Why is "Confusing emission spectra (bright lines on dark) with absorption spectra (dark lines on continuous)" a common mistake in Structure of the atom?

Why it happens: Both involve line spectra at the same frequencies, so students remember 'lines' but not which appears bright and which dark. How to avoid it: Emission = electrons drop DOWN and give OUT light → BRIGHT lines on a DARK background. Absorption = light passes THROUGH a cool gas and frequencies are removed → DARK lines on a CONTINUOUS (rainbow) background. Source: IB Physics Theme E subject reports

Why is "Forgetting to convert electronvolts to joules (or nm to m) before using E = hf" a common mistake in Structure of the atom?

Why it happens: Energy levels are quoted in eV and wavelengths in nm, so students plug those numbers straight into equations that need SI units. How to avoid it: Before any photon calculation, convert energy to joules (× 1.60 × 10⁻¹⁹) and wavelengths to metres (× 10⁻⁹). Skipping this gives an answer wrong by a factor of about 10¹⁹. Source: IB Physics Theme E subject reports

Why is "Mixing up the nucleon number A and proton number Z, or miscounting neutrons" a common mistake in Structure of the atom?

Why it happens: The two numbers sit next to the symbol and students grab the wrong one, or forget that neutrons must be found by subtraction. How to avoid it: Z (bottom) = protons; A (top) = protons + neutrons. Neutrons are always N = A − Z (a subtraction) — never read A as the neutron number. Source: IB Physics Theme E subject reports

Why is "Thinking atomic energy levels are continuous rather than discrete" a common mistake in Structure of the atom?

Why it happens: Everyday experience with energy suggests it can take any value, so students assume electron energies form a smooth range. How to avoid it: Remember that atoms produce line spectra (only certain frequencies), which is direct evidence that the electron energies — and the jumps between them — are discrete, taking only fixed allowed values. Source: IB Physics Theme E subject reports

Why is "Listing the alpha-scattering observations without linking them to the correct conclusions" a common mistake in Structure of the atom?

Why it happens: Students memorise the observations and conclusions as two separate lists and do not pair them up as the question requires. How to avoid it: Pair each one: most straight through → mostly empty space; a few large-angle back-scatters → tiny, dense, positive nucleus holding most of the mass. State the observation AND its conclusion together. Source: IB Physics Theme E subject reports

Why is "Subtracting negative energy levels incorrectly when finding a transition energy" a common mistake in Structure of the atom?

Why it happens: Energy levels are negative, so students find the gap by adding magnitudes wrongly or dropping a minus sign. How to avoid it: The photon energy is E₂ − E₁ (higher minus lower). For example (−1.5) − (−6.0) = +4.5 eV, not 7.5 eV. Take the higher (less negative) value minus the lower (more negative) value to get a positive gap. Source: IB Physics Theme E subject reports

IB DP Physics (100452)

Structure of the atom

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Detailed Study Notes

Detailed notes on Nuclear and quantum physics for IB DP Physics, covering key concepts, explanations, examples, and exam-focused revision points.

E.1 Structure of the atom — IB Physics SL Study Notes (Theme E: Nuclear and Quantum Physics)

How we know the atom has a tiny, dense, positive nucleus (the Geiger–Marsden–Rutherford alpha-scattering experiment), how to read nuclear notation and isotopes, and why atoms have discrete energy levels — evidence for which comes from the sharp bright and dark lines of emission and absorption spectra and the photon relation hf = E₂ − E₁.

At a glance

Nuclear model: a tiny, dense, positive nucleus of protons and neutrons, with electrons in the (mostly empty) space around it.
Geiger–Marsden–Rutherford: fire alpha particles at thin gold foil. Most pass straight through, some deflect through small angles, a very few (< 1 in 8000) bounce back > 90°.
Those three observations → the atom is mostly empty space, the nucleus is tiny, and the nucleus is dense, positive and holds most of the mass.
Nuclear notation $^{A}_{Z}\text{X}$ : $A$ = nucleon (mass) number, $Z$ = proton number, and neutrons $N = A - Z$ .
Isotopes are atoms of the same element (same $Z$ ) with different numbers of neutrons (different $A$ ).
Atomic energy levels are discrete (quantised) — an electron can only have certain allowed energies, not any value in between.
A photon is emitted or absorbed when an electron changes level: $hf = E_2 - E_1$ , with $E = hf = \dfrac{hc}{\lambda}$ and $h = 6.63\times10^{-34}\,\text{J s}$ .
Emission spectrum = bright coloured lines on a dark background; absorption spectrum = dark lines on a continuous (rainbow) background. Both are evidence for discrete energy levels.

What you’ll learn

Mapped to the 100452 subject guide (2025-onwards).

Describe the nuclear model of the atom and explain how the Geiger–Marsden–Rutherford alpha-scattering experiment provides evidence for it.
Use nuclear notation ᴬ_Z X to state the numbers of protons, neutrons and electrons in an atom or ion, and identify isotopes.
Explain that atomic energy levels are discrete and describe how emission and absorption line spectra are produced.
Distinguish an emission spectrum (bright lines on dark) from an absorption spectrum (dark lines on continuous) and use line spectra as evidence for discrete energy levels.
Apply E = hf = hc/λ and hf = E₂ − E₁ to calculate photon energies, frequencies and wavelengths for transitions, converting correctly between eV and joules.

The nuclear model of the atom

A tiny dense positive nucleus of protons and neutrons, surrounded by electrons in mostly empty space.

Everything in Theme E starts from one picture: the nuclear model of the atom.

At the centre sits a nucleus — a tiny, incredibly dense, positively charged ball made of protons (positive) and neutrons (no charge). Protons and neutrons are together called nucleons.
Around the nucleus, occupying almost all of the atom's volume, are the electrons — light, negatively charged particles.
A neutral atom has equal numbers of protons and electrons, so the positive and negative charges cancel and the atom is electrically neutral overall.

The scale is astonishing. If an atom were the size of a sports stadium, the nucleus would be about the size of a pea at the centre, and the electrons would be like tiny specks somewhere out in the stands. The atom is almost entirely empty space, yet more than 99.9% of its mass is packed into that tiny central nucleus, because protons and neutrons are roughly 1800 times more massive than an electron.

Particle	Charge	Relative mass	Where it is
Proton	$+1$ (i.e. $+e$ )	$\approx 1$	In the nucleus
Neutron	$0$	$\approx 1$	In the nucleus
Electron	$-1$ (i.e. $-e$ )	$\approx \tfrac{1}{1800}$	Around the nucleus

This model did not come from just looking down a microscope — atoms are far too small to see. It came from a clever scattering experiment, which is where the real physics (and the exam marks) begins.

Nucleus = protons + neutrons (nucleons); tiny, dense, positive, holds almost all the mass.
Electrons surround the nucleus in mostly empty space.
A neutral atom has equal numbers of protons and electrons.

The Geiger–Marsden–Rutherford alpha-scattering experiment

Alpha particles fired at thin gold foil: three observations lead to three conclusions about the nucleus.

