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E.1 Structure of the atomIB Physics SL: Revision notes

Section 1

The Geiger–Marsden–Rutherford experiment

Alpha particles were fired at a very thin gold foil in a vacuum and detected by flashes on a zinc sulfide screen at different angles.

  • Most passed straight through: the atom is mostly empty space.
  • A few were deflected through large angles and a very few bounced back: the positive charge and almost all the mass are concentrated in a tiny, dense nucleus.

Rutherford replaced the 'plum-pudding' model (positive charge spread through the atom) with the nuclear model: a small positive nucleus with electrons far outside it. An alpha particle is turned back only when it heads almost directly at a nucleus and feels a large electrostatic repulsion.

Key termsalpha particlenucleusnuclear model
Common mistake

Electrons did not cause the large deflections: an electron is about 7300 times lighter than an alpha particle.

Section 2

Nuclear notation

A nuclide is written ZAX^{A}_{Z}\mathrm{X}, where A is the nucleon number (protons + neutrons), Z is the proton number and X is the chemical symbol. The number of neutrons is N = A − Z. A neutral atom has Z electrons.

Example: 2656Fe^{56}_{26}\mathrm{Fe} has 26 protons, 30 neutrons and, when neutral, 26 electrons. Isotopes have the same Z but different A.

Key termsnucleon numberproton numbernuclide

Section 3

Discrete energy levels and photons

Electrons in an atom can only have certain discrete energy levels. When an electron moves down a level, a photon is emitted; to move up, a photon of exactly the right energy is absorbed (or energy is gained in a collision).

The photon energy equals the difference between the levels: ΔE = hf = hc/λ. Energies are often given in electronvolts: 1 eV = 1.60 × 10⁻¹⁹ J. Larger gaps give higher frequencies and shorter wavelengths.

Key termsenergy levelphotonelectronvolt
Exam tip

Convert eV to J before using E = hc/λ, or use hc = 1.99 × 10⁻²⁵ J m and divide by 1.60 × 10⁻¹⁹ at the end.

Section 4

Emission and absorption spectra

A hot, low-pressure gas gives an emission spectrum: bright lines on a dark background. White light passed through a cooler gas gives an absorption spectrum: dark lines on a continuous spectrum. The dark lines appear at the same wavelengths as the element's emission lines, because the same energy gaps are involved.

Because only certain lines exist, spectra are direct evidence for discrete energy levels. A continuous range of energies would give a continuous spectrum. At room temperature nearly all atoms are in the ground state, so absorption lines start from the ground state.

Key termsemission spectrumabsorption spectrumground state

Section 5

Spectra and chemical composition

Each element has its own set of energy levels, so its line pattern is a unique fingerprint. Matching the wavelengths of lines in a spectrum to laboratory spectra identifies the elements present, for example in stars, flames or lamps.

In a star, the hot interior emits a continuous spectrum and the cooler outer layers absorb; the absorption lines reveal the composition of those outer layers.

Key termsspectral fingerprint
Common mistake

Missing lines do not prove an element is absent: atoms may be ionised or not in the right starting level at that temperature.

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