Summary to follow. 11 syllabus statements (5 HL) · 22 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
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Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 11 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Geiger–Marsden–Rutherford experiment
Geiger–Marsden–Rutherford experiment
A beam of alpha particles was directed at a very thin gold foil in a vacuum, and the number of alpha particles scattered through different angles was counted by the flashes they made on a zinc sulfide screen. Almost all passed through with little or no deflection; a very small fraction were deflected through large angles, some through more than 90°.
Alpha particle
A helium nucleus: two protons and two neutrons bound together. Charge +2e (+3.2 × 10⁻¹⁹ C); mass about 6.64 × 10⁻²⁷ kg, roughly 7300 times the mass of an electron.
Nuclear model of the atom
Rutherford's conclusion from the scattering results: the positive charge and almost all the mass of an atom are concentrated in a nucleus that occupies a tiny fraction of the atom's volume, with the electrons outside it. A large deflection happens only when an alpha particle passes very close to a nucleus, where the electric repulsion is very strong.
Plum-pudding (Thomson) model
The model the scattering experiment overturned: positive charge spread evenly through the whole volume of the atom, with electrons embedded in it. A spread-out charge exerts only weak forces on a passing alpha particle, so the model predicts small deflections only, never deflections greater than 90°.
Students often think Most of the alpha particles were deflected through large angles or bounced back from the foil. In fact Almost all of them passed through the gold foil with little or no deflection. Only a very small fraction were deflected through large angles, and fewer still through more than 90°.
Students often think The large deflections were caused by alpha particles colliding with the electrons in the gold atoms. In fact No. An alpha particle has about 7300 times the mass of an electron, so an electron can deflect it only very slightly, just as a table-tennis ball cannot turn back a bowling ball.
Nucleon and nucleon number (A)
Nucleon and nucleon number (A)
A nucleon is a proton or a neutron. The nucleon number A is the total number of protons and neutrons in a nucleus. It has no unit.
Proton number (Z)
The number of protons in a nucleus. It fixes the charge of the nucleus (+Ze) and identifies the element; in a neutral atom it also equals the number of electrons. It has no unit.
Neutron number (N)
The number of neutrons in a nucleus: N = A − Z.
Nuclear notation ᴬ_Z X
A nuclide is written with the nucleon number A as a superscript and the proton number Z as a subscript, both to the left of the chemical symbol X; for example ⁴⁰₁₉K is potassium-40, with 19 protons and 21 neutrons.
Isotopes
Nuclei of the same element (same proton number Z) with different numbers of neutrons, and therefore different nucleon numbers A. Neutral atoms of isotopes have the same number of electrons and the same chemistry.
Students often think The nucleon number A is the number of neutrons in the nucleus. In fact The total number of protons and neutrons in the nucleus. The number of neutrons is A − Z.
Students often think The number of neutrons in a nucleus is always equal to the number of protons. In fact No. Only some light nuclei (such as helium-4, carbon-12 and oxygen-16) have N = Z. Heavier nuclei have more neutrons than protons; for potassium-40, N = 21 and Z = 19.
Discrete energy levels
Discrete energy levels
The electrons in an atom can have only certain definite energies, called energy levels. The lowest is the ground state; the higher ones are excited states. Level energies are often quoted in electronvolts and are negative, with zero energy for an electron that has just left the atom.
Emission (line) spectrum
The light from a hot, low-pressure gas of one element, spread out by wavelength: bright lines at a set of separate wavelengths on a dark background. Each line comes from photons of one energy, emitted when electrons move from a higher level to a lower one.
Absorption spectrum
The spectrum of white light (all wavelengths) after it has passed through a cooler gas: a continuous spectrum crossed by dark lines. The dark lines are at the wavelengths of photons whose energy exactly equals the difference between two levels of the gas atoms, and they coincide with lines in the same element's emission spectrum.
Students often think Each spectral line corresponds to one energy level, and the photon energy equals the energy of the level the electron starts from (or ends in). In fact No. The photon energy equals the DIFFERENCE between the two levels involved in the transition.
Students often think An atom's energy levels are equally spaced, like the rungs of a ladder, so its spectral lines are equally spaced too. In fact No. The spacing between levels generally changes; in hydrogen, for example, the levels crowd together towards the top, so the spectral lines are not evenly spaced.
Photon
Photon
A quantum (packet) of electromagnetic energy. A photon is emitted or absorbed whole; its energy depends on its frequency only (E = hf), not on the brightness of the light.
Atomic transition
A change in the energy level of an electron in an atom. When the electron moves to a lower level the atom emits one photon carrying away the energy lost; when the atom absorbs a photon whose energy exactly equals the difference between two levels, the electron moves to the higher level.
Students often think An atom absorbs any photon whose energy is at least as large as the gap to a higher level, and the electron keeps the extra energy. In fact No (for transitions between bound levels). A photon is absorbed whole or not at all, and only if its energy exactly equals the difference between the electron's level and a higher level.
Students often think An electron can absorb any amount of energy and move to a position between two allowed levels. In fact No. In an atom the electron's energy can take only the allowed level values; there are no in-between energies.
Photon energy and frequency (E = hf)
Photon energy and frequency (E = hf)
The energy of the photon emitted or absorbed in a transition equals the difference between the two energy levels: E = hf, so f = ΔE/h. h is the Planck constant, 6.63 × 10⁻³⁴ J s. With c = fλ, the wavelength is λ = hc/ΔE, so a larger energy difference gives a higher frequency and a shorter wavelength. Frequency in hertz (Hz).
