4 higher-level sections hidden.
Educerie · IB Diploma · Physics
Theme E Nuclear and quantum physics · E.1 Structure of the atom
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Describe the Geiger–Marsden–Rutherford experiment, its results and what each result shows about the atom | SL, HL | "Outline how the results of the experiment provide evidence for a nucleus" (3 to 4 marks), qualitative only |
| Read and write nuclear notation (A top left, Z bottom left of the symbol X), and find the numbers of protons, neutrons and nucleons | SL, HL | Paper 1A, or 1 to 2 marks inside a longer question |
| Explain how emission and absorption spectra show that atomic energy levels are discrete | SL, HL | "Explain why the spectrum consists of lines" (2 to 3 marks) |
| Describe photons being emitted and absorbed when electrons change level | SL, HL | "Describe how the line is produced" (2 marks) |
| Use E = hf with an energy level diagram to find a photon's energy, frequency or wavelength | SL, HL | "Calculate the wavelength of the photon emitted in the transition" (2 to 3 marks) |
| Explain how spectra reveal chemical composition, in a laboratory and in a star | SL, HL | Paper 1B data on spectral lines, or a 2 mark "explain" |
| Use R = R₀∛A and show that nuclear density is the same for all nuclei | HL only | "Show that the density of a nucleus is independent of A" (3 marks) |
| Calculate the distance of closest approach in a head-on collision, by energy conservation | HL only | "Determine the distance of closest approach" (2 to 3 marks) |
| Explain why scattering deviates from Rutherford's prediction at high energies | HL only | "Suggest why the number of alpha particles scattered is less than predicted" (2 marks) |
| Use the Bohr model: E = −13.6/n² eV and mvr = nh/2π | HL only | "Calculate the wavelength of the photon emitted when…" (3 marks); "Outline how the Bohr model accounts for…" (2 marks) |
Before you start
You need the electric force between charges (like charges repel, and the force grows as they get closer), kinetic energy from A.3, and the wave equation c = fλ from C.2. HL students also need electric potential energy, Ep = kq₁q₂/r, from D.2, and centripetal force from A.2. Powers of ten matter here: an atom is about 10⁻¹⁰ m across and a nucleus about 10⁻¹⁵ m to 10⁻¹⁴ m.
1The idea in one paragraph
An atom is almost entirely empty. Nearly all its mass and all its positive charge sit in a nucleus between a ten-thousandth and a hundred-thousandth of the atom's width, and the electrons occupy the space around it. We know about the nucleus because alpha particles fired at thin gold foil occasionally bounced back, which only a tiny, heavy, charged centre can do. We know about the electrons' arrangement from light. An electron in an atom can only have certain energies, called energy levels; when it drops from one to a lower one it gives out a photon whose energy equals the gap, E = hf. Because the gaps are fixed, each element gives out and takes in light at its own set of wavelengths, which is why a spectrum shows lines, and why a spectrum can tell you what a gas, or a star, is made of.
2The Geiger–Marsden–Rutherford experiment
In 1909 Hans Geiger and Ernest Marsden, working in Ernest Rutherford's laboratory, fired alpha particles at a very thin sheet of gold. An alpha particle is a helium nucleus: positively charged (+2e) and about 7300 times as massive as an electron. At the time the accepted picture of the atom was J. J. Thomson's plum pudding model: a sphere of positive charge spread evenly through the whole atom, with electrons dotted through it. Figure 1 shows the apparatus.
Each part is there for a reason, and "explain the purpose of…" questions ask exactly this.
- A radioactive source in a lead block with a narrow channel gives a narrow beam of alpha particles all moving the same way.
- The chamber is evacuated. Alpha particles are stopped by a few centimetres of air, and collisions with air molecules would deflect them too.
- The gold foil is extremely thin. Gold can be beaten to less than a micrometre, so an alpha particle meets few enough atoms that a big deflection is the work of a single encounter.
- A zinc sulfide screen gives a tiny flash of light, a scintillation, where an alpha particle hits it. The screen and its microscope could be moved round the foil to count the flashes at each angle.
