Pen colours: blue = objects and paths, red = laws, forces and key steps, green = answers and results, pencil = construction lines. Boxes: Why? = the reasoning, Mistake = what students get wrong, yellow note = exam tip, dashed box = a derivation step by step.
What is inside an atom? Nobody can see it, so physicists built models and tested them with experiments. Two pieces of evidence changed the model: the scattering of alpha particles showed that the positive charge sits in a tiny nucleus, and the line spectra of gases showed that an atom can only have certain energies. This topic follows that story. Old models are not thrown away: each one still explains some things well.
From Thomson to Rutherford to Bohr: each new model came from an experiment the old one could not explain.
Quantity
Symbol and unit
Meaning
proton number
Z (no unit)
number of protons in the nucleus
nucleon number
A (no unit)
number of protons + neutrons
photon energy
E, J or eV
E = hf = hc/λ; 1 eV = 1.60 × 10⁻¹⁹ J
Planck constant
h = 6.63 × 10⁻³⁴ J s
links a photon’s energy to its frequency
distance of closest approach HL
d, m
nearest a head-on particle gets to a nucleus
nuclear radius HL
R = R₀A^⅓, R₀ = 1.2 × 10⁻¹⁵ m
size of a nucleus (about 10⁻¹⁵ m)
Sizes to keep in mind: an atom is about 10⁻¹⁰ m across; a nucleus is about 10⁻¹⁵ m, so the nucleus is about 100 000 times smaller than the atom. 1 fm (femtometre) = 10⁻¹⁵ m.
Learn it word for word. Nucleon: a proton or a neutron. Proton number Z: the number of protons in the nucleus. Nucleon number A: the number of protons and neutrons in the nucleus. Isotopes: nuclei of the same element (same Z) with different numbers of neutrons (different A).
Nuclear notation:AZX · X = chemical symbol · number of neutrons N = A − Z A neutral atom has Z electrons. (You do not need to remember chemical symbols: they are given.)
Worked example 1A gold nucleus is written 19779Au. (a) How many protons, neutrons and nucleons does it have? (b) How many electrons are in a neutral gold atom and in the ion Au³⁺? (c) 19579Au is another nucleus. What is the same and what is different?
(a) Protons Z = 79; nucleons A = 197; neutrons 197 − 79 = 118.
(b) Neutral atom: 79 electrons. Au³⁺ has lost 3 electrons: 76 electrons (the nucleus does not change).
(c) Same Z = 79, so it is the same element: they are isotopes. It has 2 fewer neutrons (116).
In 1909–1911 Geiger and Marsden, working with Rutherford, fired alpha particles (helium nuclei, charge +2e, about 7000 times the mass of an electron) at a very thin gold foil in a vacuum, and counted how many came off at each angle.
The vacuum stops the alpha particles being slowed or scattered by air. The foil is very thin so that each alpha particle is scattered by (at most) one nucleus.
What they saw 1. Most alpha particles went straight through the foil, undeflected or with very small deflections. 2. A few were deflected through large angles. 3. A very small number (about 1 in 8000) came almost straight back (deflection greater than 90°).
Each path is drawn from the real inverse-square repulsion. Only a particle that passes very close to the nucleus is turned through a large angle.
Why gold, and why alpha particles? Gold can be hammered into a foil only a few hundred atoms thick, and its nucleus has a large charge (79e), which gives strong repulsion. Alpha particles are massive and fast: electrons in the atom are far too light to push them aside, so any large deflection must come from something heavy and positive.
In the Thomson model the positive charge is spread through the whole atom. An alpha particle passing through feels only small, spread-out forces, which can explain the small deflections, but never a bounce back: the force is never large enough. To turn a fast, heavy alpha particle around, the force must be enormous, and the electric force F = kQq/r² is enormous only when r is tiny. So the positive charge must be packed into a very small region.
Why the Thomson atom fails: spread-out charge never gives a large force. A tiny nucleus gives a huge force when the alpha particle comes close.