Electronvolt (eV)
The energy transferred when a charge of e moves through a potential difference of 1 V: 1 eV = 1.60 × 10⁻¹⁹ J. Atomic energy levels are usually quoted in eV and must be converted to joules before using E = hf with h in J s.
Students often think Energy values in electronvolts can be used directly in E = hf with h in joule seconds. In fact No. h = 6.63 × 10⁻³⁴ J s, so the energy must be in joules: multiply eV by 1.60 × 10⁻¹⁹ J eV⁻¹ first.
Students often think The energy difference between two negative levels is found by adding their sizes, ignoring the signs. In fact Subtract: ΔE = E_upper − E_lower. For −3.03 eV and −5.14 eV, ΔE = (−3.03) − (−5.14) = 2.11 eV.
Spectroscopic identification of elements
Spectroscopic identification of elements
Each element has its own set of energy levels and therefore its own set of emission (and absorption) wavelengths. Matching the lines in a spectrum, emission or absorption, to the laboratory lines of known elements identifies the elements present in the source or in the gas the light has passed through, for example in the outer layers of a star.
Students often think Dark lines appear at the wavelengths of elements that are absent, because those colours are missing from the light. In fact No. Dark lines show elements that ARE present: their atoms, in the cooler outer layers, absorb exactly those wavelengths from the light passing through.
Students often think An element whose lines are strong must be the most abundant element in the star. In fact No. The strength of a line depends on the temperature and on how many atoms are in the right energy level, as well as on how much of the element there is. The presence of the lines shows only that the element is there.
Nuclear radius (R = R₀A^(1/3)) HL
Nuclear radius (R = R₀A^(1/3))
The radius R of a nucleus of nucleon number A is R = R₀A^(1/3), where R₀ = 1.2 × 10⁻¹⁵ m (the Fermi radius, the radius of a single nucleon). Because R³ is proportional to A, the volume of a nucleus is proportional to the number of nucleons it contains. Unit: m.
Nuclear density
The mass of a nucleus divided by its volume. Mass is proportional to A and volume ⁴⁄₃πR³ = ⁴⁄₃πR₀³A is also proportional to A, so the density is the same for all nuclei, about 2 × 10¹⁷ kg m⁻³ — about 10¹⁴ times the density of water. Unit: kg m⁻³.
Students often think The radius of a nucleus is proportional to the number of nucleons it contains. In fact No. The VOLUME is proportional to A, so the radius is proportional to A^(1/3): R = R₀A^(1/3). A nucleus with 8 times as many nucleons has only twice the radius.
Students often think The proton number Z is used in R = R₀A^(1/3), because Z is the number that identifies the nucleus. In fact The nucleon number A, because the volume depends on the number of protons AND neutrons.
Deviations from Rutherford scattering HL
Deviations from Rutherford scattering
Rutherford's prediction of the number of alpha particles scattered at each angle assumes that only the electric repulsion acts. At high alpha-particle energies the particles come close enough to the nucleus for the short-range strong nuclear force to act as well, and the measured numbers depart from the prediction. The energy at which this begins gives an estimate of the size of the nucleus.
Strong nuclear force
The attractive force between nucleons that holds a nucleus together. It is much stronger than the electric force at separations of about 10⁻¹⁵ m, but its range is short: it is negligible at distances of more than a few times 10⁻¹⁵ m.
Students often think Gravitational attraction between the alpha particle and the massive nucleus becomes important when they are very close. In fact No. The gravitational force between an alpha particle and a nucleus is smaller than the electric force by a factor of about 10³⁵; it is utterly negligible at every distance.
Students often think The strong nuclear force is long-range, reaching well beyond the nucleus, because it is the strongest force. In fact No. It is a short-range force, negligible beyond a few times 10⁻¹⁵ m, so it acts only when particles are practically touching the nucleus.
Distance of closest approach HL
Distance of closest approach
The smallest separation between a charged particle and a nucleus in a head-on collision. At that point the particle is momentarily at rest, so (neglecting the recoil of the nucleus) all its initial kinetic energy has become electric potential energy: E_k = kqQ/d, giving d = kqQ/E_k. For an alpha particle and a nucleus of proton number Z, q = 2e and Q = Ze. It is an upper limit for the radius of the nucleus. Unit: m.
Students often think An alpha particle carries a charge of +e, like a proton. In fact +2e = +3.20 × 10⁻¹⁹ C, because it contains two protons.
Students often think The charge of a nucleus is found from its nucleon number, Ae. In fact The proton numbers: the charge of a nucleus is Ze, because neutrons are uncharged. For an alpha particle Q = 2e, for gold Q = 79e.
Bohr energy levels of hydrogen HL
Bohr energy levels of hydrogen
In the Bohr model the electron in a hydrogen atom can have only the energies E = −13.6/n² eV, where n = 1, 2, 3, …: −13.6 eV, −3.40 eV, −1.51 eV, … The levels get closer together as n increases and approach 0 eV, the energy of an electron that has just been freed.
Ionization energy
The minimum energy needed to remove an electron from an atom, taking it from its level to 0 eV. For hydrogen in its ground state it is 13.6 eV; from level n it is 13.6/n² eV.
Students often think The energy of level n is −13.6/n eV. In fact No. The energy depends on 1/n²: E₂ = −13.6/4 = −3.40 eV, not −13.6/2 = −6.80 eV.
Students often think An electron in any level of a hydrogen atom needs 13.6 eV to be removed from the atom. In fact No. 13.6 eV is the energy needed from the ground state, n = 1. From a higher level the electron is already nearer to E = 0, so less is needed: 3.40 eV from n = 2, 1.51 eV from n = 3.
