IB Physics HL E1 Structure of the Atom — Complete Cheatsheet
Every formula, definition and exam trap for IB Physics Topic E.1 — Rutherford scattering, atomic spectra, nuclear radius, density and the Bohr model. Hand-built by an IBO-experienced Singapore tutor.
Topic: Structure of the Atom (Quantum & Nuclear)Syllabus: Topic E.1 (SL + HL)Read time: ~18 minutesLast updated: Apr 2026
Topic E.1 — Structure of the Atom — opens the IB Physics quantum & nuclear block, and it is one of the most reliably examined nuclear topics across Paper 1, Paper 2 and Paper 3. The SL portion (Geiger–Marsden, nuclear notation, atomic spectra) returns every year as conceptual MCQs, while the HL extension (nuclear radius and density, deviations from Rutherford, the Bohr model) carries the heavier calculation marks.
This cheatsheet condenses every formula, definition, trick and trap for Topic E.1 SL + HL onto one page you can revise from before any mock or final paper. Scroll to the bottom for the printable PDF, the full notes pack, and the gated tutorial library used by Photon Academy students in Singapore.
§1 — Structure of Matter & the Nuclear Model E.1 SL+HL
Everything around us is built from about a hundred elements, and every atom is made of just three particles: protons, neutrons and electrons. In the nuclear model, protons and neutrons (together called nucleons) sit in a tiny, dense central nucleus; the electrons occupy discrete energy levels around it. Almost all the mass and all the positive charge is in the nucleus — the rest of the atom is essentially empty space.
Isotopes: same $Z$, different $A$ — the same element (identical chemistry) but a different number of neutrons. Unified atomic mass unit: $1\,\text{u} = \tfrac{1}{12}$ of the mass of a ${}^{12}_{\phantom{0}6}$C atom $= 1.66 \times 10^{-27}$ kg. Handy rest masses: proton $1.007\,276$ u, neutron $1.008\,665$ u, electron $0.000\,549$ u.
The nuclear model. The nucleus is ~100 000× smaller than the atom — if the nucleus were a football, the atom would be about 30 km across.
TrickIn any nuclear reaction, conserve $A$ (top numbers) and $Z$ (bottom numbers) separately. A $\gamma$-photon carries no $A$ and no $Z$.
Why a new model was neededA classical accelerating charge radiates energy, so an orbiting electron should spiral into the nucleus. It doesn't — which is the first hint that electron energies are quantised (§3, §7), not continuous.
From the Photon question bank
What examiners actually test on this concept:
Nuclide notation from particle countsNucleon vs proton numberDefinition of the unified mass unitIsotopes & molar mass
The unified atomic mass unit is almost-guaranteed recall.frequent $1\,\text{u} = \tfrac{1}{12}$ the mass of a carbon-12 atom. Distractors offer "average of proton and neutron mass" or an oxygen-16 standard — both wrong.
"Given protons/neutrons/electrons, write ${}^{A}_{Z}\text{X}$." $A =$ protons + neutrons, $Z =$ protons only. Extra or missing electrons change the charge, never the nuclear notation.
Equal-mass sample MCQs. For equal masses, the substance with the smallest molar mass has the most atoms/molecules ($N \propto 1/M$). Build molar mass from the nucleon numbers first.
§2 — Evidence for the Nucleus: Geiger–Marsden–Rutherford E.1 SL+HL
Setup: a beam of positive $\alpha$ particles of known energy is fired at a very thin gold foil in a vacuum; a movable detector counts how many scatter to each angle.
Observation
Conclusion
Most $\alpha$ pass almost straight through
The atom is mostly empty space
A small fraction deflect through large angles
Positive charge & mass are concentrated in a tiny nucleus
~1 in 8000 bounce back (beyond $90^\circ$)
The nucleus is extremely small, dense and positively charged
The number scattered at each angle follows an inverse-square law of repulsion from the nucleus (Coulomb's law) — quantitative agreement is what made the nuclear model convincing. These results disproved the "plum-pudding" model, in which positive charge was spread through the whole atom.
Geiger–Marsden setup. Rare large-angle and back-scattering events reveal a tiny, dense, positive nucleus surrounded by empty space.
TrapThe scattering experiment does NOT prove discrete electron energy levels — only "empty space" and "concentrated positive nucleus." Evidence for energy levels comes from spectra (§3). Any answer pairing Geiger–Marsden with "energy levels" is wrong.