Rutherford’s conclusions • Most of the mass of the atom and all of the positive charge are concentrated in a very small volume: the nucleus. • The atom is mostly empty space (most alpha particles pass straight through). • The electrons orbit the nucleus, far away from it.
Worked example 2Compare the electric field at the surface of a gold nucleus (radius 7.0 × 10⁻¹⁵ m) with the field at the surface of a Thomson gold atom of the same charge (radius 1.4 × 10⁻¹⁰ m).
E = kQ/r² for both, with the same Q, so the ratio is (ratom/rnucleus)².
The same charge squeezed into a nucleus gives a field (and a force) hundreds of millions of times larger: enough to send an alpha particle back.
“Explain how the results led to the nuclear model” (3 marks)✔ Most alpha particles pass straight through → the atom is mostly empty space. ✔ A few are deflected through large angles (some more than 90°) → a very large repulsive force, which needs a small, positively charged region. ✔ Only a small fraction bounce back → the nucleus is very small and contains most of the mass. The Thomson model explains only the small deflections: its spread-out charge cannot give the large force.
Mistake Saying “the alpha particles bounce off the gold atoms” or “hit the nucleus”. They never touch it (for these energies): they are repelled by the electric force. And the experiment tells us about the positive charge and the mass, not about neutrons (they were found only in 1932).
Fire an alpha particle head-on at a nucleus. As it gets closer it slows down, because the electric force pushes back. It stops for an instant at the distance of closest approach d, then comes straight back. Treat the nucleus as fixed (it is much heavier) and use conservation of energy: the kinetic energy far away becomes electric potential energy at d.
Kinetic energy turns into electric potential energy. Their sum stays equal to the starting kinetic energy.
Distance of closest approach (head-on): Ek = k q1q2/d → for an alpha particle on a nucleus of proton number Z: d = k(2e)(Ze)/Ek k = 8.99 × 10⁹ N m² C⁻² · Ek in joules · e = 1.60 × 10⁻¹⁹ C
Worked example 3An alpha particle of kinetic energy 6.0 MeV is fired head-on at a gold nucleus (Z = 79). (a) Find the distance of closest approach. (b) The radius of the gold nucleus is about 7.0 × 10⁻¹⁵ m. Does the alpha particle reach the nucleus?
Charges: alpha q₁ = 2e, gold q₂ = 79e. Ek = 6.0 × 10⁶ × 1.60 × 10⁻¹⁹ J.
(a) d = k(2e)(79e)/Ek = 158ke/(6.0 × 10⁶) (one e cancels with the e in “eV”) = 158 × 8.99 × 10⁹ × 1.60 × 10⁻¹⁹/6.0 × 10⁶ = 3.8 × 10⁻¹⁴ m.
(b) 3.8 × 10⁻¹⁴ m is about 5 times the nuclear radius: it does not reach the nucleus. Only the electric force acts, so Rutherford’s model works.
Worked example 4A proton and an alpha particle are each accelerated from rest through a potential difference of 4.0 MV, then fired head-on at a tin nucleus (Z = 50). Compare their distances of closest approach.
Energy gained = qV. Proton: Ek = e × 4.0 MV = 4.0 MeV. Alpha (charge 2e): Ek = 8.0 MeV.
Alpha: d = k(2e)(50e)/(8.0 × 10⁶ e) = 1.8 × 10⁻¹⁴ m: the same! The alpha has twice the energy but also twice the charge, so the two effects cancel.
What changes the distance of closest approach?From d = kq₁q₂/Ek: • more energy → closer (d ∝ 1/Ek); • bigger nuclear charge Z → further away (d ∝ Z); • a different isotope of the same element: same Z, so the same d (neutrons have no charge).
Mistake Forgetting that an alpha particle has charge 2e, using the mass number A instead of Z for the nuclear charge, or leaving the energy in MeV. Change MeV to joules: × 10⁶ × 1.60 × 10⁻¹⁹.