Quantization of angular momentum (Bohr condition) HL
Quantization of angular momentum (Bohr condition)
Bohr's postulate that the electron's angular momentum in a circular orbit is a whole-number multiple of h/2π: mvr = nh/2π, where n = 1, 2, 3, … Only orbits satisfying it are allowed, and this is what makes the orbit radii and the energies discrete. Angular momentum unit: kg m² s⁻¹ (J s).
Principal quantum number (n)
The whole number n = 1, 2, 3, … that labels the allowed orbits and energy levels in the Bohr model; it sets the angular momentum (nh/2π) and the energy (−13.6/n² eV). n = 1 is the ground state.
Students often think In the Bohr model the angular momentum is a whole-number multiple of h, so the 2π can be dropped. In fact No. It is mvr = nh/2π: the angular momentum is a whole-number multiple of h/2π, not of h.
Students often think The angular momentum in the nth orbit is n²h/2π, matching the n² in the energy formula. In fact No. The angular momentum is proportional to n: mvr = nh/2π. It is the orbit radius that is proportional to n², and the energy to 1/n².
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement about the results of the Geiger–Marsden–Rutherford alpha-particle scattering experiment is correct?
Answer and reasoning
Most of the alpha particles were deflected through large angles, and only a few passed straight through. — A student who remembers the dramatic rebounds as the main result picks this. The reverse happened: nearly all passed through with little or no deflection, and large deflections were very rare.
Nearly all the alpha particles passed through almost undeflected; a very few were deflected by more than 90°. — Almost every alpha particle went through the thin gold foil almost undeflected, and a very small fraction were deflected through large angles, some through more than 90°. Both parts matter: the rarity shows the nucleus is tiny, and the large deflections show its charge and mass are concentrated.
The few alpha particles that bounced back had collided with the electrons in the gold atoms. — A student who pictures electrons as the particles an alpha particle can hit picks this. An electron has about 1/7300 of an alpha particle's mass, so it cannot turn one back; only the massive, positive nucleus can.
The few alpha particles that bounced back had struck the hard outer surface of a gold nucleus. — A student who thinks a deflection needs contact picks this. The alpha particles were turned back by electric repulsion, which acts at a distance; they stopped and reversed before reaching the nucleus.
Working No calculation is needed. The results: almost all alpha particles passed through the foil with little or no deflection; a very small fraction were deflected through large angles, some through more than 90°.
2 A nucleus of potassium-40 is written in nuclear notation as ⁴⁰₁₉K. Which statement about this nucleus is correct?
Answer and reasoning
It contains 40 neutrons. — A student who thinks the nucleon number counts only neutrons picks this. A = 40 counts protons and neutrons together; the neutrons number 40 − 19 = 21.
It contains 19 neutrons. — A student who assumes a nucleus always has equal numbers of protons and neutrons picks this. That holds only for some light nuclei; here N = 40 − 19 = 21.
It contains 19 nucleons. — A student who reads the lower number as the nucleon number picks this. In nuclear notation the lower number is the proton number; the nucleon number is the upper number, 40.
It contains 21 neutrons. — In ᴬ_Z X the upper number is the nucleon number A = 40 and the lower number is the proton number Z = 19, so the number of neutrons is N = A − Z = 40 − 19 = 21.
Working A = 40 (upper number), Z = 19 (lower number). Protons = 19; nucleons = 40; neutrons N = A − Z = 40 − 19 = 21.
3 The emission spectrum of a hot, low-pressure gas of one element consists of bright lines at a set of separate wavelengths. Why is this evidence that the atoms have discrete energy levels?
Answer and reasoning
Each line has the photon energy of one allowed level, so each line marks one energy level. — A student who links one line to one level picks this. A photon's energy equals the difference between two levels, so each line marks a pair of levels, and there can be more lines than levels.
The lines are equally spaced in frequency, just as the atom's energy levels are equally spaced. — A student who pictures energy levels as evenly spaced rungs picks this. Spectral lines are not equally spaced, and neither are the levels; in hydrogen they crowd together at higher energies.
Each line is emitted as an electron jumps up from one allowed level to a higher level. — A student who links emission with excitation picks this. Moving up needs energy; a photon is emitted when the electron moves down, and the line's photon energy is the energy lost.
Each line has a single photon energy, equal to the difference between two of the atom's allowed energies. — Each line contains photons of one frequency and therefore one energy, E = hf. A fixed set of photon energies means the atom can change its energy only by fixed amounts, the differences between a fixed set of allowed levels.
4 Which statement correctly describes how an atom emits a photon?
Answer and reasoning
An electron moves to a lower energy level and one photon carries away the energy lost. — A downward transition releases energy equal to the difference between the levels, and it is carried away by a single photon of that energy.
An electron moves to a higher energy level and one photon carries away the energy gained. — A student who links emission with the electron jumping up picks this. Moving to a higher level requires energy to be absorbed; emission happens when the electron moves down.
An electron leaves the atom altogether, and that electron is itself the photon that is detected. — A student who confuses the electron with the photon picks this. The electron stays in the atom in a lower level; the photon is a separate quantum of electromagnetic energy, with no mass or charge.
An electron in a high level gives out photons steadily for as long as it stays there. — A student who pictures an excited atom glowing continuously picks this. One transition produces one photon; an electron sitting in a level emits nothing until it makes a transition down.
