E.1 Structure of the atom

Syllabus
First assessment 2025
Topic
—
Level
SL

Learning objectives

Interpret Rutherford Scattering

Set up the evidence

In the Geiger–Marsden–Rutherford experiment, alpha particles were directed at a thin gold foil and detected around the foil. Most particles passed through without deflection, some were deflected, and a very small number scattered backwards.

Infer the nuclear model

The results show that an atom is mostly empty space. The rare large deflections require a small, dense, positively charged nucleus that contains most of the atom’s mass; the positive charge cannot be spread uniformly through the whole atom.

Keep the conclusion qualitative

For SL, focus on linking each observation to the model: many undeflected particles imply empty space, while rare back-scattering implies a concentrated repulsive centre. Do not treat the experiment as evidence that electrons occupy fixed-radius orbits.

Common trap

Do not say that all alpha particles are deflected. The dominant observation is that most pass through essentially undeflected; the large-angle events are rare but decisive.

E.1.1 Exam Analysis

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Read Nuclear Notation

Read the symbol

Nuclear notation is written as ZAX{}^{A}_{Z}X. The chemical symbol XX identifies the element, the proton number ZZ is written below, and the nucleon number AA is written above.

Count the nucleus

The nucleus contains ZZ protons and N=A−ZN=A-Z neutrons. The number of electrons is not encoded by AA and ZZ; for a neutral atom it equals ZZ, while an ion has gained or lost electrons.

Compare nuclides

Atoms of the same element have the same ZZ. Isotopes have the same ZZ but different AA, so they contain different numbers of neutrons.

Common trap

Do not use the electron count as the proton number for an ion, and do not confuse AA with the number of neutrons. Subtract ZZ from AA to find the neutron number.

E.1.2 Exam Analysis

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Read Spectral Evidence

Emission lines

An excited gas emits light at particular frequencies, producing bright spectral lines rather than a continuous spread of frequencies. Each line corresponds to a permitted energy difference between atomic states.

Absorption lines

When continuous light passes through a cooler gas, the atoms remove the same frequencies they can emit. The resulting dark lines therefore occur at specific, repeatable wavelengths.

Infer discrete levels

Because only particular photon energies are emitted or absorbed, the atom’s energy states are discrete rather than continuous. The spectrum is evidence for quantized atomic energy levels.

Common trap

Do not treat every visible line as a separate element without considering transitions. A spectrum is evidence of allowed energy differences; the pattern, not simply the number of lines, carries the information.

E.1.3 Exam Analysis

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Model Atomic Transitions

Emission

When an electron moves from a higher atomic energy level to a lower one, the atom emits one photon. The photon energy equals the level difference: Eγ=ΔEE_\gamma=\Delta E.

Absorption

An atom can absorb a photon only when its energy matches an allowed upward transition. The electron then moves to the higher level, so the spectrum records the same allowed energy differences in reverse.

Read a transition diagram

For each downward arrow, calculate the energy gap between its initial and final levels. A larger gap produces a higher-frequency photon and a shorter wavelength; a smaller gap produces a lower-frequency photon and a longer wavelength.

Common trap

Do not use the absolute energy of one level as the photon energy. A photon is associated with the difference between two levels, and emission requires a downward transition.

E.1.4 Exam Analysis

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Calculate Photon Energy

Use the photon relation

Photon energy depends on frequency. For an atomic transition, first take the positive magnitude of the energy-level difference, then convert units consistently before finding frequency or wavelength.

E_\gamma=hf=\frac{hc}{\lambda}=|E_i-E_f|

Worked example — hydrogen transition

A transition with Eγ=1.89 eVE_\gamma=1.89\,\mathrm{eV} has energy (1.89)(1.60×10−19)=3.02×10−19 J(1.89)(1.60\times10^{-19})=3.02\times10^{-19}\,\mathrm{J}. Hence f=E/h=(3.02×10−19)/(6.63×10−34)=4.56×1014 Hzf=E/h=(3.02\times10^{-19})/(6.63\times10^{-34})=4.56\times10^{14}\,\mathrm{Hz} and λ=c/f=6.58×10−7 m\lambda=c/f=6.58\times10^{-7}\,\mathrm{m}.

Connect energy to a level gap

For an atomic transition, use Eγ=∣Ei−Ef∣E_\gamma=|E_i-E_f|. Take the magnitude of the energy difference, then convert units consistently before finding frequency or wavelength.

Predict the wavelength

A larger energy gap gives a higher-frequency photon and a shorter wavelength. Therefore the longest wavelength comes from the smallest non-zero transition energy.

Common trap

Do not carry a negative sign from bound-state energies into photon energy. The photon energy is positive and equals the magnitude of the level difference.

E.1.5 Exam Analysis

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Identify Elements from Spectra

Treat a spectrum as a fingerprint

Each element has a characteristic set of emission and absorption wavelengths because its allowed energy differences are unique. The pattern can therefore identify the chemical species producing or absorbing the light.

Use comparison evidence

Record the observed spectral lines and compare their wavelengths or frequencies with laboratory spectra of known elements. Matching several characteristic lines supports the identification.

Apply it to stars

Light from a star can contain absorption lines produced by cooler gases in its atmosphere. Comparing those lines with known spectra reveals which elements are present, even when the source cannot be sampled directly.

Common trap

Do not identify an element from one broad colour alone. The evidence is the set of matching spectral lines and their wavelengths.

E.1.6 Exam Analysis

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Retrieve the SL Atomic Model

Retrieve the evidence chain

Rutherford scattering supports a small positive nucleus; nuclear notation separates protons, neutrons and electrons; line spectra show discrete energy differences; and Eγ=hf=hc/λE_\gamma=hf=hc/\lambda connects transitions to photons.

Check the model

When reading a spectrum, identify the transition, use the energy difference rather than an absolute level, and compare characteristic lines with known spectra to identify elements.