This is the single most-examined experiment in E.1. Learn the set-up, the three observations, and the three conclusions — examiners want all three linked correctly.

The set-up. A narrow beam of alpha particles (helium nuclei, charge $+2e$ , fast-moving) is fired at a very thin sheet of gold foil in a vacuum. A movable fluorescent detector (a screen that flashes when an alpha particle hits it) is placed all around the foil to record the angle through which each alpha particle is deflected.

The three key observations:

Most alpha particles pass straight through the foil, barely deflected at all.
Some are deflected through small angles.
A very few — fewer than about 1 in 8000 — are deflected through large angles (more than 90°), effectively bouncing back towards the source.

The three conclusions (match each to its evidence):

Most pass straight through → the atom is mostly empty space.
A few are deflected through large angles → there is a tiny, concentrated region of charge and mass — the nucleus — that they occasionally come close to.
The rare backwards deflections → that nucleus is positively charged (it repels the positive alpha particles) and very dense, containing most of the atom's mass (only something massive can reverse a fast alpha particle).

Rutherford famously said the back-scattering was "as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you." The only way to repel a fast, positive, massive alpha particle straight back is to have a small region that is positive, massive and dense — the nucleus.

Alpha scattering: most particles miss every nucleus and pass straight through the empty space; a few pass close to a nucleus and deflect a little; the rare head-on approach to a tiny, dense, positive nucleus is repelled straight back.

Set-up: alpha beam → thin gold foil → detector records deflection angle.
Observations: most pass straight through; some small-angle; very few (< 1 in 8000) bounce back > 90°.
Conclusions: atom mostly empty space; nucleus tiny; nucleus dense, positive and holds most of the mass.

See the full worked example for structure of the atom →

Nuclear notation and isotopes

ᴬ_Z X tells you the nucleons (A) and protons (Z); neutrons N = A − Z; isotopes share Z but differ in N.

Every nucleus is described by two numbers, written around the element symbol as $^{A}_{Z}\text{X}$ :

$Z$ — the proton number (also called atomic number). It is written at the bottom and tells you how many protons are in the nucleus. $Z$ defines which element the atom is.
$A$ — the nucleon number (also called mass number). It is written at the top and tells you the total number of protons and neutrons in the nucleus.

From these you can find everything else:

$\text{number of neutrons } N = A - Z$

For a neutral atom, the number of electrons equals the number of protons ( $= Z$ ).

Worked micro-example. Carbon-12 is written $^{12}_{6}\text{C}$ :

protons $= Z = 6$
neutrons $= A - Z = 12 - 6 = 6$
electrons (neutral atom) $= 6$

Ions. If the atom is charged, the electron count changes but the nucleus does not. For example $^{16}_{8}\text{O}^{2-}$ still has 8 protons and $16-8 = 8$ neutrons, but it has gained 2 electrons, so it has $8 + 2 = 10$ electrons.

Isotopes. These are atoms of the same element — so the same proton number $Z$ — but with different numbers of neutrons, and therefore a different nucleon number $A$ .

Isotope	$Z$ (protons)	$A$ (nucleons)	$N = A - Z$ (neutrons)
$^{12}_{6}\text{C}$	6	12	6
$^{14}_{6}\text{C}$	6	14	8

Both are carbon (both have $Z = 6$ ), but carbon-14 has two extra neutrons. Because they have the same number of electrons, isotopes of an element are chemically identical; they differ only in mass (and, for some, in stability).

Z (bottom) = protons; A (top) = protons + neutrons; N = A − Z.
Neutral atom: electrons = protons = Z. Ions change electron count, not the nucleus.
Isotopes: same Z, different N (and different A) — same element, different mass.

See the full worked example for structure of the atom →

Discrete (quantised) atomic energy levels

An electron in an atom can only have certain allowed energies — the levels are discrete, not continuous.

In an atom, an electron cannot have any amount of energy it likes. It is only allowed to occupy certain specific, discrete energy values called energy levels. This idea — that energy is quantised (comes in fixed, separated steps) — is one of the great surprises of quantum physics, and the whole of atomic spectra depends on it.

A few key features you must be comfortable with:

The lowest energy level is called the ground state; the electron sits here normally. Higher allowed levels are the excited states.
Energy levels are usually drawn as horizontal lines on a diagram, with the ground state at the bottom and higher levels above it. The levels get closer together as they rise, and the top line ( $E = 0$ ) represents the electron being just free of the atom (ionised).
The energies are conventionally written as negative numbers (e.g. $-6.0\,\text{eV}$ ). A more negative value means the electron is more tightly bound to the nucleus. Moving up to a less negative level therefore requires the electron to gain energy.
An electron can be pushed up to a higher level (excited) by absorbing energy, and it will later drop back down, releasing energy. Crucially, because the levels are fixed, only certain sized energy jumps are possible — and this is exactly why atoms emit and absorb only certain colours of light.

Only certain electron energies are allowed. A downward jump releases a photon (emission); an upward jump requires a photon to be absorbed. The size of the jump fixes the photon's frequency and colour.

The next section shows exactly how the size of an energy jump determines the frequency (and colour) of the light involved — the relation $hf = E_2 - E_1$ .

Electrons in atoms have only discrete (quantised) allowed energies.
Ground state = lowest level; excited states are higher; E = 0 is ionisation.
Levels are drawn as horizontal lines, usually with negative energies (more negative = more tightly bound).

From energy levels to photons: deriving hf = E₂ − E₁ (first principles)

Energy conservation for a single transition gives the photon relation hf = E₂ − E₁, and hence f and λ.

You are not asked to memorise this as a black box — it drops straight out of conservation of energy applied to a single electron transition, and understanding that makes every spectral calculation feel obvious.

Step 1 — a photon carries a fixed packet of energy. Light is emitted and absorbed in tiny packets called photons. The energy of one photon depends only on its frequency $f$ (or wavelength $\lambda$ ):

$E = hf = \frac{hc}{\lambda}$

where $h = 6.63\times10^{-34}\,\text{J s}$ is the Planck constant and $c = 3.00\times10^{8}\,\text{m s}^{-1}$ is the speed of light. (The second form uses $c = f\lambda$ .)

Step 2 — apply conservation of energy to a transition. Suppose an electron drops from a higher energy level $E_2$ to a lower level $E_1$ . Energy cannot vanish, so the energy the electron loses is carried away by a single emitted photon:

$\text{energy lost by electron} = \text{energy of photon}$ $E_2 - E_1 = hf$

Rearranging gives the frequency of the emitted light directly:

$\boxed{f = \frac{E_2 - E_1}{h}}$

and, since $\lambda = c/f$ , the wavelength is:

$\lambda = \frac{hc}{E_2 - E_1}$

The same relation runs backwards for absorption. To lift an electron up from $E_1$ to $E_2$ , a photon of exactly energy $E_2 - E_1$ must be absorbed — a photon with too little or too much energy simply passes by. This "all or nothing" rule is why absorption is also fussy about frequency.