How big is a nucleus? Scattering experiments answer this. Fire faster and faster particles: the distance of closest approach shrinks. When the particle just reaches the surface of the nucleus, the scattering changes (next section). The distance of closest approach at that moment is (about) the nuclear radius. Experiments with many nuclei (and with high-energy electrons) give one simple rule:
Nuclear radius:R = R₀A^⅓ · R₀ = 1.2 × 10⁻¹⁵ m (data booklet) · A = nucleon number
Doubling the number of nucleons does not double the radius: R grows as the cube root of A.
Worked example 5(a) Find the radius of a gold-197 nucleus and of a helium-4 nucleus (an alpha particle). (b) How many times larger is the radius of uranium-238 than that of aluminium-27?
(a) Gold: R = 1.2 × 10⁻¹⁵ × 197^⅓ = 1.2 × 10⁻¹⁵ × 5.82 = 7.0 × 10⁻¹⁵ m. Helium: R = 1.2 × 10⁻¹⁵ × 4^⅓ = 1.9 × 10⁻¹⁵ m.
(b) RU/RAl = (238/27)^⅓ = 8.81^⅓ = 2.07. Nearly 9 times the nucleons, only about twice the radius.
Why every nucleus has the same density Mass of a nucleus ≈ A × u (u = 1.66 × 10⁻²⁷ kg). Volume = (4/3)πR³ = (4/3)π(R₀A^⅓)³ = (4/3)πR₀³A. Both mass and volume are proportional to A, so A cancels in the density: ρ = Au/((4/3)πR₀³A) = 3u/(4πR₀³). The density is the same for every nucleus. Nucleons are packed like balls in a bag: adding nucleons adds volume in proportion.
Worked example 6Estimate the density of nuclear matter and compare it with water (1000 kg m⁻³). What would the mass of a teaspoon (5.0 cm³) of nuclear matter be?
That is about 2 × 10¹⁴ times the density of water.
Teaspoon: m = ρV = 2.3 × 10¹⁷ × 5.0 × 10⁻⁶ = 1.1 × 10¹² kg (about a billion tonnes). Only neutron stars are this dense.
Mistake Thinking a big nucleus is denser. All nuclei have the same density (about 2 × 10¹⁷ kg m⁻³); a big nucleus is simply bigger. Also: R ∝ A^⅓, not R ∝ A.
Rutherford worked out how many alpha particles should be scattered at each angle, assuming that only the electric force acts. At low energies the measurements agree perfectly. But when the alpha particles are given more and more energy, there is a point where the measurements stop agreeing.
Above an energy E₀, fewer alpha particles are scattered back than Rutherford predicts.
Why the deviations happen With more energy, the alpha particle gets closer to the nucleus. Above E₀ the distance of closest approach is about the size of the nucleus, so the alpha particle comes within the range of the strong nuclear force (about 10⁻¹⁵ m). Now there is a second force, not just the electric repulsion, so Rutherford’s formula (electric force only) no longer works. The deviations are evidence for the strong nuclear force, and for its short range: it acts only when the particles are almost touching.
Using E₀ to measure the nucleus At E₀ the alpha particle just reaches the nuclear surface, so the nuclear radius ≈ the distance of closest approach at this energy: R ≈ d = k(2e)(Ze)/E₀ (energy conservation, as before)
Worked example 7Alpha particles fired at gold (Z = 79) show deviations from Rutherford scattering when their energy is above about 26 MeV. Estimate the radius of the gold nucleus and compare it with R₀A^⅓ (A = 197).
At 26 MeV the alpha particle just reaches the nucleus: R ≈ d = k(2e)(79e)/E₀.
d = 158 × 8.99 × 10⁹ × 1.60 × 10⁻¹⁹/26 × 10⁶ = 8.7 × 10⁻¹⁵ m.
R₀A^⅓ = 7.0 × 10⁻¹⁵ m. The estimate is a little larger because at contact the centres are Rgold + Ralpha = 7.0 + 1.9 = 8.9 × 10⁻¹⁵ m apart: very good agreement.
Worked example 8The gold foil is replaced by an aluminium foil (Z = 13). The alpha particle energy is slowly increased from a low value. Will the deviations from Rutherford scattering start at a lower or a higher energy than for gold? Explain.
d = k(2e)(Ze)/Ek: for the same energy, the alpha particle gets closer to an aluminium nucleus (smaller Z, weaker repulsion).