5 In a sodium atom, an electron makes a transition from an energy level at −3.03 eV to one at −5.14 eV. What is the frequency of the emitted photon? Take h = 6.63 × 10⁻³⁴ J s and 1 eV = 1.60 × 10⁻¹⁹ J.
7.31 × 10¹⁴ Hz — A student who takes the photon energy as the energy of the starting level uses 3.03 eV and gets 7.31 × 10¹⁴ Hz. The photon carries the difference between the two levels, 2.11 eV.
1.97 × 10¹⁵ Hz — A student who adds the sizes of the two negative levels uses 3.03 + 5.14 = 8.17 eV and gets 1.97 × 10¹⁵ Hz. The difference is (−3.03) − (−5.14) = 2.11 eV.
3.18 × 10³³ Hz — A student who puts 2.11 eV straight into f = E/h without converting to joules gets 3.18 × 10³³ Hz. h is in J s, so the energy must be in joules: 2.11 eV = 3.38 × 10⁻¹⁹ J.
Working ΔE = E_upper − E_lower = (−3.03) − (−5.14) = 2.11 eV. In joules: 2.11 × 1.60 × 10⁻¹⁹ = 3.376 × 10⁻¹⁹ J. f = ΔE/h = 3.376 × 10⁻¹⁹ J / 6.63 × 10⁻³⁴ J s = 5.09 × 10¹⁴ Hz.
6 Two discharge tubes, P and Q, each contain a single gas. The emission spectrum of each gas shows exactly four bright lines in the visible region, but none of the wavelengths of the lines from P matches any of the wavelengths of the lines from Q. A student claims that P and Q contain the same element. Which statement is correct?
Answer and reasoning
The claim is correct, because the number of lines is set by the number of electrons in an atom of the element. — A student who counts lines as electrons picks this. Lines come from transitions between energy levels, and one electron can make many transitions, so the number of lines does not identify an element; the wavelengths do, and here none of them match.
P and Q may contain the same element, because a gas at a higher temperature emits its lines at shorter wavelengths. — A student who transfers 'hotter means bluer' from a glowing solid to a gas picks this. The line wavelengths are fixed by the atom's energy levels and do not shift with temperature, so lines at different wavelengths mean different elements.
The claim is wrong, because an element is identified by the wavelengths of its lines, and none match. — Each element has its own set of energy levels, so the wavelengths of its lines form a unique pattern. Two gases whose lines share no wavelength cannot be the same element, however many lines each shows.
The claim is correct if the two tubes glow with the same overall colour, since that colour identifies an element. — A student who identifies elements by the blended colour of the glow picks this. The eye merges the lines into one colour, and different elements can look alike; the spectrometer shows that no line wavelengths coincide, so the gases are different elements.
7 A nucleus of lead-208 has proton number 82 and nucleon number 208. Using R = R₀A^(1/3) with R₀ = 1.2 × 10⁻¹⁵ m, what is the radius of this nucleus? HL
Answer and reasoning
2.5 × 10⁻¹³ m — A student who takes the radius as proportional to A calculates R₀ × 208 = 2.5 × 10⁻¹³ m. The volume, not the radius, is proportional to A, so the radius goes as the cube root of A.
5.2 × 10⁻¹⁵ m — A student who puts the proton number into the formula calculates R₀ × 82^(1/3) = 5.2 × 10⁻¹⁵ m. The size depends on all the nucleons, so A = 208 must be used.
7.1 × 10⁻¹⁵ m — R = R₀A^(1/3) = 1.2 × 10⁻¹⁵ m × 208^(1/3) = 1.2 × 10⁻¹⁵ m × 5.925 = 7.1 × 10⁻¹⁵ m.
8.3 × 10⁻¹⁴ m — A student who reads A^(1/3) as A ÷ 3 calculates R₀ × 208/3 = 8.3 × 10⁻¹⁴ m. A^(1/3) is the cube root of A, 5.925.
Working R = R₀A^(1/3) = (1.2 × 10⁻¹⁵ m)(208)^(1/3) = (1.2 × 10⁻¹⁵ m)(5.925) = 7.1 × 10⁻¹⁵ m.
8 In alpha-particle scattering, the numbers scattered through large angles agree with Rutherford's prediction, which assumes that only the electric repulsion acts, at low energies. At kinetic energies of a few tens of MeV the measured numbers fall below the prediction. What is the reason? HL
Answer and reasoning
Gravitational attraction between the alpha particles and the very massive nucleus becomes significant. — A student who thinks gravity matters inside the nucleus picks this. The gravitational force is about 10³⁵ times smaller than the electric force at any distance and has no measurable effect.
The strong nuclear force, reaching far beyond the nucleus, now outweighs the electric force. — A student who thinks the strong force is long-range picks this. If it reached far beyond the nucleus, low-energy alpha particles would deviate too; it acts only within a few femtometres, which only fast alpha particles reach.
The alpha particles get close enough to the nucleus to feel the short-range strong nuclear force. — At higher energies the distance of closest approach becomes comparable with the nuclear radius. There the short-range strong nuclear force acts on the alpha particle as well as the electric force, so Rutherford's electric-only prediction no longer applies.
The alpha particles now travel close to the speed of light, so Rutherford's calculation fails. — A student who assumes 'high energy' means relativistic picks this. At a few tens of MeV an alpha particle (rest energy about 3700 MeV) moves at only about 0.1c; relativistic effects are tiny.
