Rutherford is an evidence-to-model story. Do not just say “gold foil experiment”; state what happened to the alpha particles and what each result forced Rutherford to conclude about the atom.

Geiger and Marsden fired alpha particles at thin gold foil and observed their scattering angles.

Most alpha particles passed straight through, showing the atom is mostly empty space.

Some alpha particles were deflected through small angles, showing positive charge affects their paths.

A very small number were deflected through angles greater than 90 degrees or bounced back.

Large deflections showed that nearly all the mass and positive charge are concentrated in a tiny dense nucleus.

The nuclear model replaced the Thomson plum-pudding model of diffuse positive charge.

Match each Rutherford observation to its conclusion.

Match

Reasons

0/5

Describe Rutherford’s scattering experiment and explain the conclusions drawn from the observations.

Listing the observations without linking them to the nuclear model.

Describe Rutherford’s scattering experiment and explain the conclusions drawn from the observations.

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Nuclear notation is bookkeeping. Z tells you the element because it counts protons. A tells you the number of nucleons. The missing count, neutrons, is A-Z.

Standard nuclear notation writes a nuclide as A over Z beside the chemical symbol X.

Z is the proton number; it defines the element.

A is the nucleon number: total protons plus neutrons.

Neutron number is A - Z.

The nucleus contains protons and neutrons; electrons occupy the region around the nucleus.

A typical atom is about 10^-10 m across, while a nucleus is about 10^-15 m across.

Match each nuclear-notation symbol or structure cue to its meaning.

Match

Reasons

0/6

For a nuclide written in nuclear notation, identify A, Z, and the number of neutrons.

Confusing nucleon number with neutron number.

For a nuclide written in nuclear notation, identify A, Z, and the number of neutrons.

Choose

Spectral lines are the evidence, and discrete energy levels are the conclusion. Each line is not an energy level by itself; it is a transition between two levels.

Atoms emit and absorb only specific photon energies.

Specific photon energies produce sharp spectral lines rather than a continuous spectrum.

This is evidence that electrons in atoms have discrete allowed energy levels.

A spectral line corresponds to an electron transition between two energy levels.

The photon energy equals the energy difference between the levels: ΔE = hf.

Match each spectrum cue to the energy-level conclusion.

Match

Reasons

0/5

Explain how spectral lines provide evidence for discrete atomic energy levels.

Saying lines are just colours without linking to quantized energy differences.

Explain how spectral lines provide evidence for discrete atomic energy levels.

Choose

Atomic spectra are produced by transitions. Upward transitions require absorption of exactly the right photon energy. Downward transitions release a photon with energy equal to the drop.

An electron absorbs a photon when the photon energy matches the gap to a higher allowed energy level.

Absorption removes those photon frequencies from transmitted light, producing dark lines in an absorption spectrum.

An electron emits a photon when it transitions from a higher energy level to a lower energy level.

Emission produces bright lines at photon energies equal to energy-level differences.

Energy is conserved: photon energy equals the magnitude of the level difference.

Match each atomic-transition cue to emission or absorption.

Match

Reasons

0/5

Describe how emission and absorption spectra are produced by atomic transitions.

Mixing up upward and downward transitions.

Describe how emission and absorption spectra are produced by atomic transitions.

Choose

This is the calculation card for line spectra. First find the energy gap, then choose frequency or wavelength form. Be precise with units: eV is useful for level diagrams, but SI frequency calculations need joules.

For an atomic transition, photon energy equals the magnitude of the energy-level difference: ΔE.

Photon energy is E = hf, where h is Planck’s constant and f is frequency.

Using c = fλ gives E = hc/λ.

If energy levels are given in electronvolts, convert to joules for SI calculations unless the answer is requested in eV.

Emission and absorption use the same energy-gap equation; only the direction of the transition changes.

Assemble the photon-energy equations.

Formula

Target formula ΔE = hf = hc/λ

ΔE

magnitude of energy-level difference

J or eV

h

Planck constant

J s

f

photon frequency

Hz

c

speed of light

m s^-1

λ

photon wavelength

m

1Find the magnitude of the energy-level difference.ΔE = |E_high - E_low|

2Use the photon-energy equation.ΔE = hf

3Combine with c = fλ.ΔE = hc/λ

4Convert eV to J when using SI constants.1 eV = 1.60 × 10^-19 J

Build the formula

Calculate the frequency or wavelength of a photon emitted in an atomic transition.

Using the lower energy level itself instead of the difference between levels.

Calculate the frequency or wavelength of a photon emitted in an atomic transition.

Choose

Spectra act like fingerprints because each element has a unique set of energy levels. In exams, the key phrase is “pattern of lines,” not a single colour.

Each element has a unique set of allowed energy-level differences.

Therefore each element has a unique emission and absorption line spectrum.

Matching the pattern of observed spectral lines to laboratory reference spectra identifies elements present.

Absorption spectra from starlight can reveal elements in a star’s atmosphere.

A reliable identification uses several matching lines, not just one isolated line.

Match each spectrum-analysis cue to its composition meaning.

Match

Reasons

0/5

Explain how emission or absorption spectra can be used to determine chemical composition.

Saying the brightest line alone identifies the element.

Explain how emission or absorption spectra can be used to determine chemical composition.

Choose

This retrieval card ties the SL E.1 story together: scattering evidence builds the nuclear atom, while spectral evidence builds the discrete energy-level model.

Rutherford scattering showed the atom is mostly empty space with a tiny dense positively charged nucleus.

Nuclear notation uses Z for proton number and A for nucleon number; neutron number is A-Z.

Atoms are about 10^-10 m across and nuclei about 10^-15 m across.

Emission and absorption spectra are produced by electron transitions between discrete atomic energy levels.

Photon energy equals the energy-level difference: ΔE = hf = hc/λ.

Each element has a unique spectral-line pattern, so spectra provide information about chemical composition.

Match each core E.1 cue to its model statement.

Match

Reasons

0/7

Summarize the core E.1 model of atomic structure and spectra.

Treating Rutherford scattering and spectra as disconnected facts.

Summarize the core E.1 model of atomic structure and spectra.

Choose

IB Physics SL E.1 Structure of the atom Notes

Explain Rutherford experiment

Explain Nuclear notation

Explain Atomic energy levels

Atomic transitions

Explain Photon energy

Explain Spectra and composition

Retrieve the Core E.1 Structure of the atom Model