What they saw, and what each result means. Learn these as three pairs.
| Observation | What it shows |
|---|---|
| Almost all alpha particles passed straight through, or were deflected by a degree or two | The atom is mostly empty space |
| A small fraction were deflected through large angles | Somewhere there is a very strong electric field, so the positive charge must be concentrated in a tiny volume |
| A very few, far fewer than one in a thousand, bounced back by more than 90° | The concentrated charge must also carry most of the atom's mass; a light object cannot turn a heavy one round |
Figure 2 compares what the plum pudding model predicts with what Rutherford's model predicts.
In the plum pudding model the positive charge is spread over a sphere about 10⁻¹⁰ m across, so its field is never strong anywhere, and the electrons are far too light to deflect an alpha particle. Every alpha particle should go almost straight through. The backward bounces were impossible in that model. Rutherford said it was as if a shell fired at tissue paper had come back and hit you.
In 1911 Rutherford proposed the nuclear model: all the positive charge and nearly all the mass in a tiny central nucleus, with electrons far outside it. An alpha particle that passes far from every nucleus feels almost nothing. One that passes close is pushed away along a curved path. One heading straight at a nucleus slows, stops, and is driven back. The paths bend away from the nucleus because both are positive. Rutherford's calculations of how many alpha particles should arrive at each angle matched Geiger and Marsden's counts, which is what turned an idea into an accepted model.
How small? The results put the nucleus at around 10⁻¹⁴ m or less, against 10⁻¹⁰ m for the atom. If the nucleus were a bead 1 cm across, the atom would be hundreds of metres wide, and almost all of that is empty.
3Inside the nucleus: nuclear notation
A nucleus is made of protons and neutrons, together called nucleons. Every nucleus is written in nuclear notation:
Nuclear notation: the nucleon number A is written top left of the chemical symbol X, and the proton number Z bottom left, as in ⁵⁶₂₆Fe
- The proton number Z is the number of protons. It fixes the charge of the nucleus, +Ze, and it fixes the element.
- The nucleon number A is the total number of protons plus neutrons. It fixes the mass of the nucleus, roughly.
- The number of neutrons is N = A − Z. It is not written in the notation; you work it out.
Figure 3 labels one.
An iron nucleus ⁵⁶₂₆Fe has 26 protons, 56 nucleons and 56 − 26 = 30 neutrons. A neutral iron atom also has 26 electrons, one for each proton. An alpha particle is ⁴₂He: 2 protons and 2 neutrons. You do not need to recall chemical symbols; the question gives them.
Two nuclei with the same Z and different A are isotopes of the same element: same number of protons, different number of neutrons. Chlorine-35 and chlorine-37 (³⁵₁₇Cl and ³⁷₁₇Cl) are both chlorine, because both have 17 protons. Isotopes are central to E.3.
4Energy levels, photons and E = hf
Now the electrons. The evidence here comes from light, and it says something strange: an electron bound in an atom cannot have any energy it likes. It can only have one of a fixed set of values, the atom's energy levels. The energy is discrete, or quantized.
An energy level diagram, Figure 4, draws the allowed energies as horizontal lines. Three conventions matter.
- Zero is the top. An energy of 0 means the electron has just been freed from the atom. A bound electron has less energy than that, so every level is negative.
- The lowest level is the ground state. The atom sits there unless something gives it energy. Levels above it are excited states.
- Energies are usually in electronvolts. One electronvolt, 1 eV, is the energy an electron gains when it is accelerated through a potential difference of 1 V: 1 eV = 1.60 × 10⁻¹⁹ J.
When an electron drops from a higher level to a lower one, the atom emits one photon, a single packet of electromagnetic energy. Energy is conserved, so the photon carries away exactly the difference between the two levels. A photon's energy is fixed by its frequency:
E = hf, where h = 6.63 × 10⁻³⁴ J s is the Planck constant
So a transition between two levels gives a photon of one exact frequency, and so one exact wavelength, λ = c/f. A large drop gives a high-energy, high-frequency, short-wavelength photon; a small drop gives a low-frequency, long-wavelength one.
Worked example. In hydrogen the levels n = 3 and n = 2 are at −1.51 eV and −3.40 eV. Find the wavelength of the photon emitted when an electron falls from n = 3 to n = 2.