Step 3 — the electronvolt. Atomic energies are tiny in joules, so we use a friendlier unit, the electronvolt (eV):

$1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$

This is the energy gained by one electron accelerated through a potential difference of 1 volt. You must convert eV to joules before putting numbers into $E = hf$ (or your frequency will be out by a factor of about $10^{19}$ ).

Full worked route (learn this order). For a transition of size $\Delta E$ given in eV:

Convert: $\Delta E\,(\text{J}) = \Delta E\,(\text{eV}) \times 1.60\times10^{-19}$ .
Frequency: $f = \Delta E / h$ .
Wavelength: $\lambda = c/f$ (or $\lambda = hc/\Delta E$ in one line).

For example, a jump of $\Delta E = 4.5\,\text{eV}$ gives $\Delta E = 4.5 \times 1.60\times10^{-19} = 7.2\times10^{-19}\,\text{J}$ , so $f = 7.2\times10^{-19} / 6.63\times10^{-34} = 1.09\times10^{15}\,\text{Hz}$ and $\lambda = 3.00\times10^{8} / 1.09\times10^{15} = 2.76\times10^{-7}\,\text{m} = 276\,\text{nm}$ (ultraviolet).

Photon energy: E = hf = hc/λ, with h = 6.63×10⁻³⁴ J s.
Conservation of energy for a transition gives hf = E₂ − E₁, so f = (E₂ − E₁)/h.
Always convert eV → J (×1.60×10⁻¹⁹) before using E = hf.

See the full worked example for structure of the atom →

Emission and absorption line spectra

Emission = bright lines on dark; absorption = dark lines on continuous — both prove energy levels are discrete.

Because the energy jumps in an atom are fixed, the photons an atom emits or absorbs have only certain frequencies — so we see sharp lines rather than a smooth spread of colour. Each element has its own unique pattern of lines, like a barcode or fingerprint.

Emission line spectrum (bright lines on a dark background).

The gas is given energy (e.g. heated or a high voltage passed through it), which excites electrons up to higher levels.
The electrons then drop back down, and each downward jump releases a photon of energy $E_2 - E_1$ .
Only the allowed jumps occur, so only certain frequencies of light come out. Viewed through a spectrometer this appears as a set of bright coloured lines on a dark background — each line is one transition.

Absorption line spectrum (dark lines on a continuous background).

White light (containing all visible frequencies — a continuous spectrum) is passed through a cool gas.
Photons whose energy exactly matches an allowed jump $E_2 - E_1$ are absorbed, lifting electrons up.
Those particular frequencies are removed from the beam, so the continuous rainbow now has dark lines where the missing frequencies should be — at exactly the same positions as the bright lines in that element's emission spectrum.

The same element gives bright emission lines and dark absorption lines at identical frequencies. Because both show sharp lines rather than a smooth spread, they are direct evidence that atomic energy levels are discrete.

Why this is the evidence for discrete energy levels. If electrons could have any energy, atoms would emit and absorb a continuous range of colours. Instead we see only a few specific lines. The only explanation is that the energy jumps — and therefore the electron energies themselves — are discrete. Line spectra are the fingerprint of quantised energy levels, and are used to identify elements in stars and distant galaxies.

Emission: excited electrons drop down → bright coloured lines on a dark background.
Absorption: white light through a cool gas → dark lines removed from a continuous spectrum.
The dark absorption lines sit at exactly the same frequencies as the bright emission lines.

See the full worked example for structure of the atom →

Quick recap

The nuclear model: a tiny, dense, positive nucleus of protons and neutrons, with electrons in the mostly empty surrounding space.
Geiger–Marsden–Rutherford alpha scattering: most pass straight through, some deflect a little, a very few bounce back — showing the atom is mostly empty space with a tiny, dense, positive nucleus holding most of the mass.
Nuclear notation ᴬ_Z X: A = nucleons, Z = protons, N = A − Z neutrons; isotopes share Z but differ in N.
Atomic energy levels are discrete (quantised); the ground state is the lowest level and E = 0 is ionisation.
A transition emits or absorbs one photon with hf = E₂ − E₁, where E = hf = hc/λ; convert eV → J (×1.60×10⁻¹⁹) before calculating.
Emission spectra (bright lines on dark) and absorption spectra (dark lines on continuous) are direct evidence that energy levels are discrete.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

Nucleus = protons + neutrons; tiny, dense, positive, holds almost all the mass. Electrons surround it in mostly empty space.
Alpha scattering — three observations: most straight through; some small-angle; very few (< 1 in 8000) bounce back > 90°.
Alpha scattering — three conclusions: atom mostly empty space; nucleus tiny; nucleus dense, positive, most of the mass.
Nuclear notation ᴬ_Z X: A = nucleon number, Z = proton number, neutrons N = A − Z. Isotopes: same Z, different N.
Atomic energy levels are discrete (quantised); ground state = lowest; E = 0 = ionised.
Photon: E = hf = hc/λ, with h = 6.63×10⁻³⁴ J s. Transition: hf = E₂ − E₁, so f = (E₂ − E₁)/h.
1 eV = 1.60×10⁻¹⁹ J — always convert eV → J before using E = hf.
Emission = bright lines on dark; absorption = dark lines on continuous. Both prove discrete energy levels.

How it’s examined

E.1 appears across all three papers. Paper 1A (MCQ): identifying which alpha-scattering observation supports which conclusion, reading numbers of protons/neutrons/electrons from nuclear notation, spotting isotopes, and matching a described transition to an emission or absorption line — no calculator, so photon numbers are chosen to be clean. Paper 1B (data-based): interpreting an energy-level diagram or a spectrum, and ranking transitions by photon energy or wavelength. Paper 2: structured questions (3–6 marks) that ask you to describe the alpha-scattering experiment and explain its conclusions, and calculation chains that go from an energy jump (in eV) to a photon frequency and wavelength using hf = E₂ − E₁, sometimes as part of a longer extended-response question. Command terms: state, identify, describe, explain, calculate, determine, show that. Examiner reports repeatedly flag: describing observations without linking them to conclusions, confusing emission with absorption spectra, muddling A and Z (or forgetting N = A − Z), and forgetting to convert eV to joules (a factor-of-10¹⁹ error). Always show full working and quote units and sensible significant figures — method marks are awarded even if the final arithmetic slips.

Sources: IB Diploma Programme Physics Guide (first assessment 2025) — Theme E: Nuclear and quantum physics (E.1 Structure of the atom); IB Physics Data Booklet (2025); IB Physics subject reports and specimen papers (2023–2025). Last reviewed 2026-07-21.

Take this whole topic with you

Step-by-step worked examples — Structure of the atom

Step-by-step solutions to past-paper-style questions on structure of the atom, written exactly the way a tutor would explain them at the board.

