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IB Physics HL/Notes/E.1 Structure of the atom

IB Physics HLE.1 Structure of the atomNotes

Explain Rutherford experiment

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.

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Reasons
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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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Explain Nuclear notation

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.

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Reasons
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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.

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Explain Atomic energy levels

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.

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Reasons
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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.

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Atomic transitions

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.

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Reasons
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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.

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Explain Photon energy

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

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.

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Explain Spectra and composition

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.

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Reasons
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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.

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Track Nuclear radius

The one-third power is the clue. If radius grows as A^(1/3), then volume grows as A, matching the way nuclear mass grows with nucleon number. That is why nuclear densities are similar.

HL uses the nuclear radius model R = R0A^(1/3).
A is nucleon number and R0 is an empirical constant of order 10^-15 m.
Because R^3 is proportional to A, nuclear volume is proportional to A.
Nuclear mass is also approximately proportional to A.
Density = mass/volume is therefore approximately constant for different nuclei.

Assemble the nuclear-radius density argument.

Formula
Target formula R = R0A^(1/3)
R
nuclear radius
m
R0
nuclear radius constant
m
A
nucleon number
ρ
nuclear density
kg m^-3
V
nuclear volume
m^3
1Use the empirical radius model.R = R0A^(1/3)
2Cube the radius scaling.V ∝ R^3 ∝ A
3Nuclear mass scales with nucleon number.m ∝ A
4Divide mass by volume.ρ = m/V ≈ constant

Use the nuclear-radius relation to explain why nuclear density is approximately constant.

Stating R is proportional to A rather than A^(1/3).

Use the nuclear-radius relation to explain why nuclear density is approximately constant.

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High-energy scattering deviations

This HL point is about model limits. Rutherford’s electrostatic model works while alpha particles stay outside the nuclear-force region. Higher energy particles get closer, so deviations appear.

Rutherford scattering at lower alpha-particle energies can be explained by electrostatic repulsion between alpha particles and the positive nucleus.
Higher energy alpha particles can approach closer to the nucleus before turning back.
At sufficiently small separations, the alpha particles begin to encounter nuclear-force effects.
The observed scattering then deviates from the purely Coulomb/Rutherford prediction.
The deviation provides evidence about the finite size of the nucleus and the onset of the strong nuclear interaction region.

Match each high-energy scattering cue to its model meaning.

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Reasons
0/5

Explain why Rutherford scattering results change when higher energy alpha particles are used.

Saying higher energy particles are deflected less without mentioning closer approach and model limits.

Explain why Rutherford scattering results change when higher energy alpha particles are used.

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Closest approach

Closest approach is an energy conversion calculation. The alpha particle starts with kinetic energy; at the turning point, that energy has become electric potential energy between the alpha particle and nucleus.

In a head-on alpha scattering event, the alpha particle slows as electric potential energy increases.
At the distance of closest approach, the alpha particle is momentarily at rest in the simple model.
Energy conservation gives initial kinetic energy equal to electrostatic potential energy at the turning point.
For alpha charge +2e and nuclear charge +Ze, E_k = k(2e)(Ze)/d.
Therefore d = k(2e)(Ze)/E_k, assuming only electric repulsion is considered.

Assemble the closest-approach equation.

Formula
Target formula d = k(2e)(Ze)/E_k
d
distance of closest approach
m
k
Coulomb constant
N m^2 C^-2
e
elementary charge
C
Z
proton number of target nucleus
E_k
initial kinetic energy of alpha particle
J
1Identify the charges.q_alpha = +2e, q_nucleus = +Ze
2At closest approach, kinetic energy has become electric potential energy.E_k = k(2e)(Ze)/d
3Rearrange for distance.d = k(2e)(Ze)/E_k
4State the model condition.head-on scattering; electric repulsion only

Use energy conservation to derive the distance of closest approach for an alpha particle scattered head-on by a nucleus.

Using gravitational or force-balance reasoning instead of electric potential energy.