That is red light, and it is the bright red line of a hydrogen lamp. A shortcut that saves time: λ = hc/ΔE, and with ΔE in eV, hc = 1.243 × 10⁻⁶ eV m, so λ = 1.243 × 10⁻⁶ ÷ 1.89 = 6.58 × 10⁻⁷ m. Show which method you use.
Absorption is the same story in reverse. An electron in a lower level can take in a photon and jump to a higher level, but only if the photon's energy exactly matches a gap. A photon with a little too much or too little energy is not absorbed at all; it passes through. (A photon with more energy than it takes to free the electron is the exception: it can ionise the atom, and the freed electron keeps the extra as kinetic energy.)
Counting lines. From level n an electron can reach the ground state by many routes, one jump or several. Every pair of levels gives a possible line, so an atom excited to its fourth level can give up to 4 × 3 ÷ 2 = 6 different wavelengths. Draw every arrow downwards and count them rather than trusting a formula in the exam.
5Emission and absorption spectra
A spectrum is light spread out by wavelength, using a prism or a diffraction grating (C.3). Figure 5 shows the three kinds you must recognise.
A continuous spectrum contains every wavelength over a range. A hot, dense object such as a lamp filament or the body of a star gives one, because its atoms are so crowded that their levels merge.
An emission line spectrum is a set of bright lines on a dark background. It comes from a hot gas at low pressure, such as a discharge tube, where collisions keep lifting electrons to excited levels and each electron falls back emitting photons. Each line is one transition. The lines are sharp because the levels are sharp: this is the evidence that the energy levels are discrete. If an electron could have any energy, the gas would give a continuous smear.
An absorption line spectrum is a continuous spectrum crossed by dark lines. It appears when white light passes through a cooler gas. The gas absorbs exactly the photons whose energies match its gaps, so those wavelengths are missing. The absorbed energy is re-emitted soon afterwards, but in all directions and sometimes as several smaller photons, so little of it continues in the original direction, and the line looks dark.
Put the two line spectra of one element side by side and the dark lines sit at exactly the same wavelengths as the bright ones, because both come from the same gaps. An absorption spectrum from a cool gas usually shows fewer lines than the emission spectrum, because at low temperature almost every atom is in its ground state, so only jumps from the ground state are seen.
6What spectra tell us, and how the model of the atom was built
Chemical composition. Every element has its own set of energy levels, so every element has its own pattern of lines, as individual as a fingerprint. Measure the wavelengths in a spectrum, match them against laboratory spectra, and you know which elements are present. This works on something you can never touch. The dark lines in sunlight come from gases in the Sun's outer layers absorbing the continuous spectrum from below. In 1868 astronomers found a line in sunlight that matched no known element; the element was named helium, after the Greek for the Sun, and was found on Earth only decades later. The same lines, shifted a little in wavelength by the Doppler effect (C.5), also tell astronomers how fast a star or galaxy moves towards or away from us.
Evidence and models. The picture you now hold was built in steps, and each step was forced by an experiment the old model could not explain.
- Thomson's discovery of the electron (1897) showed that atoms have parts, which gave the plum pudding model.
- Geiger and Marsden's backward-scattered alpha particles (1909) could not happen in a plum pudding, which gave Rutherford's nuclear model (1911).
- The nuclear model could not explain line spectra, or why orbiting electrons did not spiral into the nucleus. Bohr's model of hydrogen (1913) added fixed energy levels (section 10).
- Chadwick's discovery of the neutron (1932) explained why nuclei are heavier than their protons alone.
- The modern quantum model replaced definite orbits with regions where the electron is likely to be found.
The guide asks in what ways older models are still valid. The small, dense nucleus of 1911 is exactly right. Energy levels are exactly right, even though Bohr's circular orbits are not. You use the older models every day because they give the right answer for the question being asked, and a model is judged by that, not by being the latest.
7HLThe size and density of the nucleus
SL students can skip to section 11.
