrum tells you which elements are present in a gas that light has passed through, such as the outer layers of a star or a gas sample in a lab.

To identify an element, all of its expected lines (for the conditions) must be present. One matching line is not enough evidence. For example, hydrogen’s visible emission lines are at about 656, 486, 434 and 410 nm. They come from drops down to n = 2. A gas showing only one of those wavelengths is not proven to be hydrogen.

Background: the dark lines in sunlight are absorption lines from gases in the Sun’s outer layers. Helium was first identified from a line in the Sun’s spectrum, before it was found on Earth.

Worked example 1: lines from the model atom

Question. Atoms of the Figure 1 atom are excited into the n = 3 state. (a) List every photon energy they can emit as they return to the ground state. (b) Find each wavelength and say which part of the spectrum it is in. (c) Find the frequency of the visible photon.

1. Possible transitions from n = 3. Directly 3 → 1, or in two steps 3 → 2 then 2 → 1. So three different photons are possible.
2. Energies. 3 → 1: E = −2.0 − (−12.0) = 10.0 eV. 3 → 2: E = −2.0 − (−4.0) = 2.0 eV. 2 → 1: E = −4.0 − (−12.0) = 8.0 eV.
3. Wavelengths (λ ≈ 1240 ÷ E). 10.0 eV → 124 nm (ultraviolet). 2.0 eV → 620 nm (visible). 8.0 eV → 155 nm (ultraviolet).
4. Frequency of the visible photon. f = c/λ = (3.00 × 10⁸ m/s) ÷ (620 × 10⁻⁹ m) = 4.84 × 10¹⁴ Hz.
5. Check with joules. 2.0 eV × 1.60 × 10⁻¹⁹ J/eV = 3.2 × 10⁻¹⁹ J, and f = E/h = 3.2 × 10⁻¹⁹ J ÷ 6.63 × 10⁻³⁴ J·s = 4.83 × 10¹⁴ Hz. The two routes agree to within rounding.

Interpretation. A single atom emits either one photon (3 → 1) or two photons (3 → 2 → 1). In each case the total energy given out is 10.0 eV, which is exactly the energy the atom lost. A large number of atoms produce all three lines, so you would see a single red-orange line by eye, plus two ultraviolet lines that need a detector.

Worked example 2: which photons are absorbed?

Question. A beam containing photons of 8.0 eV, 9.0 eV, 10.0 eV and 13.0 eV passes through a cool gas of the Figure 1 atom. All the atoms are in the ground state. Which photons can be absorbed, and what happens to each atom that absorbs one?

1. Rule. A ground-state atom absorbs a photon only if −12.0 eV + E_photon is the energy of an allowed level, or if E_photon is at least the 12.0 eV binding energy.
2. 8.0 eV. −12.0 + 8.0 = −4.0 eV, which is level n = 2. Absorbed; the atom goes to n = 2.
3. 9.0 eV. −12.0 + 9.0 = −3.0 eV. There is no level at −3.0 eV. Not absorbed; the photon passes through.
4. 10.0 eV. −12.0 + 10.0 = −2.0 eV, which is n = 3. Absorbed; the atom goes to n = 3.
5. 13.0 eV. This is more than the 12.0 eV binding energy. Absorbed, and the atom is ionized. The freed electron has kinetic energy 13.0 − 12.0 = 1.0 eV.

Check and interpretation. The 8.0 eV photon has wavelength 1240 ÷ 8.0 = 155 nm, which matches the dark line at 155 nm in Figure 2. Notice the 9.0 eV photon carries more energy than the 8.0 eV photon, yet it is not absorbed. “More energy” is not enough; the energy must match a gap (or exceed the binding energy). If the gas were hot, so some atoms sat in n = 2, those atoms could also absorb 2.0 eV and 3.0 eV photons, and any photon of 4.0 eV or more would ionize them.

Common misconceptions

• “An atom absorbs any photon with enough energy to reach a level.” The photon energy must equal the gap. A photon with a little extra is not absorbed for a bound-to-bound jump. Only ionization accepts any energy above a threshold.
• “The electron emits the photon.” Treat the atom (electron plus nucleus) as the system. The photon’s energy comes from the change in the system’s interaction energy.
• “Dark lines mean the gas emits darkness” or “the absorbed light is gone for good.” The atoms re-emit, but in all directions. Along the original beam the light at those wavelengths is much weaker, so the line looks dark.
• “Absorption and emission spectra of an element have exactly the same lines.” Every absorption line matches an emission line, but a cool gas absorbs mainly from the ground state, so it shows fewer lines.
• “A negative energy means the atom has no energy.” The negative sign comes from the choice of zero (electron free and at rest). A more negative level is more tightly bound.
• Mixing up which end is bigger. A larger energy gap gives a higher frequency and a shorter wavelength.
• Calling one matching line proof of an element. You need the whole pattern of lines expected under those conditions.

Where this leads

Topic 15.4 looks at the opposite kind of spectrum: the continuous glow of a hot, dense object, and how Planck’s quantum idea explained it. See the blackbody radiation study guide. For this topic, try the practice questions, then use the revision notes and the topic checklist to consolidate. You can look back at the Bohr model guide for where the energy levels come from.