Use energy conservation to derive the distance of closest approach for an alpha particle scattered head-on by a nucleus.

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Explain Bohr hydrogen levels

Bohr energy levels are negative because zero is chosen for a free electron at infinity. Transitions use differences between levels, not the level value alone.

The Bohr model for hydrogen has discrete energy levels labelled by principal quantum number n = 1, 2, 3, ...
The energy of level n is E_n = -13.6/n^2 eV.
n = 1 is the ground state and has the most negative energy.
As n increases, energy approaches 0 eV, corresponding to an electron far from the atom.
A photon is emitted or absorbed when the electron changes level, with photon energy equal to the difference between levels.
The Bohr model predicts the hydrogen spectrum but does not work generally for multi-electron atoms.

Assemble the Bohr hydrogen level calculation.

Formula
Target formula E_n = -13.6/n^2 eV
E_n
energy of hydrogen level n
eV
n
principal quantum number
ΔE
transition energy magnitude
eV or J
h
Planck constant
J s
1Calculate the energy of each hydrogen level.E_n = -13.6/n^2 eV
2Find the transition energy magnitude.ΔE = |E_final - E_initial|
3Link the gap to photon energy.ΔE = hf
4State the model scope.hydrogen spectrum; limited for multi-electron atoms

Use the Bohr model to calculate the energy difference for a hydrogen transition.

Using the final level energy as the photon energy.

Use the Bohr model to calculate the energy difference for a hydrogen transition.

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Use Bohr angular momentum

The Bohr model does not allow arbitrary orbital radii. Quantized angular momentum selects allowed orbits, which then correspond to discrete energies.

Bohr proposed that electrons in hydrogen can occupy only certain circular orbits.
The allowed orbits satisfy angular momentum quantization: mvr = nh/(2π).
n is a positive integer called the principal quantum number.
An electron in an allowed orbit has a fixed energy and does not radiate continuously in the model.
Radiation is emitted or absorbed only when the electron transitions between allowed orbits.
This model was a successful step for hydrogen but is not a complete modern description of atoms.

Assemble Bohr angular momentum quantization.

Formula
Target formula mvr = nh/2π
m
electron mass
kg
v
electron orbital speed
m s^-1
r
orbit radius
m
n
principal quantum number
h
Planck constant
J s
1Write orbital angular momentum.L = mvr
2Apply Bohr quantization.mvr = nh/2π
3State allowed n values.n = 1, 2, 3, ...
4Connect to allowed orbits.only certain r and energy values are allowed

State Bohr’s angular momentum quantization condition and explain its consequence for hydrogen.

Saying electrons can orbit at any radius.

State Bohr’s angular momentum quantization condition and explain its consequence for hydrogen.

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Retrieve the Core E.1 Structure of the atom Model

Review

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.

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Reasons
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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.

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Retrieve the HL E.1 Structure of the atom Model

Review

HL E.1 adds quantitative model checks. Radius scaling explains density; closest approach uses energy conservation; high-energy scattering shows model limits; Bohr explains hydrogen spectra through quantized levels and angular momentum.

Nuclear radius follows R = R0A^(1/3), so volume and mass both scale approximately with A and nuclear density is roughly constant.
Higher energy alpha particles approach closer to the nucleus and can show deviations from Rutherford’s purely electrostatic scattering prediction.
For head-on alpha scattering, closest approach uses energy conservation: E_k = k(2e)(Ze)/d.
The Bohr hydrogen energy levels are E_n = -13.6/n^2 eV.
Bohr quantized angular momentum with mvr = nh/(2π), allowing only certain orbits.
The Bohr model predicts hydrogen spectra but is not a complete model for multi-electron atoms.

Match each HL E.1 cue to the correct model or condition.

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Summarize the HL extensions to E.1 Structure of the atom.

Writing equations without their physical assumptions or model limits.

Summarize the HL extensions to E.1 Structure of the atom.

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