From the Photon question bank
What examiners actually test on this concept:
Interpret α-scattering dataPlum-pudding vs nuclear modelValid vs invalid conclusions
The most repeated trap in the whole topic:frequent the experiment gives no evidence for energy levels. Only empty space + dense positive nucleus are valid conclusions.
Know the historical logic. Large-angle back-scatter is what disproved the plum-pudding model and established the nuclear model — a recurring conceptual MCQ.
Learn the observation → conclusion pairing as a table. "Most pass straight → empty space; ~1 in 8000 rebound → tiny dense nucleus" is worth marks verbatim.
Electrons in an atom can only occupy discrete (quantised) energy levels. An electron moves between levels only by absorbing or emitting a photon whose energy exactly matches the gap. Because the levels are fixed and unique to each element, only specific photon energies — and therefore specific wavelengths — appear.
Transition energy:$\Delta E = E_{\text{upper}} - E_{\text{lower}}$ (absorbed on the way up, emitted on the way down)
Emission vs absorption
Emission spectrum: a hot, excited gas emits — you see bright discrete lines on a dark background.
Absorption spectrum: white light passes through a cooler gas — you see dark lines on a continuous spectrum, at the same wavelengths as that element's emission lines.
Chemical fingerprint: each element has a unique line pattern, so spectra identify the composition of stars and galaxies. (Helium was found in the Sun's spectrum before it was found on Earth.)
Same element, two spectra. The dark absorption lines fall at exactly the wavelengths of the bright emission lines.
Energy-level diagrams
Energy levels are drawn as horizontal lines, most negative at the bottom (ground state). A downward arrow emits a photon; an upward arrow absorbs one. An electron freed from the atom ($E \ge 0$) is ionised — do not confuse this with merely excited.
Hydrogen energy levels. Gold = photon emitted (electron falls); navy dashed = photon absorbed (electron rises). Levels bunch up towards $n=\infty$.
TrickFor $n$ levels there are $\dfrac{n(n-1)}{2}$ possible downward transitions — and so that many spectral lines. The longest wavelength comes from the smallest energy gap ($\lambda \propto 1/\Delta E$).
TrapFor a two-step cascade, the reciprocals of wavelength add: $\dfrac{1}{\lambda_{\text{total}}} = \dfrac{1}{\lambda_1} + \dfrac{1}{\lambda_2}$ (because energies add). Adding the wavelengths themselves is wrong.
From the Photon question bank
What examiners actually test on this concept:
Count transitions n(n−1)/2Longest λ = smallest gapExcited vs ionisedEM region from λEmission/absorption fingerprint
Energy-level diagrams are examined almost every session.very frequent Read values carefully and mind the negative signs when forming $\Delta E$.
"Longest wavelength emitted" = smallest gap; "how many lines?" = $n(n-1)/2$. Both are recurring one-mark MCQs.
Distinguish excited from ionised, and absorption (up) from emission (down). After finding $E$ from $E=hc/\lambda$, convert J ↔ eV to match a level in the data table, and identify the EM region from $\lambda$.
§4 — Nuclear Radius & Density E.1 HL only
Nuclei behave like hard spheres of tightly-packed nucleons at essentially constant density. The radius grows only as the cube root of the nucleon number:
Because volume $V = \tfrac{4}{3}\pi R_0^3 A \propto A$ and mass $\propto A$, the factor of $A$ cancels — density is the same for every nucleus (~$10^{14}$ times denser than ordinary matter; the density of a neutron star).
Cube-root scaling. $8\times$ the nucleons ($2^3$) gives $2\times$ the radius — and identical packing density.
Trick — ratio methodTo compare two nuclei: $\dfrac{R_Y}{R_X} = \left(\dfrac{A_Y}{A_X}\right)^{1/3}$. E.g. $A_X = 2 \to A_Y = 16$: $(16/2)^{1/3} = 8^{1/3} = 2$, so $R_Y = 2R_X$. To find $A$ from a measured $R$: cube it — $A = (R/R_0)^3$.
TrapNever say density increases with $A$. Nuclear density is (approximately) constant for all nuclei — this is a distractor in nearly every HL radius question.
From the Photon question bank
What examiners actually test on this concept:
Apply R = R₀A^(1/3)Density constant across nuclidesSolve for A from RRadius/volume scaling
$R = R_0 A^{1/3}$ is the single most-examined HL formula in the E.1 bank.very frequent Expect either "scale $R$ when $A$ changes" (cube root) or "find $A$ from a measured $R$" (cube and rearrange).