The aluminium nucleus is also smaller, but the effect of Z is larger, so the alpha particle reaches the nuclear surface at a much lower energy.
Deviations start at a lower energy for aluminium.
“Outline why the deviation occurs” (1) · “how E₀ gives the nuclear radius” (2)✔ At closest approach the alpha particle is within the range of the strong nuclear force of the nucleus. ✔ The nuclear radius is (approximately) the distance of closest approach at E₀. ✔ Find it from energy conservation: the initial kinetic energy E₀ becomes electric potential energy, E₀ = kQq/r, so r = kQq/E₀. And the reverse: if the closest approach is much larger than the nucleus, the strong force does not act and Rutherford’s model holds.
Heat a gas at low pressure (or pass an electric current through it) and it glows. Pass that light through a diffraction grating or a prism and you do not get a rainbow: you get a few bright lines of particular wavelengths. Shine white light through a cool gas instead and you get a rainbow with dark lines, at exactly the same wavelengths.
How the two kinds of line spectrum are made.Hydrogen: the dark absorption lines are at exactly the same wavelengths as the bright emission lines.
Learn it word for word. Emission spectrum: the set of wavelengths emitted by an element (bright lines on a dark background). Absorption spectrum: a continuous spectrum with dark lines at the wavelengths absorbed by an element.
Spectra are fingerprints Every element has its own set of lines, different from every other element. So the lines tell us which elements are present: the chemical composition of a gas, a flame, or a star thousands of light years away. Helium was discovered in the spectrum of the Sun before it was found on Earth: its lines matched no known element.
Mistake Thinking spectral lines tell us about the nucleus, or about mass–energy. They come from electrons moving between energy levels, and they tell us the chemical composition.
Why only certain wavelengths? Because an atom can only have certain energies: its energy is discrete (quantised). Each allowed energy is an energy level. When an electron moves (makes a transition) from a higher level to a lower one, the atom loses exactly the difference in energy, carried away by one photon. A photon is a packet of light energy.
Photon energy:E = hf = hc/λ · h = 6.63 × 10⁻³⁴ J s · c = 3.00 × 10⁸ m s⁻¹ In a transition:Ephoton = ΔE = Eupper − Elower 1 eV = 1.60 × 10⁻¹⁹ J (the energy an electron gains through 1 V)
The energy levels of hydrogen. The energies are negative: the electron is trapped (bound). At 0 eV it is free.
For hydrogen these levels follow E = −13.6/n² eV (HL: the Bohr model explains why). This formula is only for hydrogen. Other atoms have their own levels, which a question gives you in a diagram or a table.
Learn it word for word. Ground state: the lowest energy level of the atom. Excited state: any energy level above the ground state.
Worked example 9Find the wavelength of the photon emitted when the electron in a hydrogen atom falls from n = 3 (−1.51 eV) to n = 2 (−3.40 eV). What colour is it?
ΔE = −1.51 − (−3.40) = 1.89 eV = 1.89 × 1.60 × 10⁻¹⁹ = 3.02 × 10⁻¹⁹ J.
This is the red line of hydrogen in the spectrum above.
Spectra are the evidence for discrete energy levels Each line has one wavelength, so one photon energy (E = hc/λ). If an atom could have any energy, it could emit photons of any energy, and we would see a continuous spectrum. Only certain lines appear, so only certain energy differences are possible: the energy levels are discrete.
Energy in eV is your friendhc = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ = 1.99 × 10⁻²⁵ J m. A quick check: λ (in nm) ≈ 1240/E (in eV). For 1.89 eV: 1240/1.89 ≈ 656 nm. Use it to check answers, but show the full method in the exam.
A favourite Paper 1A question gives a diagram of energy levels and some arrows, and asks which arrow gives the longest wavelength, the highest frequency, or the order of the wavelengths. One rule answers them all:
Bigger energy gap → bigger photon energy → higher frequency → SHORTER wavelength. E = hf, so f ∝ ΔE. E = hc/λ, so λ ∝ 1/ΔE. The longest arrow gives the shortest wavelength.
