An electron drops from a level at -3.0\ eV to one at -6.8\ eV. Find the energy of the emitted photon. [2 marks]

SEAB Original-style practice: An atom has energy levels at -10.4\ eV, -5.5\ eV and -1.6\ eV. Take h = 6.63 × 10^-34\ J s, c = 3.00 × 10^8\ m s^-1, 1\ eV = 1.60 × 10^-19\ J. (a) Find the energy of the photon emitted in a transition from -1.6\ eV to -5.5\ eV. (b) Find its wavelength.

(a) Photon energy: hf = E_i - E_f = (-1.6) - (-5.5) = 3.9\ eV. In joules: 3.9 × 1.60 × 10^-19 = 6.24 × 10^-19\ J. (b) Wavelength: λ = hcE = (6.63 × 10^-34)(3.00 × 10^8)6.24 × 10^-19 = 1.989 × 10^-256.24 × 10^-19 = 3.19 × 10^-7\ m. The emitted photon has wavelength about 319\ nm (ultraviolet). Markers reward the photon energy as the difference between levels, conversion to joules, and the wavelength from λ = hc/E with units.

SingaporePhysicsSyllabus dot point

How do discrete atomic energy levels produce the characteristic line spectra of atoms?

Explain discrete energy levels in atoms, relate transitions to emitted or absorbed photon energies, and account for line emission and absorption spectra

A focused answer to the H2 Physics learning outcome on atomic energy levels. Discrete energy levels, the photon-transition relation hf = E_i - E_f, and how line emission and absorption spectra arise as evidence of quantisation.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to explain that atoms have discrete (quantised) energy levels, to relate transitions between levels to the energy of an emitted or absorbed photon, and to account for line emission and absorption spectra. Line spectra are direct evidence that energy in an atom is quantised.

The answer

Discrete energy levels

An electron in an atom can occupy only certain allowed energies, called energy levels, and not the values in between. The levels are usually drawn as horizontal lines on an energy-level diagram, with the lowest (most negative) being the ground state and higher levels being excited states. The energies are negative because the electron is bound to the atom, with zero taken when the electron is just free.

Transitions and photon energy

When an electron moves from a higher level $E_i$ to a lower level $E_f$ , the atom emits a photon whose energy is exactly the difference between the levels:

hf = E_i - E_f

Conversely, an atom can absorb a photon only if its energy exactly matches a gap between two levels, lifting an electron to a higher level. Because the levels are discrete, only specific photon energies (and so specific frequencies and wavelengths) are emitted or absorbed.

Line emission spectra

When atoms are excited (by heating or an electric discharge), their electrons drop back down through the levels, emitting photons at specific wavelengths. Passing this light through a prism or grating reveals a series of bright lines on a dark background, the line emission spectrum. Each element has a unique pattern of lines, a spectral fingerprint set by its energy levels.

Line absorption spectra

When continuous (white) light passes through a cool gas, atoms absorb the photons whose energy matches their level gaps, removing those wavelengths from the transmitted light. The result is a continuous spectrum crossed by dark lines, the line absorption spectrum. The dark absorption lines fall at exactly the same wavelengths as the bright emission lines of the same element, because both arise from the same set of energy-level differences.

Worked example

A hydrogen atom has energy levels $E_n = -\dfrac{13.6}{n^2}\ \text{eV}$ . Find the wavelength of the photon emitted when an electron falls from $n = 3$ to $n = 2$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ , $c = 3.00 \times 10^8\ \text{m s}^{-1}$ , $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ .

Step 1: Find the two energy levels

E_3 = -\frac{13.6}{9} = -1.51\ \text{eV}, \qquad E_2 = -\frac{13.6}{4} = -3.40\ \text{eV}

Step 2: Find the photon energy

hf = E_3 - E_2 = (-1.51) - (-3.40) = 1.89\ \text{eV} = 1.89 \times 1.60 \times 10^{-19} = 3.02 \times 10^{-19}\ \text{J}

Step 3: Find the wavelength

\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{3.02 \times 10^{-19}} = \frac{1.989 \times 10^{-25}}{3.02 \times 10^{-19}} = 6.59 \times 10^{-7}\ \text{m}

Step 4: Interpret

The emitted photon has wavelength about $659\ \text{nm}$ , in the red part of the visible spectrum. This is the first line of the Balmer series, seen as a bright red line in hydrogen's emission spectrum.