9 An alpha particle of kinetic energy 5.0 MeV moves head-on towards a gold-197 nucleus (proton number 79). Only the electric repulsion acts, and the recoil of the nucleus is negligible. The radius of a nucleus is R₀A^(1/3), where R₀ = 1.2 × 10⁻¹⁵ m. Take k = 8.99 × 10⁹ N m² C⁻² and e = 1.60 × 10⁻¹⁹ C. What is the distance of closest approach? HL
Answer and reasoning
4.5 × 10⁻¹⁴ m — At closest approach all the kinetic energy has become electric potential energy: E_k = kqQ/d, with q = 2e and Q = 79e. d = kqQ/E_k = (8.99 × 10⁹)(2 × 1.60 × 10⁻¹⁹)(79 × 1.60 × 10⁻¹⁹)/(8.0 × 10⁻¹³ J) = 4.5 × 10⁻¹⁴ m, about six times the nuclear radius.
2.3 × 10⁻¹⁴ m — A student who gives the alpha particle a charge of +e instead of +2e gets half the correct distance. An alpha particle has two protons, so q = 2e.
2.3 × 10⁻¹³ m — A student who uses the nucleon numbers for the charges (4e and 197e) gets 2.3 × 10⁻¹³ m. Only protons are charged, so the charges are 2e and 79e.
7.0 × 10⁻¹⁵ m — A student who thinks the alpha particle turns back only on touching the nucleus gives the nuclear radius, R₀ × 197^(1/3) = 7.0 × 10⁻¹⁵ m. Electric repulsion stops a 5.0 MeV alpha particle about six nuclear radii away.
Working E_k = 5.0 MeV = 5.0 × 10⁶ × 1.60 × 10⁻¹⁹ J = 8.0 × 10⁻¹³ J. At closest approach E_k = kqQ/d with q = 2e, Q = 79e. d = kqQ/E_k = (8.99 × 10⁹ N m² C⁻²)(3.20 × 10⁻¹⁹ C)(1.264 × 10⁻¹⁷ C)/(8.0 × 10⁻¹³ J) = 3.636 × 10⁻²⁶ J m/8.0 × 10⁻¹³ J = 4.5 × 10⁻¹⁴ m. (Nuclear radius R₀A^(1/3) = 7.0 × 10⁻¹⁵ m, so the alpha particle never reaches the nucleus.)
10 In the Bohr model, the energy levels of hydrogen are given by E = −13.6/n² eV. Which statement about these energy levels is correct? HL
Answer and reasoning
The levels become more closely spaced as n increases, converging on E = 0, at which the electron is free. — E₁ = −13.6 eV, E₂ = −3.40 eV, E₃ = −1.51 eV, E₄ = −0.85 eV: the gaps shrink as n grows, and as n → ∞ the energy approaches zero, the energy of an electron just freed from the nucleus.
The levels are equally spaced in energy, because each increase of n by one adds the same energy to the atom. — A student who thinks of levels as evenly spaced ladder rungs picks this. The 1/n² dependence makes the gaps shrink rapidly: 10.2 eV from n = 1 to 2, but only 1.89 eV from n = 2 to 3.
The electron needs 13.6 eV to be removed from the atom whichever level it occupies at the start. — A student who has memorized '13.6 eV' as the ionization energy of hydrogen picks this. That value applies only to the ground state; from n = 2 the electron needs 0 − (−3.40) = 3.40 eV, and from higher levels less still.
The n = 1 level is the highest energy level of the atom, because 13.6 eV is the largest of the values. — A student who compares the sizes of the numbers and ignores the minus sign picks this. −13.6 eV is the most negative, so lowest, energy: n = 1 is the ground state, and every other level lies above it.
Working E_n = −13.6/n² eV: E₁ = −13.6 eV, E₂ = −13.6/4 = −3.40 eV, E₃ = −13.6/9 = −1.51 eV, E₄ = −13.6/16 = −0.850 eV. Gaps: 10.2 eV, 1.89 eV, 0.66 eV, … shrinking towards zero; as n → ∞, E → 0, the energy of a free electron at rest. The ground state (n = 1) is the lowest, most negative, level.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
12 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 What did Rutherford conclude about the structure of the atom from the results of the Geiger–Marsden experiment?
Answer and reasoning
Its nucleus consists of protons and neutrons, which the experiment first detected. — A student who credits the whole modern picture of the atom to this experiment picks this. The scattering depends only on the nucleus's charge, mass and size; the neutron was not discovered until 1932.
It is a solid ball, and most alpha particles passed through the gaps between the atoms. — A student who pictures atoms as solid balls picks this. The foil was many atoms thick, so the alpha particles had to go through the atoms themselves; they passed because each atom is almost entirely empty of mass and concentrated charge.
Its positive charge and nearly all its mass are in a tiny central nucleus. — Only a small, massive, concentrated positive charge can repel an alpha particle strongly enough to turn it back, and the fact that so few were turned back shows that this nucleus fills only a tiny fraction of the atom.
Its electrons are heavy enough to turn back any alpha particles that hit them. — A student who thinks electrons caused the large deflections picks this. An electron has about 1/7300 of the mass of an alpha particle, far too little to reverse its motion.
2 Chlorine-35 and chlorine-37 are two isotopes of chlorine. A chlorine-35 nucleus contains 17 protons. Which statement about a chlorine-37 nucleus is correct?
Answer and reasoning
It contains 17 protons and 20 neutrons. — Isotopes of one element have the same proton number, so chlorine-37 also has 17 protons. The nucleon number counts protons and neutrons together: chlorine-35 has 35 − 17 = 18 neutrons and chlorine-37 has 37 − 17 = 20, two more.