Six possible downward transitions between four levels. Measure each arrow: the longer it is, the shorter the wavelength.
Worked example 10Use the energy levels in the diagram (E1 = −12.0 eV, E2 = −6.5 eV, E3 = −3.6 eV, E4 = −1.8 eV). (a) Which transition gives the longest wavelength, and which the shortest? (b) Put all six in order of wavelength. (c) Which lines are visible (400 nm to 700 nm)?
Gaps: A (4→3) 1.8 eV · B (3→2) 2.9 eV · C (4→2) 4.7 eV · D (2→1) 5.5 eV · E (3→1) 8.4 eV · F (4→1) 10.2 eV.
(a) Smallest gap A → longest wavelength: A. Largest gap F → shortest wavelength: F.
(b) λ = hc/ΔE gives: A 691 nm, B 429 nm, C 264 nm, D 226 nm, E 148 nm, F 122 nm. Order (longest first): A, B, C, D, E, F.
(c) Only A (691 nm, red) and B (429 nm, violet) are visible; the rest are ultraviolet.
Worked example 11Photons emitted when an atom goes from level E1 to the ground state E0 have wavelength 600 nm, and from E2 to E0 they have wavelength 400 nm. Find the energy gap E2 − E1 in eV, and the wavelength of the photon emitted from E2 to E1.
Energies add, wavelengths do not: 400 + 600 is not the answer. (Check: 1/λ = 1/400 − 1/600 gives 1200 nm.)
How many lines?From level n (with all the levels below it) the electron can reach the ground state by many routes. The number of different lines possible is n(n − 1)/2: 4 levels give 4 × 3/2 = 6 lines (as in the diagram); 5 levels give 10 lines.
Mistake Thinking the longest arrow gives the longest wavelength. It is the other way round: the longest arrow (biggest energy) gives the shortest wavelength. Also: subtracting wavelengths instead of energies.
An electron can move up to a higher level only if the atom gains exactly the right energy. There are two ways to give it that energy, and they behave differently.
A photon must be the exact energy of a gap. An electron can give part of its energy.
Absorbing a photon: the photon energy must equal an energy gap exactly; the whole photon is absorbed or not at all. Hit by an electron: the atom can take part of the electron’s kinetic energy (exactly one gap); the electron leaves with the rest.
Learn it word for word. Ionisation energy: the minimum energy needed to remove an electron from an atom in its ground state (to n = ∞, energy 0). For hydrogen: 13.6 eV.
Worked example 12Hydrogen atoms are in the ground state. What happens if (a) a photon of 11.0 eV, (b) an electron of 11.0 eV, (c) a photon of 14.0 eV meets an atom?
(a) The gaps from n = 1 are 10.2 eV, 12.09 eV, …: none is 11.0 eV, so the photon is not absorbed (it passes through).
(b) The atom takes 10.2 eV (n = 1 → 2); the electron leaves with 0.8 eV.
(c) 14.0 eV is more than the ionisation energy (13.6 eV): the atom is ionised and the freed electron leaves with 14.0 − 13.6 = 0.4 eV of kinetic energy.
Worked example 13Find the ionisation energy of a hydrogen atom in the state n = 2, and the minimum frequency of a photon that can ionise it.
From −3.40 eV to 0: 3.40 eV.
f = E/h = 3.40 × 1.60 × 10⁻¹⁹/6.63 × 10⁻³⁴ = 8.2 × 10¹⁴ Hz.
Why are the absorption lines dark? The absorbed photons excite the atoms, and the atoms quickly fall back and emit photons of the same wavelengths. So why are the lines missing? Because the re-emitted photons go off in all directions, not along the original beam. Very few of them reach the observer, so those wavelengths are much dimmer: dark lines.
“Explain, with reference to energy levels, the formation of an absorption spectrum” (3 marks)✔ White light (a continuous spectrum) passes through a cool gas. ✔ Photons with energy equal to the difference between two energy levels are absorbed, and electrons move to higher levels (E = hf = ΔE). ✔ The photons are re-emitted in all directions, so these wavelengths are missing (dark lines) in the transmitted light.