Question type:

1Reading nuclear notation and spotting isotopes
Getting startedIdentify & classify• nuclear notation, isotopes, AO2
Question
An atom is written as ³⁵₁₇Cl. (a) State the number of protons, neutrons and electrons in the neutral atom. (b) A second atom is ³⁷₁₇Cl. Explain why ³⁵₁₇Cl and ³⁷₁₇Cl are isotopes. (3 marks)
Step-by-step solution
1. Step 1
  The bottom number is the proton number $Z$ and the top number is the nucleon number $A$ . Protons $= Z = 17$ ; neutrons $= A - Z = 35 - 17 = 18$ .
  $N = A - Z = 35 - 17 = 18$
2. Step 2
  A neutral atom has equal numbers of protons and electrons, so electrons $= Z = 17$ .
3. Step 3
  Both atoms have the same proton number ( $Z = 17$ , so both are chlorine) but different nucleon numbers ( $35$ and $37$ ), meaning different numbers of neutrons ( $18$ and $20$ ). Same $Z$ , different $N$ → they are isotopes.
Answer
(a) 17 protons, 18 neutrons, 17 electrons. (b) Both have Z = 17 (same element, chlorine) but different A (35 vs 37) so different neutron numbers (18 vs 20) — same protons, different neutrons = isotopes.
Examiner tip
Mark scheme: (1) protons and neutrons correct (N = A − Z); (2) electrons = protons for a neutral atom; (3) isotopes = same Z, different N. The classic slip is quoting A (35) as the neutron number instead of subtracting.
2Photon energy from its frequency
Getting startedDirect calculation• photon energy, AO2
Question
A photon of green light has a frequency of 5.0 × 10¹⁴ Hz. Calculate its energy in (a) joules and (b) electronvolts. (h = 6.63 × 10⁻³⁴ J s; 1 eV = 1.60 × 10⁻¹⁹ J) (3 marks)
Step-by-step solution
1. Step 1
  Use the photon energy relation $E = hf$ with $f = 5.0\times10^{14}\,\text{Hz}$ .
  $E = hf = (6.63\times10^{-34})(5.0\times10^{14}) = 3.32\times10^{-19}\,\text{J}$
2. Step 2
  Convert to electronvolts by dividing by $1.60\times10^{-19}\,\text{J eV}^{-1}$ .
  $E = \frac{3.32\times10^{-19}}{1.60\times10^{-19}} = 2.07\,\text{eV}$
Answer
(a) E ≈ 3.3 × 10⁻¹⁹ J; (b) E ≈ 2.1 eV.
Examiner tip
Mark scheme: (1) E = hf used correctly; (2) 3.32 × 10⁻¹⁹ J; (3) 2.07 eV. Watch the powers of ten on h — a common error is 6.63 × 10⁻³⁴ mistyped as 10⁻³⁴ vs 10⁻¹⁹.
3Converting between electronvolts and joules
Getting startedDirect calculation• electronvolt, unit conversion, AO2
Question
(a) Convert an energy of 6.3 eV into joules. (b) Convert a photon energy of 4.8 × 10⁻¹⁹ J into electronvolts. (1 eV = 1.60 × 10⁻¹⁹ J) (2 marks)
Step-by-step solution
1. Step 1
  To go from eV to J, multiply by $1.60\times10^{-19}$ .
  $E = 6.3 \times 1.60\times10^{-19} = 1.01\times10^{-18}\,\text{J}$
2. Step 2
  To go from J to eV, divide by $1.60\times10^{-19}$ .
  $E = \frac{4.8\times10^{-19}}{1.60\times10^{-19}} = 3.0\,\text{eV}$
Answer
(a) 1.01 × 10⁻¹⁸ J; (b) 3.0 eV.
Examiner tip
Mark scheme: (1) eV → J by multiplying; (2) J → eV by dividing. Remembering the direction (multiply to get the bigger joule figure into eV, divide the other way) prevents the most common conversion slip in photon questions.
4Frequency of an emitted photon from a transition
Building confidenceMulti-step problem• energy levels, hf = E₂ − E₁, AO2
Question
In an atom, an electron falls from an energy level at −1.5 eV to the ground state at −6.0 eV. Calculate (a) the energy of the emitted photon in joules and (b) its frequency. (h = 6.63 × 10⁻³⁴ J s; 1 eV = 1.60 × 10⁻¹⁹ J) (4 marks)
Step-by-step solution
1. Step 1
  The photon energy equals the size of the energy jump: $\Delta E = E_2 - E_1 = (-1.5) - (-6.0) = 4.5\,\text{eV}$ .
  $\Delta E = -1.5 - (-6.0) = 4.5\,\text{eV}$
2. Step 2
  Convert this energy to joules before using $E = hf$ .
  $\Delta E = 4.5 \times 1.60\times10^{-19} = 7.2\times10^{-19}\,\text{J}$
3. Step 3
  Rearrange $hf = E_2 - E_1$ to make frequency the subject: $f = \Delta E / h$ .
  $f = \frac{7.2\times10^{-19}}{6.63\times10^{-34}} = 1.09\times10^{15}\,\text{Hz}$
Answer
(a) ΔE = 7.2 × 10⁻¹⁹ J; (b) f ≈ 1.09 × 10¹⁵ Hz.
Examiner tip
Mark scheme: (1) ΔE = 4.5 eV (correct subtraction of the negatives); (2) convert to 7.2 × 10⁻¹⁹ J; (3) f = ΔE/h; (4) 1.09 × 10¹⁵ Hz. Subtracting two negative energies trips students up — the gap is a positive 4.5 eV, not 7.5 eV.
5Emission or absorption? Ranking lines on an energy-level diagram
Building confidenceGraph or diagram• spectra, energy levels, AO3
Question
An atom has energy levels at E₁ = −6.0 eV, E₂ = −3.4 eV and E₃ = −1.5 eV. (a) An electron moves from E₁ up to E₃. Is a photon emitted or absorbed? (b) Of the three possible downward transitions (E₃→E₂, E₃→E₁, E₂→E₁), which produces the spectral line of longest wavelength? Justify your answer. (3 marks)
Step-by-step solution
1. Step 1
  Moving up a level (from $E_1 = -6.0\,\text{eV}$ to $E_3 = -1.5\,\text{eV}$ ) means the electron gains energy, so a photon must be absorbed.
2. Step 2
  Find the energy gap for each downward transition: $E_3\to E_2 = (-1.5)-(-3.4) = 1.9\,\text{eV}$ ; $E_2\to E_1 = (-3.4)-(-6.0) = 2.6\,\text{eV}$ ; $E_3\to E_1 = (-1.5)-(-6.0) = 4.5\,\text{eV}$ .