Common traps

Treating energy levels as continuous: Atomic energies are quantised; only the discrete allowed levels exist, which is why spectra show lines rather than a continuous band.
Getting the sign of the level difference wrong: Emission gives off a photon of energy $E_i - E_f$ (a positive value, higher minus lower); be careful with negative level energies.
Forgetting that absorption requires an exact match: A photon is absorbed only if its energy equals a gap between two levels; otherwise it passes through.
Confusing emission and absorption spectra: Emission shows bright lines on a dark background; absorption shows dark lines on a continuous background, at the same wavelengths.
Using $hf$ without converting eV to joules: Convert level differences from eV to joules before finding frequency or wavelength.

Examples in context

Example 1. Identifying elements in stars. The dark absorption lines in a star's spectrum match the emission lines of known elements, so astronomers can identify which elements are present in a star's outer layers without ever sampling it. This is how helium was first discovered in the Sun's spectrum.

Example 2. Sodium street lamps. A sodium lamp glows with a distinctive yellow light because excited sodium atoms emit strongly at two close wavelengths around $589\ \text{nm}$ , corresponding to a specific energy-level transition. The colour is a direct signature of sodium's quantised energy levels.

Try this

Q1. Explain what is meant by a discrete energy level in an atom. [2 marks]

Cue. An allowed (quantised) electron energy; the electron can occupy only these specific energies, not values in between.

Q2. An electron drops from a level at $-3.0\ \text{eV}$ to one at $-6.8\ \text{eV}$ . Find the energy of the emitted photon. [2 marks]

Cue. $hf = E_i - E_f = (-3.0) - (-6.8) = 3.8\ \text{eV}$ (about $6.1 \times 10^{-19}\ \text{J}$ ).

Q3. Explain why each element has a unique line spectrum. [2 marks]

Cue. Each element has its own unique set of energy levels, so the differences between levels (and hence the emitted or absorbed photon energies and wavelengths) form a distinctive pattern.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksAn atom has energy levels at

-10.4\ \text{eV}

-5.5\ \text{eV}

and

-1.6\ \text{eV}

. Take

h = 6.63 \times 10^{-34}\ \text{J s}

c = 3.00 \times 10^8\ \text{m s}^{-1}

1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}

. (a) Find the energy of the photon emitted in a transition from

-1.6\ \text{eV}

-5.5\ \text{eV}

. (b) Find its wavelength.

Show worked answer →

(a) Photon energy: $hf = E_i - E_f = (-1.6) - (-5.5) = 3.9\ \text{eV}$ . In joules: $3.9 \times 1.60 \times 10^{-19} = 6.24 \times 10^{-19}\ \text{J}$ .

(b) Wavelength: $\lambda = \dfrac{hc}{E} = \dfrac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{6.24 \times 10^{-19}} = \dfrac{1.989 \times 10^{-25}}{6.24 \times 10^{-19}} = 3.19 \times 10^{-7}\ \text{m}$ .

The emitted photon has wavelength about $319\ \text{nm}$ (ultraviolet).

Markers reward the photon energy as the difference between levels, conversion to joules, and the wavelength from $\lambda = hc/E$ with units.

Original4 marks(a) Explain what is meant by discrete energy levels in an atom. (b) Explain how a line absorption spectrum is produced and why the dark lines correspond exactly to lines in the emission spectrum of the same element.

Show worked answer →

(a) Discrete energy levels means an atom's electrons can occupy only certain allowed (quantised) energies, not any value in between. Energy is quantised within the atom.

(b) When white light (a continuous spectrum) passes through a cool gas, atoms absorb photons whose energy exactly matches the difference between two energy levels, lifting electrons to higher levels. These specific wavelengths are removed from the transmitted light, appearing as dark lines.

The dark absorption lines correspond exactly to the bright emission lines of the same element because both arise from the same set of energy-level differences: absorption raises electrons across a gap, emission releases a photon when an electron falls across the same gap.

Markers reward quantised allowed energies for part (a), absorption of matching photons producing dark lines, and the explanation that emission and absorption share the same energy-level differences.