Rutherford’s model has a serious problem. An electron in a circular orbit is accelerating (centripetal acceleration), and classical physics says that an accelerating charge radiates electromagnetic energy. It would lose energy, spiral into the nucleus in a tiny fraction of a second, and every atom would collapse. Atoms are obviously stable, so something is missing.
Left: the classical prediction. Right: Bohr’s allowed orbits, where the electron does not radiate.
Bohr’s condition (data booklet): mvr = nh/2π, n = 1, 2, 3, … mvr is the angular momentum of the electron. Only orbits in which the angular momentum is a whole-number multiple of h/2π are allowed. In these orbits the electron does not radiate.
Discrete angular momentum → discrete orbits → discrete energies Bohr’s condition allows only certain values of mvr. Combined with the electric force that keeps the electron in orbit, this allows only certain radii (r ∝ n²). Each radius has its own energy, so only certain energies are allowed: E = −13.6/n² eV. The electron radiates only when it jumps from one allowed orbit to another, and the photon carries the energy difference. This explains the line spectrum of hydrogen exactly.
Show that r ∝ n² and E = −13.6/n² eV
The electric force provides the centripetal force: ke²/r² = mv²/r, so mv² = ke²/r. (1)
Bohr’s condition: mvr = nh/2π, so v = nh/(2πmr). (2)
Put (2) into (1): m × n²h²/(4π²m²r²) = ke²/r. Cancel one r and one m: r = n² × h²/(4π²kme²), so r ∝ n².
For n = 1: r₁ = h²/(4π²kme²) = (6.63 × 10⁻³⁴)²/(4π² × 8.99 × 10⁹ × 9.11 × 10⁻³¹ × (1.60 × 10⁻¹⁹)²) = 5.3 × 10⁻¹¹ m (the Bohr radius).
Total energy = kinetic + electric potential: E = ½mv² − ke²/r = ½ke²/r − ke²/r (using (1)) = −ke²/(2r).
Put in r = n²r₁: E = −ke²/(2r₁) × 1/n² = −13.6 eV/n² (with accurate constants). The energy is negative because the electron is bound.
Important: E = −13.6/n² eV is for HYDROGEN onlyThe number 13.6 eV comes from the charge of one proton and one electron, so this formula works only for the hydrogen atom. • Other one-electron ions (He⁺, Li²⁺) still have levels that go as 1/n², but the number is different: the question gives it (for example E = −K/n², see worked example 15). • For atoms with more electrons (helium, sodium, mercury …) there is no formula: use the energy-level diagram or the values in the question.
Worked example 14For the electron in a hydrogen atom in the state n = 3, find (a) the orbit radius, (b) its speed, (c) its energy. (Bohr radius r₁ = 5.3 × 10⁻¹¹ m.)
(a) r = n²r₁ = 9 × 5.3 × 10⁻¹¹ = 4.8 × 10⁻¹⁰ m.
(b) v = nh/(2πmr) = 3 × 6.63 × 10⁻³⁴/(2π × 9.11 × 10⁻³¹ × 4.77 × 10⁻¹⁰) = 7.3 × 10⁵ m s⁻¹ (one third of the speed in n = 1, 2.2 × 10⁶ m s⁻¹: v ∝ 1/n).
(c) E = −13.6/3² = −1.51 eV.
Worked example 15In a hydrogen-like ion (one electron) the energy levels are E = −K/n². A photon of 40.8 eV is needed to move the electron from n = 1 to n = 2. Find K and the photon energy needed from n = 2 to n = 3.
ΔE(1→2) = K(1 − 1/4) = 3K/4 = 40.8 eV, so K = 54.4 eV (this is helium with one electron removed, He⁺).
ΔE(2→3) = K(1/4 − 1/9) = 54.4 × 5/36 = 7.56 eV.