Constant density is a near-permanent distractor. Since $V \propto A$ and mass $\propto A$, $\rho$ is independent of $A$. State this explicitly for the mark.
Electron-diffraction sizing sometimes pairs with this ($\theta_{\min} \approx \lambda/D$): if the nuclear diameter doubles, the first-minimum angle halves.
§5 — Closest Approach & Deviations from Rutherford E.1 HL only
Fire an $\alpha$ particle head-on at a nucleus. As it approaches it slows, converting all its kinetic energy into electric potential energy at the distance of closest approach $d$, where it momentarily stops before being repelled back.
Head-on closest approach. At $d$ the $\alpha$ is momentarily at rest: all its KE is now Coulomb PE.
Deviations from the Rutherford formula
Low-energy $\alpha$: only Coulomb repulsion acts — the Rutherford (inverse-square) formula fits perfectly.
High-energy $\alpha$: the particle gets close enough to enter the range of the strong nuclear force, so fewer large-angle deflections occur than Rutherford predicts.
These deviations are evidence for the strong nuclear force and let physicists estimate the nuclear radius (the closest approach at the onset energy is roughly $R$).
TrapDeviations appear only at HIGH energy (close approach), never at low energy. Also: $d$ is measured from the nucleus centre — if $d \gg R$, pure Rutherford scattering still holds.
From the Photon question bank
What examiners actually test on this concept:
KE → PE closest approachCharges 2e and ZeMeV → J conversionDeviations ⇒ strong force + radius
Two marks are routinely lost by forgetting the $\alpha$ charge is $2e$ and by not converting MeV → J before substituting into $d = k(2e)(Ze)/E_k$.
"Why do deviations occur?"frequent Correct: evidence for the strong nuclear force + a way to estimate nuclear radius, at HIGH energy. "Deviations at low energy" is the standard wrong option.
§6 — Hydrogen Spectrum & the Rydberg Formula E.1 HL only
The visible hydrogen lines fit a simple formula found by Balmer, later generalised by Rydberg. Every hydrogen line is a transition down to a fixed lower level $n_1$:
Hydrogen series. Transitions ending on $n_1=1$ (Lyman) are UV; on $n_1=2$ (Balmer) are visible; on $n_1=3$ (Paschen) are IR.
TrickVisible lines ⇒ Balmer ⇒ $n_1 = 2$. The Rydberg formula and the Bohr energy levels (§7) describe the same lines — you can get any wavelength from either.
From the Photon question bank
What examiners actually test on this concept:
Rydberg 1/λ = R(1/n₁²−1/n²)Identify the seriesConvert to energy via E = hc/λ
Know the series by their lower level: Lyman (1, UV), Balmer (2, visible), Paschen (3, IR). "Which series is visible?" → Balmer.
Watch the algebra of fractions with a common denominator, and give the final wavelength to sensible significant figures.
§7 — The Bohr Model of Hydrogen E.1 HL only
Bohr fixed the "spiralling electron" problem by postulating that only orbits with quantised angular momentum are allowed, and that an electron radiates only when it jumps between them:
Angular momentum:$m_e v r = \dfrac{nh}{2\pi} = n\hbar, \quad n = 1, 2, 3, \ldots$
Energy levels:$E_n = -\dfrac{13.6}{n^{\,2}}\,\text{eV}$
Orbital scaling:$r_n \propto n^2$ and, from $m_evr = nh/2\pi$, $v_n \propto \dfrac{1}{n}$ (higher levels orbit slower)
Sketch of the derivation: equate the Coulomb force to the centripetal force, $\dfrac{e^2}{4\pi\varepsilon_0 r^2} = \dfrac{m_e v^2}{r}$; substitute $v$ from the quantisation rule to get $r_n = \dfrac{\varepsilon_0 n^2 h^2}{\pi m_e e^2}$; then $E_n = \text{KE} + \text{PE} = -\dfrac{m_e e^4}{8\varepsilon_0^2 h^2 n^2}$ — which is the $-13.6/n^2$ eV result and has the same form as the Rydberg formula.
$n$
$E_n$
State
1
$-13.6$ eV
Ground
2
$-3.40$ eV
1st excited
3
$-1.51$ eV
2nd excited
$\infty$
$0$ eV
Ionised
Bohr orbits. Only orbits with $L = nh/2\pi$ are allowed; radius grows as $n^2$.