It contains 19 protons and 18 neutrons. — A student who thinks isotopes differ in proton number gives the two extra nucleons to the protons. The proton number defines the element: a nucleus with 19 protons would be potassium, not chlorine. Chlorine-37 has 17 protons and 37 − 17 = 20 neutrons.
It contains 17 protons and 18 neutrons. — A student who thinks isotopes differ only in their electrons treats the two nuclei as identical and gives chlorine-37 the 18 neutrons of chlorine-35. Atoms that differ only in their electrons are ions; isotopes differ in their nuclei. A chlorine-37 nucleus has 37 − 17 = 20 neutrons.
It contains 17 protons and 37 neutrons. — A student who takes the nucleon number to be the number of neutrons picks this. A = 37 counts protons and neutrons together, so chlorine-37 has 37 − 17 = 20 neutrons (and chlorine-35 has 35 − 17 = 18, not 35).
Working Isotopes of one element have the same proton number, so Z = 17 for both. Chlorine-35: N = A − Z = 35 − 17 = 18 neutrons. Chlorine-37: N = A − Z = 37 − 17 = 20 neutrons, two more than chlorine-35.
3 White light, containing all visible wavelengths, passes through a cool gas of a single element and is then examined with a spectrometer. What is observed?
Answer and reasoning
Dark lines at wavelengths that differ from all of the element's emission lines. — A student who thinks absorption is the 'opposite' of emission picks this. The same energy differences are involved, so absorption lines fall at the wavelengths of emission lines.
Dark lines at wavelengths that match some of the element's emission lines. — Atoms absorb only photons whose energy equals a difference between two of their levels. Those are the energies of photons emitted in the reverse transitions, so the dark lines coincide with emission lines; a cool gas absorbs mainly from the ground state, so only some emission lines appear.
A dark band across every wavelength shorter than the one for the smallest energy gap. — A student who thinks an atom absorbs any photon with at least enough energy picks this. A photon is absorbed only if its energy exactly matches a level difference, so separate dark lines appear, not a band.
A complete spectrum with no dark lines, as the atoms re-emit all the light they absorb. — A student who thinks re-emitted photons continue along the beam picks this. The atoms re-emit in all directions, so the light travelling on towards the spectrometer is depleted at the absorbed wavelengths.
4 An atom has energy levels at −10.0 eV (the ground state), −5.0 eV and −3.0 eV. An atom in its ground state is struck by a photon of energy 3.0 eV. What happens?
Answer and reasoning
The photon is absorbed, and the electron moves up to the −3.0 eV energy level. — A student who thinks a photon's energy must equal the energy of a level matches 3.0 eV to the −3.0 eV level. Absorption needs the photon energy to equal a DIFFERENCE between levels: reaching −3.0 eV from −10.0 eV needs 7.0 eV.
The photon is not absorbed, as it matches neither gap from the ground state, 15.0 or 13.0 eV. — A student who adds the sizes of the negative levels finds gaps of 10.0 + 5.0 = 15.0 eV and 10.0 + 3.0 = 13.0 eV. The photon is indeed not absorbed, but the gaps are wrong: they are (−5.0) − (−10.0) = 5.0 eV and (−3.0) − (−10.0) = 7.0 eV.
The photon is not absorbed, as it matches neither gap from the ground state, 5.0 or 7.0 eV. — From the ground state a photon can be absorbed only if its energy exactly equals an upward gap, (−5.0) − (−10.0) = 5.0 eV or (−3.0) − (−10.0) = 7.0 eV, or if it is at least 10.0 eV, enough to ionize the atom. 3.0 eV is neither, so the photon is not absorbed and passes on.
The photon is absorbed, and the electron moves up to an energy of −7.0 eV. — A student who thinks an electron can take any amount of energy adds 3.0 eV to −10.0 eV. −7.0 eV is not an allowed level; the electron can only be at −10.0, −5.0 or −3.0 eV, so it cannot take 3.0 eV.
Working Photon energies that can be absorbed from the ground state: exactly −5.0 − (−10.0) = 5.0 eV or −3.0 − (−10.0) = 7.0 eV for excitation, or at least 0 − (−10.0) = 10.0 eV for ionization. 3.0 eV is neither, so the photon is not absorbed. (The atom is in the ground state, so the 2.0 eV gap between −5.0 eV and −3.0 eV is not available either, and it does not equal 3.0 eV.)
5 An atom has only three energy levels: −1.00 eV, −3.00 eV and −7.00 eV. What is the energy of the emitted photon that has the SHORTEST wavelength?
Answer and reasoning
2.00 eV — A student who thinks a longer wavelength means more energy picks the smallest energy drop, (−1.00) − (−3.00) = 2.00 eV, for the shortest wavelength. λ = hc/E, so the smallest photon energy gives the LONGEST wavelength.
6.00 eV — Emission needs a downward transition. The possible photon energies are 2.00 eV (−1.00 to −3.00), 4.00 eV (−3.00 to −7.00) and 6.00 eV (−1.00 to −7.00). λ = hc/E, so the largest photon energy, 6.00 eV, gives the shortest wavelength.
3.00 eV — A student who takes the photon energy as the energy of the level the electron starts from sees that emission can start only from −1.00 eV or −3.00 eV, and picks the larger, 3.00 eV. The photon carries the DIFFERENCE between two levels; the largest is (−1.00) − (−7.00) = 6.00 eV.
10.0 eV — A student who adds the sizes of the negative levels gets 4.00 eV, 10.0 eV and 8.00 eV for the three transitions and picks the largest, 3.00 + 7.00 = 10.0 eV. The differences are 2.00, 4.00 and 6.00 eV; no transition in this atom can emit a photon of more than 6.00 eV.