Where the Bohr model works and where it failsIt works only for atoms with one electron: hydrogen, and ions such as He⁺ and Li²⁺ (an atom with all but one electron removed). It cannot explain atoms with more electrons, the different brightness of the lines, or why the condition is true. Later quantum theory (the electron as a wave) explains it, but the energy levels it predicts for hydrogen are still correct.
“Explain why Bohr introduced his condition” (3 marks)✔ Electrons in orbit accelerate, so (classically) they radiate energy. ✔ They would therefore spiral into the nucleus: atoms would be unstable. ✔ The only orbits where electrons do not radiate are those that satisfy mvr = nh/2π. ✔ So the electrons have discrete energy levels.
Mistake Writing the condition as mvr = nh (forgetting 2π), or thinking the energy of a higher level is more negative. E = −13.6/n²: n = 1 is the lowest (most negative) energy; the levels get closer together as n increases.
Writing the explanations
E.1 has short “describe”, “outline” and “explain” questions. The markschemes look for the same key words every time.
“Describe the results of the Geiger–Marsden–Rutherford experiment” (2 marks)✔ Most alpha particles passed through the foil undeflected (or with small deflections). ✔ A very small number were deflected through large angles (more than 90°, back towards the source).
“Outline the Rutherford model of the atom” (2 marks)✔ (Most of) the mass of the atom is confined within a very small volume, the nucleus. ✔ (All) the positive charge is confined within this very small volume. ✔ Electrons orbit the nucleus (the atom is mostly empty space).
“Explain why the Thomson model cannot explain the large deflections”✔ In the Thomson model the positive charge is spread through the whole atom. ✔ So the electric force on an alpha particle is always small: only small deflections. ✔ A large deflection needs a very large force: charge concentrated in a tiny nucleus (F = kQq/r² with r very small).
“Explain how line spectra give evidence for discrete energy levels” (2 marks)✔ Only certain wavelengths (frequencies) are emitted or absorbed. ✔ Each corresponds to a photon of one energy (E = hf), equal to the difference between two levels, so only certain energy differences exist: the levels are discrete.
“Explain the formation of an absorption spectrum” (3 marks)Continuous spectrum through a cool gas · photons with energy equal to a difference in energy levels absorbed, electrons excited · re-emitted in all directions → dark lines.
HL: “Explain why Bohr introduced his condition” (3 marks)accelerating electrons radiate · they would spiral into the nucleus · only orbits with mvr = nh/2π do not radiate · discrete energy levels.
HL: deviations from Rutherford scatteringAlpha particle within the range of the strong nuclear force · nuclear radius ≈ distance of closest approach at E₀ · found from energy conservation, E₀ = kQq/r.
Common mistakes
Ten traps in E.1 1. “Most alpha particles were deflected.” Most went straight through; only a very small number were deflected through large angles. 2. Saying the alpha particles hit or bounced off the nucleus. They are repelled by the electric force. 3. Mixing up Z and A: the proton number Z gives the charge; neutrons = A − Z. 4. HL: forgetting that the alpha particle’s charge is 2e, or leaving the energy in MeV in d = kq₁q₂/Ek. 5. HL: R ∝ A instead of R ∝ A^⅓, or thinking a big nucleus is denser (all nuclei have the same density). 6. HL: explaining deviations from Rutherford scattering without the strong nuclear force. 7. Longest arrow = longest wavelength. No: biggest energy gap = shortest wavelength. 8. Subtracting wavelengths instead of energies (E2 − E1 = hc/λ₂₁, not λ₂ − λ₁). 9. A photon of “nearly the right” energy is absorbed. A photon must match a gap exactly; an electron can give part of its energy. 10. Dark lines “because the atoms keep the energy”. The photons are re-emitted in all directions. And: using E = −13.6/n² eV for an atom that is not hydrogen; HL: mvr = nh instead of nh/2π.
Practise next
You have read the notes. Now practise, in this order:
Levelled practice: first a ★ at Level 2 in every topic (the goal), then Levels 3 and 4 for a 6 or 7. Hints and worked answers help you as you go.
IB-format practice (Paper 1A · Paper 2): timed, one attempt, and your report goes to your teacher. Do these when the practice goal is reached.