Scattering experiments at many energies and with many targets give the radius R of a nucleus. The results follow one simple rule:
R = R₀∛A, where R₀ = 1.20 × 10⁻¹⁵ m (the data booklet writes the cube root as the power 1/3)
R₀ is the radius of a single nucleon, near enough. The one-third power says that R³ is proportional to A, and R³ sets the volume. Figure 6 shows the rule plotted two ways. Panel (b) is the version to reach for in a Paper 1B question, because plotting R against ∛A turns a curve into a straight line through the origin, with gradient R₀.
| Nucleus | A | ∛A | R = 1.20 × 10⁻¹⁵ × ∛A |
|---|---|---|---|
| carbon-12 | 12 | 2.29 | 2.75 × 10⁻¹⁵ m |
| iron-56 | 56 | 3.83 | 4.59 × 10⁻¹⁵ m |
| gold-197 | 197 | 5.82 | 6.98 × 10⁻¹⁵ m |
Gold has more than sixteen times as many nucleons as carbon but is only about two and a half times as wide.
The implication for density. The mass of a nucleus is about A × u, where u is the unified atomic mass unit, 1.661 × 10⁻²⁷ kg. Its volume is (4/3)πR³.
The A cancels, so every nucleus has the same density, about 2 × 10¹⁷ kg m⁻³, some 10¹⁴ times the density of water. That tells you how nucleons pack. They behave like marbles in a bag: adding more makes the bag bigger but never squeezes the marbles closer. Nucleons keep a fixed distance from their neighbours, which is the first hint that the force holding them is short-range and repels at very close range (E.3).
8HLThe distance of closest approach
When an alpha particle heads straight at a nucleus, it slows as the electric repulsion grows, stops for an instant, and is pushed back the way it came. The point where it stops is the distance of closest approach, d. At that instant all of its kinetic energy has become electric potential energy. Figure 7 shows this in two ways.
Energy is conserved, and the nucleus is so much heavier that you can treat it as staying still:
Ek = kq₁q₂/d, so d = kq₁q₂/Ek, with q₁ = 2e for the alpha particle and q₂ = Ze for the nucleus
The guide keeps the energies low enough that only the electric force acts, so this is the whole calculation.
Worked example. A 5.0 MeV alpha particle is fired head-on at a gold nucleus (Z = 79). Find the distance of closest approach.
Compare with section 7: the gold nucleus has a radius of 7.0 × 10⁻¹⁵ m. The alpha particle stops about six nuclear radii away, so d is an upper limit on the size of the nucleus, not a measurement of it. That is exactly what Rutherford could say: the nucleus is no bigger than this.
Two checks. d gets smaller as Ek gets bigger (d ∝ 1/Ek): double the energy and the alpha particle gets twice as close. And d is larger for a nucleus with more protons, because the repulsion is stronger.
9HLWhere Rutherford's model stops working
Rutherford's calculation assumed that the only force between the alpha particle and the nucleus is the electric force between two point charges. For the alpha particles he had, with energies of a few MeV from natural sources, that was true, because they never got close enough to touch the nucleus.
Accelerators give alpha particles far more energy. As Ek rises, d shrinks, and eventually the alpha particle reaches the surface of the nucleus. Two things then change. The strong nuclear force, which acts only over about 10⁻¹⁵ m, starts to pull on it; and the alpha particle can be absorbed and cause a nuclear reaction instead of bouncing off. Either way, fewer alpha particles arrive at large angles than Rutherford's formula predicts. Figure 8 shows the shape of the result.
The energy where the deviation begins measures the nucleus. The alpha particle's centre reaches the surface when d equals the sum of the two radii. For gold, RAu + Rα = 1.20 × 10⁻¹⁵ × (∛197 + ∛4) = 8.9 × 10⁻¹⁵ m, which needs Ek = kq₁q₂/d ≈ 26 MeV. That is our estimate from the model, well above the energy of any natural alpha source, which is why Geiger and Marsden saw perfect agreement. Deviations like this are how the radius rule of section 7 was measured, and they were the first direct sign of a force other than electricity inside the nucleus.
10HLThe Bohr model of hydrogen
The nuclear model has a problem. An electron orbiting a nucleus is accelerating, and an accelerating charge should radiate energy and spiral inwards in a tiny fraction of a second. Atoms do not collapse, and they give line spectra, not a smear. In 1913 Niels Bohr kept the orbiting electron and added rules that forbid the collapse.