  $\Delta E_{3\to2} = 1.9,\quad \Delta E_{2\to1} = 2.6,\quad \Delta E_{3\to1} = 4.5\ \text{(eV)}$
3. Step 3
  Longest wavelength ↔ smallest photon energy (since $\lambda = hc/\Delta E$ ). The smallest gap is $E_3\to E_2 = 1.9\,\text{eV}$ , so this transition gives the longest-wavelength line.
Answer
(a) A photon is absorbed (electron moves up). (b) E₃→E₂ (ΔE = 1.9 eV) — the smallest energy gap, and λ = hc/ΔE means the smallest energy gives the longest wavelength.
Examiner tip
Mark scheme: (1) absorbed, because the electron gains energy moving up; (2) correct energy gaps; (3) longest λ ↔ smallest ΔE → E₃→E₂. Recognising the inverse link between energy gap and wavelength earns the reasoning mark without a full calculation.
6Wavelength of a spectral line
Building confidenceDirect calculation• spectra, wavelength, AO2
Question
An electron in an atom drops from a level at −3.4 eV to the ground state at −6.0 eV, emitting a photon. Calculate the wavelength of the emitted light and state whether it is in the visible region. (h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹; 1 eV = 1.60 × 10⁻¹⁹ J) (4 marks)
Step-by-step solution
1. Step 1
  Energy of the photon = size of the jump: $\Delta E = (-3.4)-(-6.0) = 2.6\,\text{eV}$ . Convert to joules.
  $\Delta E = 2.6 \times 1.60\times10^{-19} = 4.16\times10^{-19}\,\text{J}$
2. Step 2
  Use $\lambda = hc/\Delta E$ (combining $E = hf$ and $c = f\lambda$ ).
  $\lambda = \frac{hc}{\Delta E} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{4.16\times10^{-19}}$
3. Step 3
  Evaluate the wavelength.
  $\lambda = \frac{1.989\times10^{-25}}{4.16\times10^{-19}} = 4.78\times10^{-7}\,\text{m} = 478\,\text{nm}$
Answer
λ ≈ 4.78 × 10⁻⁷ m = 478 nm — this is in the visible region (blue light, roughly 400–700 nm).
Examiner tip
Mark scheme: (1) ΔE = 2.6 eV converted to J; (2) select λ = hc/ΔE; (3) 4.78 × 10⁻⁷ m; (4) identify visible/blue. Quoting the answer in nm and comparing with the 400–700 nm visible band earns the final mark.
7From an energy jump to the wavelength and spectral region
StretchMulti-step problem• energy levels, spectra, AO2
Question
An electron falls from a level at −0.85 eV to the ground state at −6.0 eV. Calculate (a) the photon's frequency, (b) its wavelength, and (c) state the region of the electromagnetic spectrum it belongs to. (h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹; 1 eV = 1.60 × 10⁻¹⁹ J) (5 marks)
Step-by-step solution
1. Step 1
  Energy gap: $\Delta E = (-0.85)-(-6.0) = 5.15\,\text{eV}$ . Convert to joules.
  $\Delta E = 5.15 \times 1.60\times10^{-19} = 8.24\times10^{-19}\,\text{J}$
2. Step 2
  Frequency from $f = \Delta E / h$ .
  $f = \frac{8.24\times10^{-19}}{6.63\times10^{-34}} = 1.24\times10^{15}\,\text{Hz}$
3. Step 3
  Wavelength from $\lambda = c/f$ .
  $\lambda = \frac{3.00\times10^{8}}{1.24\times10^{15}} = 2.42\times10^{-7}\,\text{m} = 242\,\text{nm}$
4. Step 4
  242 nm is shorter than the visible range (~400–700 nm), so the photon lies in the ultraviolet (UV) region.
Answer
(a) f ≈ 1.24 × 10¹⁵ Hz; (b) λ ≈ 2.42 × 10⁻⁷ m = 242 nm; (c) ultraviolet.
Examiner tip
Mark scheme: (1) ΔE = 5.15 eV → 8.24 × 10⁻¹⁹ J; (2) f from ΔE/h; (3) λ = c/f; (4) 242 nm; (5) identify UV. A large energy jump gives a short wavelength — landing below 400 nm signals ultraviolet, not visible.
8Show that a spectral line matches a transition
StretchShow that / prove• spectra, show that, AO2
Question
A red line in a star's spectrum has a wavelength of 656 nm. An atom has levels at E₂ = −3.4 eV and E₃ = −1.5 eV. Show that the photon energy for the 656 nm line is about 1.9 eV, and hence show that this line corresponds to the E₃ → E₂ transition. (h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹; 1 eV = 1.60 × 10⁻¹⁹ J) (4 marks)
Step-by-step solution
1. Step 1
  Convert the wavelength to metres: $656\,\text{nm} = 656\times10^{-9} = 6.56\times10^{-7}\,\text{m}$ . Then find the photon energy with $E = hc/\lambda$ .
  $E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3.00\times10^{8})}{6.56\times10^{-7}} = 3.03\times10^{-19}\,\text{J}$
2. Step 2
  Convert this energy to electronvolts.
  $E = \frac{3.03\times10^{-19}}{1.60\times10^{-19}} = 1.9\,\text{eV}$
3. Step 3
  The energy gap between the two levels is $E_3 - E_2 = (-1.5)-(-3.4) = 1.9\,\text{eV}$ , which matches the photon energy. So the 656 nm line is produced by an electron falling from $E_3$ to $E_2$ .
  $E_3 - E_2 = (-1.5) - (-3.4) = 1.9\,\text{eV}$
Answer
The 656 nm photon has energy ≈ 1.9 eV, which equals the E₃ → E₂ gap (−1.5 − (−3.4) = 1.9 eV), so the line corresponds to the E₃ → E₂ transition.
Examiner tip
Mark scheme: (1) nm → m and E = hc/λ; (2) 3.03 × 10⁻¹⁹ J → 1.9 eV; (3) compute E₃ − E₂ = 1.9 eV; (4) conclude the transition matches. In a 'show that' question you must state the target value is reached AND make the explicit comparison to earn full marks.