Working Downward transitions and photon energies: −1.00 → −3.00 eV: 2.00 eV; −3.00 → −7.00 eV: 4.00 eV; −1.00 → −7.00 eV: 6.00 eV. Since λ = hc/E, the largest photon energy, 6.00 eV, gives the shortest wavelength.
6 The spectrum of light from a star shows strong dark lines at exactly the wavelengths of the two bright yellow lines in the laboratory emission spectrum of sodium. What can be concluded from this evidence?
Answer and reasoning
Sodium is absent from the star, because the light at sodium's wavelengths is missing from its spectrum. — A student who reads a missing colour as a missing element picks this. The light at those wavelengths is missing BECAUSE sodium atoms absorbed it; the dark lines are evidence that sodium is present.
Nothing about sodium, because an element absorbs at different wavelengths from those it emits. — A student who thinks absorption and emission lines of an element differ picks this. The same energy differences give both, so the dark lines coinciding with sodium's emission lines identify sodium.
Sodium is the most abundant element in the star, because its absorption lines are so strong. — A student who equates strong lines with large amounts picks this. Line strength also depends on temperature and on how many atoms are in the absorbing level; the lines show that sodium is present, not that it dominates.
Sodium atoms in the star's outer layers absorb light at these wavelengths, so sodium is present in the star. — Atoms absorb exactly the wavelengths they emit, because the same level differences are involved. Dark lines at sodium's emission wavelengths show that sodium atoms lie in the cooler gas the light passes through, the star's outer layers.
7 Nuclear radii are given by R = R₀A^(1/3), and the mass of a nucleus may be taken as proportional to its nucleon number A. What is the ratio (density of a carbon-12 nucleus)/(density of a molybdenum-96 nucleus)? HL
Answer and reasoning
0.13 — A student who thinks a heavier nucleus is denser takes the ratio of the masses, 12/96 = 0.125, ignoring the volumes. The molybdenum nucleus also has 8 times the volume, so the densities are equal.
1.00 — Mass ∝ A and volume = ⁴⁄₃πR³ = ⁴⁄₃πR₀³A ∝ A, so density = mass/volume does not depend on A. Both nuclei have the same density, about 2 × 10¹⁷ kg m⁻³, and the ratio is 1.00.
64.0 — A student who takes the radius as proportional to A makes the volume ∝ A³ and the density ∝ 1/A², giving (96/12)² = 64.0. The radius goes as A^(1/3), so the volume goes as A and the density is the same for both.
0.25 — A student who forgets to cube the radius takes volume ∝ R ∝ A^(1/3), so density ∝ A^(2/3), giving (12/96)^(2/3) = 0.25. The volume is ⁴⁄₃πR³ ∝ A, so the density does not depend on A.
Working Density ρ = m/V with m ∝ A and V = ⁴⁄₃πR³ = ⁴⁄₃πR₀³A, so ρ ∝ A/A, the same for every nucleus. ρ(carbon-12)/ρ(molybdenum-96) = (12/12)/(96/96) = 1.00. (Numerically, ρ = 3 × 1.67 × 10⁻²⁷ kg/(4π × (1.2 × 10⁻¹⁵ m)³) = 2.3 × 10¹⁷ kg m⁻³ for both.)
8 Alpha particles of kinetic energy E_k are fired head-on at gold-197 nuclei (proton number 79), and their distance of closest approach is d. Alpha particles of kinetic energy 2E_k are then fired head-on at silver-107 nuclei (proton number 47). In both cases only the electric repulsion acts and nuclear recoil is negligible. What is the new distance of closest approach? HL
Answer and reasoning
0.55d — A student who writes the potential energy as kqQ/r² (the force expression) gets d ∝ √(Z/E_k) and √(47/79 × 1/2) = 0.55d. The potential energy is kqQ/r, so d ∝ Z/E_k.
0.82d — A student who thinks the alpha particle turns back when it touches the nucleus takes d as the nuclear radius, ∝ A^(1/3), and gets (107/197)^(1/3) = 0.82d. The alpha particles stop well outside the nucleus, at a distance set by charge and energy.
0.27d — A student who uses nucleon numbers for the nuclear charge gets (107/197) × ½ = 0.27d. The charge of a nucleus is Ze; the proton numbers 47 and 79 must be used.
0.30d — d = kqQ/E_k, so d ∝ Z/E_k. The new distance is d × (47/79) × (1/2) = 0.30d: the smaller nuclear charge and the doubled energy both let the alpha particle get closer.
Working At closest approach E_k = kqQ/d, so d = k(2e)(Ze)/E_k ∝ Z/E_k. d_new/d = (47/79) × (E_k/2E_k) = 0.595 × 0.5 = 0.297, so d_new = 0.30d.
9 In the Bohr model, the energy levels of hydrogen are E = −13.6/n² eV. What is the wavelength of the photon emitted when the electron makes a transition from n = 4 to n = 3? Take h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and 1 eV = 1.60 × 10⁻¹⁹ J. HL
Answer and reasoning
1.10 × 10⁻⁶ m — A student who uses 1/n instead of 1/n² gets ΔE = 13.6(⅓ − ¼) = 1.13 eV and λ = 1.10 × 10⁻⁶ m. The levels depend on 1/n²: −0.850 eV and −1.51 eV.
1.88 × 10⁻⁶ m — E₄ = −13.6/16 = −0.850 eV and E₃ = −13.6/9 = −1.51 eV, so ΔE = 0.661 eV = 1.06 × 10⁻¹⁹ J. λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(1.06 × 10⁻¹⁹) = 1.88 × 10⁻⁶ m, in the infrared.