- The electron moves in a circular orbit, held by the electric attraction of the proton.
- Only orbits whose angular momentum is a whole number of units of h/2π are allowed:
mvr = nh/2π, where n = 1, 2, 3, … is the principal quantum number
- In an allowed orbit the electron does not radiate. It emits or absorbs a photon only when it jumps between orbits, with hf equal to the energy difference.
Rule 2 is the key. Once angular momentum is restricted to whole-number steps, the radius and the energy can take only certain values too. You do not have to derive the result, but seeing where it comes from once makes it believable. Combine the centripetal force with rule 2:
Because r grows as n², the energy falls off as 1/n², and putting in the constants gives the data booklet result:
E = −13.6/n² eV
So E₁ = −13.6 eV, E₂ = −3.40 eV, E₃ = −1.51 eV, and so on up to 0 at n = ∞. These are the levels of Figure 4, now predicted rather than measured. Figure 9 draws the orbits and the levels together.
Worked example. Using the Bohr model, find the wavelength of the photon emitted when a hydrogen electron falls from n = 4 to n = 2.
That is the blue-green line of hydrogen, measured at 486 nm, so the model is right to within the rounding of the constants. The ionisation energy of hydrogen, the energy to free the electron from the ground state, is 0 − (−13.6) = 13.6 eV, which also matches experiment.
Where the Bohr model fails. The guide asks this as a nature of science question. The model works for hydrogen and for ions with a single electron (He⁺, Li²⁺). It fails for any atom with two or more electrons, because it ignores the forces between electrons. It cannot predict how bright each line is, or why some lines split into close pairs. And its central picture, an electron as a particle on a definite circular track, is wrong. E.2 gives the reason: electrons have a wavelength. If the orbit must fit a whole number of electron wavelengths, 2πr = nλ with λ = h/mv, you get mvr = nh/2π straight back. Bohr's rule turns out to be a wave condition in disguise, which is why his energies survive even though his orbits do not.
11Where marks are lost
- Saying "most alpha particles bounced back". Almost all went straight through. Only a very small fraction were deflected by large angles, and that rarity is the evidence.
- Pairing the wrong observation with the wrong conclusion. Straight through → mostly empty space. Large deflection → small, concentrated positive charge. Bouncing back → that charge carries most of the mass. Keep the three pairs separate.
- Using A for the number of neutrons. A counts protons and neutrons together. Neutrons are A − Z.
- Forgetting to convert eV to joules. E = hf needs joules when h is in J s. 1.89 eV is 3.02 × 10⁻¹⁹ J.
- Taking ΔE from the level value, not the gap. The photon's energy is the difference between two levels, not the value of the level it lands on.
- Saying photons of any energy above a gap are absorbed. For a jump between two bound levels the photon must match the gap exactly. Only photons that ionise the atom can carry more.
- Calling the distance of closest approach the radius of the nucleus (HL). It is an upper limit. The alpha particle turns round well outside the nucleus unless its energy is very high.
- Treating Bohr's orbits as correct (HL). Bohr's energies for hydrogen are right; his circular orbits are not, and the model fails for atoms with more than one electron.
12Draw it right
- Energy level diagrams: the zero at the top, labelled 0 or "ionisation"; every level negative and labelled with its value and unit (eV); the ground state at the bottom.
- Emission is an arrow pointing down between two levels; absorption is an arrow pointing up. Arrow tips must touch the levels, not float between them.
- Draw one arrow per possible transition when asked "how many lines". Count the arrows, not the levels.
- Spectra: emission as bright lines on dark; absorption as dark lines on a continuous background at the same wavelengths; a wavelength scale with units.
- Alpha paths near a nucleus: smooth curves that bend away from the nucleus, with the closest ones bending most, and a head-on path that returns along its own line. Paths never cross into the nucleus.
- HL, R against A: a curve rising ever more slowly; R against ∛A a straight line through the origin with gradient R₀.
- HL, closest approach: mark d on the diagram from the centre of the nucleus to the turning point, not to its surface.
13Try it
Marks in brackets. h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, k = 8.99 × 10⁹ N m² C⁻². Answers and marker's notes follow.