Model Answers — Structure of the atom

High-scoring sample answers for structure of the atom on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define the term isotopes, and state one physical quantity in which two isotopes of the same element differ.
Model answer
Isotopes are atoms of the same element — so they have the same number of protons (the same proton number $Z$ ) — but with different numbers of neutrons (and therefore a different nucleon number $A$ ).

Two isotopes of an element differ in their mass (nucleon number $A$ ). (They are, however, chemically identical because they have the same number of electrons.)
Why this scores
Why this scores 2/2. (1) same element / same proton number Z; (2) different number of neutrons (different A), giving different mass. Saying only 'different mass' without mentioning the same Z and different neutron number is incomplete.
Question 2
3 marks
Q (3 marks). In the Geiger–Marsden alpha-scattering experiment, alpha particles were directed at a thin gold foil. State the three main observations made about how the alpha particles behaved.
Model answer
Most of the alpha particles passed straight through the foil with little or no deflection.

A small number were deflected through small angles.

A very few (fewer than about 1 in 8000) were deflected through large angles — more than 90°, i.e. they bounced back towards the source.
Why this scores
Why this scores 3/3. One mark for each distinct observation: (1) most straight through; (2) some small-angle deflection; (3) a very few back-scattered through more than 90°. The question asks for observations, not conclusions — do not describe the nucleus here.
Question 3
3 marks
Q (3 marks). A photon has a frequency of 6.0 × 10¹⁴ Hz. Calculate its energy in (a) joules and (b) electronvolts. (h = 6.63 × 10⁻³⁴ J s; 1 eV = 1.60 × 10⁻¹⁹ J)
Model answer
(a) Using $E = hf$ :

$E = (6.63\times10^{-34})(6.0\times10^{14}) = 3.98\times10^{-19}\,\text{J}$

(b) Convert to electronvolts by dividing by $1.60\times10^{-19}$ :

$E = \frac{3.98\times10^{-19}}{1.60\times10^{-19}} = 2.49\,\text{eV}$
Why this scores
Why this scores 3/3. (1) correct use of E = hf; (2) 3.98 × 10⁻¹⁹ J with unit; (3) 2.49 eV. Both parts must carry units, and the conversion to eV is a straight division by 1.60 × 10⁻¹⁹.
Question 4
4 marks
Q (4 marks). Explain how an emission line spectrum and an absorption line spectrum are produced, and describe how their appearance differs.
Model answer
Emission spectrum: the atoms of a gas are given energy (e.g. heated or by an electrical discharge), which excites electrons to higher energy levels. When the electrons fall back down to lower levels, each downward transition releases a photon of energy $E_2 - E_1$ . Only certain jumps are allowed, so only certain frequencies are emitted, appearing as bright coloured lines on a dark background.

Absorption spectrum: white light (a continuous spectrum containing all visible frequencies) is passed through a cool gas. Photons whose energy exactly equals an allowed jump $E_2 - E_1$ are absorbed, exciting electrons upward. These frequencies are removed from the transmitted light, leaving dark lines on an otherwise continuous (rainbow) background, at exactly the same frequencies as that element's bright emission lines.
Why this scores
Why this scores 4/4. (1) emission: electrons drop down and emit photons; (2) emission appearance = bright lines on dark; (3) absorption: photons of the right energy are absorbed from white light lifting electrons up; (4) absorption appearance = dark lines on continuous. Linking each process to a transition (down for emission, up for absorption) is the assessed idea.
Question 5
5 marks
Q (5 marks). An electron in an atom makes a transition from an energy level of −1.5 eV to the ground state at −6.0 eV. Calculate (a) the energy of the emitted photon in joules, (b) its frequency and (c) its wavelength. (h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹; 1 eV = 1.60 × 10⁻¹⁹ J)
Model answer
(a) Energy of the photon — equal to the energy gap: $\Delta E = (-1.5) - (-6.0) = 4.5\,\text{eV}$ $\Delta E = 4.5 \times 1.60\times10^{-19} = 7.2\times10^{-19}\,\text{J}$

(b) Frequency — from $hf = E_2 - E_1$ , so $f = \Delta E / h$ : $f = \frac{7.2\times10^{-19}}{6.63\times10^{-34}} = 1.09\times10^{15}\,\text{Hz}$

(c) Wavelength — from $\lambda = c/f$ : $\lambda = \frac{3.00\times10^{8}}{1.09\times10^{15}} = 2.76\times10^{-7}\,\text{m} = 276\,\text{nm}\ (\text{ultraviolet})$
Why this scores
Why this scores 5/5. (1) ΔE = 4.5 eV; (2) converted to 7.2 × 10⁻¹⁹ J; (3) f = ΔE/h = 1.09 × 10¹⁵ Hz; (4) λ = c/f; (5) 276 nm with unit. The negative-level subtraction (4.5 eV, not 7.5 eV) and the eV → J conversion are the two mark-critical steps.
Question 6
4 marks
Q (4 marks). Explain how the existence of line spectra provides evidence that the energy levels of an atom are discrete rather than continuous.
Model answer
When atoms emit or absorb light, we observe only a small number of sharp, specific spectral lines — not a smooth, continuous band of all colours.

Each line corresponds to a photon of a definite frequency, and by $hf = E_2 - E_1$ this photon energy equals the difference between two electron energy levels.

If electrons could have any energy (continuous levels), transitions of every possible size would occur and the atom would emit/absorb a continuous range of frequencies — a continuous spectrum.

Because instead we see only certain fixed frequencies, the energy differences — and therefore the electron energy levels themselves — must be discrete (quantised). The unique pattern of lines is a fingerprint of that element's fixed set of levels.
Why this scores
Why this scores 4/4. (1) spectra show only specific lines, not a continuous range; (2) each line = a photon of definite energy = a difference between two levels (hf = E₂ − E₁); (3) continuous levels would give a continuous spectrum; (4) fixed lines therefore prove the levels are discrete. The logical contrast (what we would see if levels were continuous vs what we actually see) is the assessed reasoning.
Question 7
6 marks
Q (6 marks). In the Geiger–Marsden–Rutherford experiment, alpha particles were fired at a thin gold foil. For each of the three main observations, state the observation and explain the conclusion it leads to about the structure of the atom.
Model answer
Observation 1 — most alpha particles pass straight through, barely deflected. Conclusion: the atom is mostly empty space, so most alpha particles never come near anything that could deflect them.

Observation 2 — a small number are deflected through small angles. Conclusion: there is a small, concentrated region of positive charge within the atom; alpha particles that pass reasonably close to it are repelled and deflected. This concentrated region is the nucleus.

Observation 3 — a very few (fewer than about 1 in 8000) are deflected through more than 90°, bouncing back. Conclusion: the nucleus is very tiny (because such near head-on approaches are extremely rare), and it is dense, positively charged and contains most of the atom's mass — only a small, massive, positive object could repel a fast, positive alpha particle straight back.
Why this scores
Why this scores 6/6. Three observation–conclusion pairs, 2 marks each: (1a) most straight through → (1b) mostly empty space; (2a) some small deflection → (2b) small concentrated positive nucleus; (3a) very few bounce back > 90° → (3b) tiny, dense, positive nucleus holding most of the mass. Marks are lost when observations are listed without their matching conclusion.
Question 8
6 marks
Q (6 marks). An atom has energy levels E₁ = −6.0 eV, E₂ = −3.4 eV and E₃ = −1.5 eV. (a) Calculate the wavelength of the photon emitted when an electron falls from E₃ to E₂. (b) State whether a photon is emitted or absorbed in the E₁ → E₃ transition. (c) Calculate the wavelength for the E₃ → E₁ transition and state which of the two transitions, E₃→E₂ or E₃→E₁, gives the longer-wavelength line. (h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹; 1 eV = 1.60 × 10⁻¹⁹ J)
Model answer
(a) E₃ → E₂: $\Delta E = (-1.5) - (-3.4) = 1.9\,\text{eV} = 1.9 \times 1.60\times10^{-19} = 3.04\times10^{-19}\,\text{J}$ $\lambda = \frac{hc}{\Delta E} = \frac{1.989\times10^{-25}}{3.04\times10^{-19}} = 6.54\times10^{-7}\,\text{m} = 654\,\text{nm (visible, red)}$

(b) The electron moves up from $E_1$ to $E_3$ , so it gains energy — a photon is absorbed.