1.46 × 10⁻⁶ m — A student who takes the photon energy as the energy of the starting level uses 0.850 eV and gets 1.46 × 10⁻⁶ m. The photon carries the difference between the levels, 1.51 − 0.850 = 0.661 eV.
5.27 × 10⁻⁷ m — A student who adds the sizes of the two negative levels uses 0.850 + 1.51 = 2.36 eV and gets 5.27 × 10⁻⁷ m. ΔE = (−0.850) − (−1.51) = 0.661 eV.
Working E₄ = −13.6/4² = −0.850 eV; E₃ = −13.6/3² = −1.511 eV. ΔE = 0.661 eV × 1.60 × 10⁻¹⁹ J eV⁻¹ = 1.058 × 10⁻¹⁹ J. λ = hc/ΔE = (6.63 × 10⁻³⁴ J s)(3.00 × 10⁸ m s⁻¹)/(1.058 × 10⁻¹⁹ J) = 1.88 × 10⁻⁶ m (infrared).
10 In the Bohr model of hydrogen, what is the angular momentum of the electron in the n = 3 orbit? Take h = 6.63 × 10⁻³⁴ J s. HL
Answer and reasoning
3.17 × 10⁻³⁴ J s — The Bohr condition is mvr = nh/2π, so in the n = 3 orbit L = 3h/2π = 3 × 6.63 × 10⁻³⁴ J s/2π = 3.17 × 10⁻³⁴ J s.
9.50 × 10⁻³⁴ J s — A student who carries the n² of the energy formula into the angular momentum calculates 9h/2π = 9.50 × 10⁻³⁴ J s. The angular momentum is proportional to n: mvr = nh/2π = 3h/2π.
1.06 × 10⁻³⁴ J s — A student who gives every orbit the same angular momentum uses h/2π = 1.06 × 10⁻³⁴ J s. That is the value for n = 1 only; in the third orbit mvr = 3h/2π.
1.99 × 10⁻³³ J s — A student who writes the condition as mvr = nh, dropping the 2π, calculates 3h = 1.99 × 10⁻³³ J s, 2π times too large. The Bohr condition is mvr = nh/2π.
Working Bohr condition: L = mvr = nh/2π. For n = 3: L = 3 × 6.63 × 10⁻³⁴ J s/2π = 1.989 × 10⁻³³ J s/6.283 = 3.17 × 10⁻³⁴ J s.
11 In the Bohr model of hydrogen, why can the electron occupy only certain orbits, each with its own energy? HL
Answer and reasoning
Its angular momentum mvr must be n² times h/2π, to match the n² in the energy formula. — A student who carries the n² from the energy formula into the angular momentum picks this. The condition is mvr = nh/2π; the n² appears in the radius and, as 1/n², in the energy.
Each orbit can hold only a fixed number of electrons, so the electron must go where there is room. — A student who links energy levels to shell capacity picks this. Hydrogen has only one electron, yet its levels are discrete; the discreteness comes from the quantization of angular momentum.
Light can be emitted only as photons of fixed sizes, and this limits the orbits the electron takes. — A student who reverses cause and effect picks this. Photons can have any energy; it is because the orbits and energies are discrete that the photons emitted in transitions have only certain energies.
Its angular momentum mvr must be a whole number n times h/2π, which allows only certain radii. — Bohr's postulate is that angular momentum is quantized, mvr = nh/2π. Only orbits meeting this condition are allowed, so the radii are discrete, and each allowed orbit has its own energy, E = −13.6/n² eV.
Working Bohr condition: L = mvr = n(h/2π), n = 1, 2, 3, … Only radii for which this holds are allowed, and each allowed orbit has its own energy, E = −13.6/n² eV.
12 Alpha particles scattered by a thin gold foil are counted at large angles. Up to an alpha-particle kinetic energy E₀ of a few tens of MeV, the counts agree with Rutherford's prediction, which assumes that only the electric repulsion acts. Above E₀ the counts fall below the prediction. Which statement about this result is correct? HL
Answer and reasoning
Below E₀ the alpha particles rebound from the nuclear surface like balls from a wall; above E₀ they break through it, so fewer return. — A student who pictures the nucleus as a hard wall picks this. There is no surface to rebound from: at every energy the deflection is caused by electric repulsion acting at a distance. Within a few femtometres the attractive strong force acts as well, and alpha particles that get this close may be absorbed or take part in nuclear reactions, which is why the counts fall.
Above E₀ the electric repulsion no longer acts, and the strong nuclear force alone decides how the alpha particles scatter. — A student who thinks one force hands over to the other picks this. The electric repulsion acts at every separation; above E₀ the strong force acts as well, and it is this additional short-range force that makes the electric-only prediction fail.
The distance of closest approach of an alpha particle of energy E₀ gives an estimate of the radius of the gold nucleus. — Deviations begin when the alpha particles get close enough to the nucleus for the short-range strong nuclear force to act. At E₀ the distance of closest approach has become about equal to the nuclear radius, so that distance, found from energy conservation, estimates the size of the nucleus.
Above E₀ the alpha particles move at nearly the speed of light, so Rutherford's non-relativistic calculation fails. — A student who reads 'high energy' as 'relativistic' picks this. An alpha particle of a few tens of MeV has kinetic energy about 1% of its rest energy (about 3700 MeV) and moves at roughly 0.1c, so relativistic effects are negligible.
That was your twenty minutes. Real practice on E.1 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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