Q1. Outline the main observations of the Geiger–Marsden–Rutherford experiment and what each tells us about the structure of the atom. 4 marks
Q2. A nucleus is written ⁶⁵₂₉Cu. 3 marks
(a) State the number of protons, neutrons and nucleons in the nucleus. 2 marks
(b) Another isotope of copper has 34 neutrons. Write its nuclear notation. 1 mark
Q3. An atom of an invented element has four energy levels at −9.0 eV, −4.2 eV, −2.6 eV and −1.2 eV. A tube of the gas is heated so that electrons reach every level. 7 marks
(a) State the number of different wavelengths that the gas can emit. 1 mark
(b) Calculate the longest wavelength emitted. 3 marks
(c) White light is passed through the same gas at room temperature. Explain why the absorption spectrum shows only three dark lines. 3 marks
Q4. (Data-based, Paper 1B style. The star is invented; the laboratory wavelengths are real.) Absorption lines in the spectrum of a star are measured at 434.0 nm, 447.1 nm, 486.1 nm, 587.6 nm and 656.3 nm. Laboratory spectra give these lines: hydrogen 434.0, 486.1, 656.3 nm; helium 447.1, 501.6, 587.6 nm; sodium 589.0, 589.6 nm. 4 marks
(a) Identify the elements present in the star's outer layers, with a reason. 2 marks
(b) Explain why these wavelengths appear as dark lines. 2 marks
Q5 (HL). An alpha particle with kinetic energy 5.5 MeV moves head-on towards a stationary gold-197 nucleus (Z = 79). R₀ = 1.20 × 10⁻¹⁵ m. 5 marks
(a) Calculate the distance of closest approach. 3 marks
(b) Calculate the radius of the gold nucleus and comment on your two answers. 2 marks
Q6 (HL). 6 marks
(a) Using the Bohr model, calculate the wavelength of the photon emitted when an electron in hydrogen falls from n = 5 to n = 2. 3 marks
(b) Outline how the condition mvr = nh/2π leads to discrete energy levels. 2 marks
(c) State one limitation of the Bohr model. 1 mark
14In one breath
Geiger and Marsden fired alpha particles at thin gold foil in a vacuum: nearly all went straight through, so the atom is mostly empty; a few were deflected through large angles, so the positive charge is concentrated; a very few came back, so that concentrated charge carries most of the mass. That is Rutherford's nucleus, written with A top left and Z bottom left of the symbol, with Z protons, A nucleons and A − Z neutrons. Electrons in an atom can only have certain energies, all negative, with zero meaning free; a fall between levels emits one photon with hf equal to the gap, and absorbing a photon of exactly that energy lifts the electron back up. So hot low-pressure gases give bright lines, cool gases in front of white light give dark lines at the same wavelengths, the sharpness of the lines proves the levels are discrete, and the pattern identifies the element, even in a star. HL: R = R₀∛A, so every nucleus has the same density, about 2 × 10¹⁷ kg m⁻³; a head-on alpha particle stops where Ek = kq₁q₂/d, which is an upper limit on the nuclear radius; at high energies scattering falls below Rutherford's prediction because the alpha particle reaches the nucleus and feels the strong force; Bohr's mvr = nh/2π makes hydrogen's orbits and energies discrete, E = −13.6/n² eV, right for hydrogen but not for bigger atoms.
Answers
Q1. Most alpha particles passed straight through the foil with little or no deflection, so the atom is mostly empty space. A small number were deflected through large angles, so the positive charge is concentrated in a very small region, where the electric field is very strong. A very small fraction were deflected through more than 90°, back towards the source, so the concentrated region also contains most of the atom's mass. 1 for each observation paired with its correct conclusion, up to 3; 1 for stating that the nucleus is small, positive and massive, or for contrasting with the plum pudding model. "Most bounced back" scores 0 for that observation.
Q2. (a) Protons = 29; nucleons = 65; neutrons = 65 − 29 = 36. (b) A = 29 + 34 = 63, so ⁶³₂₉Cu. (a) A1 for protons and nucleons, A1 for neutrons = 36. (b) A1 for ⁶³₂₉Cu; the proton number must stay 29.