(c) E₃ → E₁: $\Delta E = (-1.5) - (-6.0) = 4.5\,\text{eV} = 7.2\times10^{-19}\,\text{J}$ $\lambda = \frac{1.989\times10^{-25}}{7.2\times10^{-19}} = 2.76\times10^{-7}\,\text{m} = 276\,\text{nm (ultraviolet)}$

The E₃ → E₂ transition (654 nm) gives the longer-wavelength line, because it has the smaller energy gap ( $\lambda = hc/\Delta E$ , so smaller ΔE → longer λ).
Why this scores
Why this scores 6/6. (a) ΔE = 1.9 eV → λ = 654 nm (2); (b) absorbed, electron moves up (1); (c) ΔE = 4.5 eV → λ = 276 nm (2) and correct comparison: smaller energy gap gives longer wavelength, so E₃→E₂ (1). Using λ = hc/ΔE consistently and keeping the inverse relationship straight are the discriminators.
Question 9
10 marks
Q (10 marks — extended response). An atom of a certain element is represented by ⁷Li₃ and has electron energy levels at E₁ = −5.4 eV (ground state), E₂ = −2.0 eV and E₃ = −0.9 eV. (a) State the number of protons, neutrons and electrons in the neutral atom. (b) The nucleus of this atom is extremely small compared with the size of the whole atom. Explain how the Geiger–Marsden alpha-scattering experiment provided evidence for such a small, dense, positive nucleus. (c) An electron falls from E₃ to E₁. Calculate the frequency and wavelength of the emitted photon, and state the region of the electromagnetic spectrum it lies in. (d) Explain why this atom produces a line spectrum rather than a continuous spectrum, and state whether a downward transition gives an emission or an absorption line. (h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹; 1 eV = 1.60 × 10⁻¹⁹ J)
Model answer
(a) Composition of ⁷Li₃ (A = 7, Z = 3):

protons $= Z = 3$

neutrons $= A - Z = 7 - 3 = 4$

electrons (neutral atom) $= 3$

(b) Evidence from alpha scattering. In the experiment, most alpha particles passed straight through the thin gold foil, showing the atom is mostly empty space. A very few were deflected through large angles (> 90°), bouncing back towards the source. The only way to repel a fast, positively charged alpha particle straight back is for the atom to contain a small, dense, positively charged region — the nucleus. Because such large deflections were so rare, this region must be tiny; because it can turn back a massive alpha particle, it must be dense and hold most of the atom's mass.

(c) E₃ → E₁ transition: $\Delta E = (-0.9) - (-5.4) = 4.5\,\text{eV} = 4.5 \times 1.60\times10^{-19} = 7.2\times10^{-19}\,\text{J}$ Frequency (from $hf = E_2 - E_1$ ): $f = \frac{\Delta E}{h} = \frac{7.2\times10^{-19}}{6.63\times10^{-34}} = 1.09\times10^{15}\,\text{Hz}$ Wavelength (from $\lambda = c/f$ ): $\lambda = \frac{3.00\times10^{8}}{1.09\times10^{15}} = 2.76\times10^{-7}\,\text{m} = 276\,\text{nm}$ This wavelength is shorter than the visible range (~400–700 nm), so the photon lies in the ultraviolet region.

(d) Why a line spectrum. The atom's electrons can only occupy discrete (quantised) energy levels, so only certain sized energy jumps are possible. Each allowed transition emits a photon of a single, definite frequency ( $hf = E_2 - E_1$ ), giving a set of separate lines rather than a continuous band of all colours. A downward transition (electron losing energy) emits a photon, so it produces an emission line — a bright line on a dark background.
Why this scores
Why this scores 10/10. (a) 3 protons, 4 neutrons, 3 electrons (2). (b) most straight through → empty space; rare large-angle back-scatter → small, dense, positive nucleus with most of the mass (3). (c) ΔE = 4.5 eV → 7.2 × 10⁻¹⁹ J, f = 1.09 × 10¹⁵ Hz, λ = 276 nm, ultraviolet (3). (d) discrete levels → only certain frequencies → separate lines, and a downward transition gives an emission line (2). This is a model IB Paper 2 extended response: the grade-9 discriminators are pairing each scattering observation with its conclusion, correctly subtracting the negative energy levels, and explaining the link between discreteness and line spectra rather than merely asserting it.

Key Formulae — Structure of the atom

The formulae you need to memorise for structure of the atom on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Photon energy (from frequency)
$E = hf$
$E$
energy of the photon (J)
$h$
Planck constant = 6.63 × 10⁻³⁴ J s
$f$
frequency of the light (Hz)
When to use
To find a photon's energy from its frequency (or the frequency from a known energy). Convert any eV energies to joules first.
Example
f = 5.0 × 10¹⁴ Hz: E = (6.63 × 10⁻³⁴)(5.0 × 10¹⁴) = 3.3 × 10⁻¹⁹ J ≈ 2.1 eV.
Photon energy (from wavelength)
$E = \frac{hc}{\lambda}$
$E$
energy of the photon (J)
$h$
Planck constant = 6.63 × 10⁻³⁴ J s
$c$
speed of light = 3.00 × 10⁸ m s⁻¹
$\lambda$
wavelength of the light (m)
When to use
To link a photon's energy directly to its wavelength (comes from combining E = hf with c = fλ). Handy for finding λ of a spectral line.
Example
ΔE = 2.6 eV = 4.16 × 10⁻¹⁹ J: λ = hc/ΔE = 1.989 × 10⁻²⁵ / 4.16 × 10⁻¹⁹ = 478 nm.
Energy-level transition (photon condition)
$hf = E_2 - E_1$
$h$
Planck constant = 6.63 × 10⁻³⁴ J s
$f$
frequency of the emitted/absorbed photon (Hz)
$E_2$
energy of the higher level (J or eV)
$E_1$
energy of the lower level (J or eV)
When to use
For any electron transition between two energy levels: the photon's energy equals the gap E₂ − E₁. Emission (drop down) or absorption (jump up).
Example
Drop from −1.5 eV to −6.0 eV: ΔE = 4.5 eV = 7.2 × 10⁻¹⁹ J, so f = ΔE/h = 1.09 × 10¹⁵ Hz.
Neutron number from nuclear notation
$N = A - Z$
$N$
number of neutrons
$A$
nucleon (mass) number — protons + neutrons
$Z$
proton number (atomic number)
When to use
To find the number of neutrons in a nucleus written as ᴬ_Z X. For a neutral atom, electrons = Z.
Example
³⁵₁₇Cl: N = A − Z = 35 − 17 = 18 neutrons.
Electronvolt–joule conversion
$1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$
$\text{eV}$
electronvolt — the energy unit used at atomic scales
$\text{J}$
joule — the SI unit of energy
When to use
To convert energies between eV and J. Multiply eV by 1.60 × 10⁻¹⁹ to get joules; divide joules by 1.60 × 10⁻¹⁹ to get eV. Always convert to J before using E = hf.
Example
6.3 eV = 6.3 × 1.60 × 10⁻¹⁹ = 1.01 × 10⁻¹⁸ J.