TrapThe Bohr model works only for hydrogen and one-electron ions (He⁺, Li²⁺). Never apply $-13.6/n^2$ to neutral helium or heavier atoms. It also can't explain line intensities or fine structure — use energy-level diagrams from the data booklet for multi-electron atoms.
From the Photon question bank
What examiners actually test on this concept:
Angular momentum L = nh/2πOnly hydrogen / one-electronv ∝ 1/n, r ∝ n²Identify Bohr equations
Angular momentum is quantised: $L = n\hbar = nh/2\pi$.frequent Ratio questions (ground vs first-excited) collapse to a ratio of $n$.
"Does $-13.6/n^2$ apply?" Only if exactly one electron remains — a favourite MCQ gives an isotope, removes one electron and asks you to check.
Scaling & sorting. $r \propto n^2$ and $v \propto 1/n$; and be ready to pick which listed equations are genuinely Bohr-model equations (energy + angular-momentum quantisation) rather than decay/other formulas.
Worked Example — Closest Approach & Bohr Transition
Question (HL Paper 2 style — 6 marks)
(a) An $\alpha$ particle with kinetic energy $E_k = 5.00$ MeV is fired head-on at a gold nucleus ($Z = 79$). Calculate the distance of closest approach $d$. [3]
(b) An electron in a hydrogen atom drops from $n = 4$ to $n = 2$. Calculate the wavelength of the emitted photon and state the spectral series. [3]
This is in the visible range with the lower level $n = 2$, so the line belongs to the Balmer series. (A1)
Examiner's note: Two classic mistakes — (i) forgetting that $E_n$ is negative when computing $\Delta E$, and (ii) using the alpha kinetic energy in MeV directly without converting to joules. Always convert to SI before plugging into $hc/\lambda$.
Common Student Questions
What does the Geiger-Marsden experiment actually prove?
Only two things: (i) the atom is mostly empty space, and (ii) positive charge and almost all the mass are concentrated in a tiny, dense nucleus. It does not prove discrete energy levels — that requires spectral line evidence. This is the IB May 2023 Paper 1 Q22 trap (correct answer: A). If a multiple choice option mentions "energy levels" alongside Geiger-Marsden, eliminate it.
Is nuclear density the same for all nuclei?
Yes. Because $R = R_0 A^{1/3}$, the nuclear volume is proportional to $A$, and the mass is also proportional to $A$, so density $\rho = m/V$ cancels the $A$ and stays constant at about $2.3 \times 10^{17}$ kg m${}^{-3}$ for every nucleus. Saying density rises with $A$ is a guaranteed mark loss in any HL E.1 question.
When does Rutherford scattering break down?
At high alpha energies — when the alpha particle gets close enough that the strong nuclear force starts to act. You then see fewer large-angle deflections than the pure Coulomb (Rutherford) prediction. This deviation gives evidence for the strong nuclear force AND lets you estimate the nuclear radius (the closest-approach distance at the energy of onset is roughly $R$).
Can I use $E_n = -13.6/n^2$ for helium or heavier atoms?
No. The Bohr model only works for hydrogen and other one-electron systems (e.g. He$^+$, Li${}^{2+}$). Applying $-13.6/n^2$ to neutral helium or heavier atoms is wrong. The IB only ever asks Bohr-level calculations on hydrogen — if a question gives you helium, expect to be using energy-level diagrams from the data booklet, not the $-13.6/n^2$ formula.
What is the difference between emission and absorption spectra?
Emission: a hot gas emits bright discrete lines on a dark background. Absorption: white light passes through a cool gas, leaving dark lines on a continuous spectrum — at the same wavelengths as the emission lines for the same element. Both arise because electrons can only occupy discrete energy levels, so only specific photon energies are absorbed or emitted. The same element therefore "fingerprints" both ways.
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy E.1 Structure of the Atom library (only available to enrolled students or via the resource library subscription) adds:
40-page Notes PDF — every concept worked through in full, with derivations and intuition.
Tutorial booklet — 30+ IB-style questions sequenced from foundation to AHL difficulty.
Tutorial Solutions — full examiner-style worked solutions with M1/A1/R1 annotations.
Practice Solutions — extra past-paper-style problems with detailed walk-throughs.
Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