Q3. (a) Four levels give 4 × 3 ÷ 2 = 6 wavelengths. (b) The longest wavelength comes from the smallest gap, between −2.6 eV and −1.2 eV: ΔE = 1.4 eV = 1.4 × 1.60 × 10⁻¹⁹ = 2.24 × 10⁻¹⁹ J. λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 2.24 × 10⁻¹⁹ = 8.9 × 10⁻⁷ m (888 nm, infrared). (c) At room temperature almost all atoms are in the ground state, −9.0 eV. Absorption can therefore only lift electrons from the ground state, to one of the three higher levels. So only three photon energies are absorbed (4.8 eV, 6.4 eV and 7.8 eV), giving three dark lines. (a) A1. (b) M1 for choosing the smallest gap, M1 for converting to joules and using λ = hc/ΔE, A1 for 8.9 × 10⁻⁷ m. (c) R1 for atoms in the ground state at room temperature, R1 for absorption only from the ground state, A1 for three transitions and so three lines. Using the largest gap gives 159 nm and scores M0 then M1 A0.
Q4. (a) Hydrogen and helium. All three hydrogen lines (434.0, 486.1, 656.3 nm) appear, and two helium lines (447.1 and 587.6 nm) appear; no line at 589.0 or 589.6 nm, so there is no evidence of sodium. (b) Cooler gas in the star's outer layers absorbs photons from the continuous spectrum below whose energy exactly equals the gap between two of its energy levels. The absorbed energy is re-emitted in all directions, so fewer photons of those wavelengths reach us, and the lines look dark. (a) A1 for hydrogen and helium, R1 for matching the wavelengths to the laboratory lines. (b) R1 for absorption of photons with energy equal to an energy level difference, R1 for re-emission in random directions. Saying sodium is present because 587.6 nm is "close to" 589.0 nm scores 0 for (a)'s reasoning mark: lines must match.
Q5 (HL). (a) Ek = 5.5 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.8 × 10⁻¹³ J. d = kq₁q₂/Ek = 8.99 × 10⁹ × (2 × 1.60 × 10⁻¹⁹)(79 × 1.60 × 10⁻¹⁹) ÷ 8.8 × 10⁻¹³ = 4.1 × 10⁻¹⁴ m. (b) R = 1.20 × 10⁻¹⁵ × ∛197 = 7.0 × 10⁻¹⁵ m. The alpha particle stops about six times further out than the nuclear radius, so it never reaches the nucleus: the closest approach is only an upper limit on its size, and only the electric force acts. (a) M1 for converting MeV to J, M1 for charges 2e and 79e in kq₁q₂/d, A1 for 4.1 × 10⁻¹⁴ m. (b) A1 for 7.0 × 10⁻¹⁵ m, R1 for a comment that d > R, so d is an upper limit or the alpha particle does not reach the nucleus. Using q = e for the alpha particle gives 2.1 × 10⁻¹⁴ m and loses the second M mark.
Q6 (HL). (a) E₅ = −13.6 ÷ 25 = −0.544 eV; E₂ = −3.40 eV; ΔE = 2.856 eV = 4.57 × 10⁻¹⁹ J. λ = hc/ΔE = 1.989 × 10⁻²⁵ ÷ 4.57 × 10⁻¹⁹ = 4.35 × 10⁻⁷ m (435 nm). (b) The condition allows angular momentum only in whole-number steps of h/2π. Combined with the electric force providing the centripetal force, this allows only certain orbit radii (r ∝ n²), and each radius has one total energy, so the energy can only take discrete values (E ∝ −1/n²). (c) Any one of: it works only for one-electron atoms or ions; it cannot predict the relative intensities of lines; it treats the electron as a particle in a definite orbit, which is inconsistent with its wave nature. (a) M1 for both energies from −13.6/n², M1 for ΔE converted and λ = hc/ΔE, A1 for 4.35 × 10⁻⁷ m. (b) R1 for quantized angular momentum giving only certain radii, R1 for each radius having one energy, so energies are discrete. (c) A1. "It is old" scores 0.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section E.1 Structure of the atom. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.