Key Definitions and Keywords — Structure of the atom

Definitions to memorise and the exact keywords mark schemes credit for structure of the atom answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Nucleus
Examiner keyword
The tiny, dense, positively charged central region of an atom, containing the protons and neutrons and almost all of the atom's mass.
Proton
Examiner keyword
A positively charged nucleon (charge +e). The number of protons (proton number Z) defines which element an atom is.
Neutron
Examiner keyword
An uncharged nucleon of almost the same mass as a proton. The number of neutrons is N = A − Z.
Nucleon
Examiner keyword
A particle found in the nucleus — either a proton or a neutron. The nucleon number A is the total number of nucleons.
Proton number (Z)
Examiner keyword
The number of protons in a nucleus (also called the atomic number). Written at the bottom in nuclear notation ᴬ_Z X; it identifies the element.
Nucleon number (A)
Examiner keyword
The total number of protons and neutrons in a nucleus (also called the mass number). Written at the top in nuclear notation ᴬ_Z X.
Isotopes
Examiner keyword
Atoms of the same element (same proton number Z) that have different numbers of neutrons, and therefore different nucleon numbers A.
Alpha particle
Examiner keyword
A fast-moving helium nucleus (2 protons + 2 neutrons, charge +2e) — the projectile fired at the gold foil in the Geiger–Marsden experiment.
Energy level
Examiner keyword
One of the discrete (fixed) energies an electron in an atom is allowed to have. Usually written as a negative value; the lowest is the ground state.
Discrete (quantised)
Examiner keyword
Taking only certain separate, fixed values rather than a continuous range. Atomic energy levels are discrete, which is why line spectra occur.
Ground state
The lowest allowed energy level of an atom, in which the electron is most tightly bound to the nucleus.
Photon
Examiner keyword
A discrete packet (quantum) of electromagnetic energy, with energy E = hf = hc/λ. Emitted or absorbed when an electron changes energy level.
Emission (line) spectrum
Examiner keyword
The set of bright coloured lines on a dark background produced when excited electrons drop to lower levels and emit photons of specific frequencies.
Absorption (line) spectrum
Examiner keyword
The set of dark lines on an otherwise continuous spectrum, produced when a cool gas absorbs photons of specific frequencies from white light passing through it.
Electronvolt (eV)
Examiner keyword
A unit of energy convenient at atomic scales: 1 eV = 1.60 × 10⁻¹⁹ J — the energy gained by one electron accelerated through a potential difference of 1 V.
Planck constant (h)
Examiner keyword
The constant of proportionality between a photon's energy and its frequency, h = 6.63 × 10⁻³⁴ J s, appearing in E = hf.

Common Mistakes and Misconceptions — Structure of the atom

The traps other students keep falling into on structure of the atom questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Confusing emission spectra (bright lines on dark) with absorption spectra (dark lines on continuous)
IB Physics Theme E subject reports
Why it happens
Both involve line spectra at the same frequencies, so students remember 'lines' but not which appears bright and which dark.
How to avoid it
Emission = electrons drop DOWN and give OUT light → BRIGHT lines on a DARK background. Absorption = light passes THROUGH a cool gas and frequencies are removed → DARK lines on a CONTINUOUS (rainbow) background.
✕Forgetting to convert electronvolts to joules (or nm to m) before using E = hf
IB Physics Theme E subject reports
Why it happens
Energy levels are quoted in eV and wavelengths in nm, so students plug those numbers straight into equations that need SI units.
How to avoid it
Before any photon calculation, convert energy to joules (× 1.60 × 10⁻¹⁹) and wavelengths to metres (× 10⁻⁹). Skipping this gives an answer wrong by a factor of about 10¹⁹.
✕Mixing up the nucleon number A and proton number Z, or miscounting neutrons
IB Physics Theme E subject reports
Why it happens
The two numbers sit next to the symbol and students grab the wrong one, or forget that neutrons must be found by subtraction.
How to avoid it
Z (bottom) = protons; A (top) = protons + neutrons. Neutrons are always N = A − Z (a subtraction) — never read A as the neutron number.
✕Thinking atomic energy levels are continuous rather than discrete
IB Physics Theme E subject reports
Why it happens
Everyday experience with energy suggests it can take any value, so students assume electron energies form a smooth range.
How to avoid it
Remember that atoms produce line spectra (only certain frequencies), which is direct evidence that the electron energies — and the jumps between them — are discrete, taking only fixed allowed values.
✕Listing the alpha-scattering observations without linking them to the correct conclusions
IB Physics Theme E subject reports
Why it happens
Students memorise the observations and conclusions as two separate lists and do not pair them up as the question requires.
How to avoid it
Pair each one: most straight through → mostly empty space; a few large-angle back-scatters → tiny, dense, positive nucleus holding most of the mass. State the observation AND its conclusion together.
✕Subtracting negative energy levels incorrectly when finding a transition energy
IB Physics Theme E subject reports
Why it happens
Energy levels are negative, so students find the gap by adding magnitudes wrongly or dropping a minus sign.
How to avoid it
The photon energy is E₂ − E₁ (higher minus lower). For example (−1.5) − (−6.0) = +4.5 eV, not 7.5 eV. Take the higher (less negative) value minus the lower (more negative) value to get a positive gap.

Structure of the atom

Detailed Study Notes

At a glance

What you’ll learn

The nuclear model of the atom

The Geiger–Marsden–Rutherford alpha-scattering experiment

Nuclear notation and isotopes

Discrete (quantised) atomic energy levels

From energy levels to photons: deriving hf = E₂ − E₁ (first principles)

Emission and absorption line spectra

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Reading nuclear notation and spotting isotopes

2Photon energy from its frequency

3Converting between electronvolts and joules

4Frequency of an emitted photon from a transition

5Emission or absorption? Ranking lines on an energy-level diagram

6Wavelength of a spectral line

7From an energy jump to the wavelength and spectral region

8Show that a spectral line matches a transition

Photon energy (from frequency)

Photon energy (from wavelength)

Energy-level transition (photon condition)

Neutron number from nuclear notation

Electronvolt–joule conversion

Nucleus

Proton

Neutron

Nucleon

Proton number (Z)

Nucleon number (A)

Isotopes

Alpha particle

Energy level

Discrete (quantised)

Ground state

Photon

Emission (line) spectrum

Absorption (line) spectrum

Electronvolt (eV)

Planck constant (h)

✕Confusing emission spectra (bright lines on dark) with absorption spectra (dark lines on continuous)

✕Forgetting to convert electronvolts to joules (or nm to m) before using E = hf

✕Mixing up the nucleon number A and proton number Z, or miscounting neutrons

✕Thinking atomic energy levels are continuous rather than discrete

✕Listing the alpha-scattering observations without linking them to the correct conclusions

✕Subtracting negative energy levels incorrectly when finding a transition energy