E.1 Structure of the atom flashcards: IB Physics SL and HL

All 75 E.1 Structure of the atom cards

1. DefinitionSL & HL

   State the observations of the Geiger–Marsden–Rutherford alpha-scattering experiment and the conclusion drawn from each.

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   A beam of alpha particles from a radioactive source in an evacuated chamber strikes a very thin gold foil; scattered alphas are counted by a movable zinc sulfide scintillation screen. Observations and conclusions: (1) the vast majority pass through with little or no deflection → the atom is mostly empty space; (2) a small fraction (about 1 in 8000) is deflected through more than 90°, a few almost straight back → the positive charge and nearly all the mass are concentrated in a tiny, dense core, the nucleus. Exam tip: mark schemes require observation AND matching conclusion — an answer that only says "most went straight through" gains one mark of two.

2. DefinitionSL & HL

   Define nucleon number (mass number), A.

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   The nucleon number A is the total number of nucleons — protons plus neutrons — in a nucleus. It is a dimensionless integer and it is the superscript in the notation ᴬ₍Z₎X. Exam tip: call them "nucleons", not "particles" — electrons are particles in the atom but are not counted. A is NOT the mass in kilograms, although the nuclear mass is approximately A × u with u = 1.66 × 10⁻²⁷ kg. A is conserved in every nuclear reaction and decay, which is the standard checking tool for balancing equations.

3. DefinitionSL & HL

   Define proton number (atomic number), Z.

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   The proton number Z is the number of protons in the nucleus. It is a dimensionless integer, written as the subscript in ᴬ₍Z₎X, and it fixes the nuclear charge as +Ze, where e = 1.60 × 10⁻¹⁹ C. Exam tip: Z determines the chemical element — change Z and you have a different element, which is why beta decay transmutes an element while gamma emission does not. In a neutral atom Z also equals the number of orbital electrons, but the definition itself must refer to protons in the nucleus; "number of electrons" is not accepted. Z is conserved as charge in nuclear equations.

4. DefinitionSL & HL

   Define neutron number N and explain the nuclide notation ᴬ₍Z₎X.

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   The neutron number N is the number of neutrons in a nucleus, a dimensionless integer, with N = A − Z. In the notation ᴬ₍Z₎X, X is the chemical symbol, the superscript A is the nucleon number and the subscript Z is the proton number; for example ²³⁵₍₉₂₎U has 92 protons, 143 neutrons and 235 nucleons. Exam tip: Z is strictly redundant because the symbol X already fixes it, but IB expects it written in decay equations so that charge balance can be checked. A nuclide means a specific nuclear species defined by a particular pair (Z, N).

5. DefinitionSL & HL

   Define isotopes.

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   Isotopes are nuclei (or atoms) of the same element that have the same number of protons, Z, but different numbers of neutrons, N, and therefore different nucleon numbers A. Examples: ¹H, ²H and ³H; ¹²C and ¹⁴C. Exam tip: the mark requires BOTH halves — "same proton number, different neutron number" — and answers saying "different mass" alone score zero. Isotopes of an element have identical chemical behaviour and identical line spectra to a very good approximation, because these depend on the electron configuration set by Z, but they differ in nuclear stability, mass and hence in diffusion or spectrometer deflection.

6. DefinitionSL & HL

   Define an emission spectrum and state what it demonstrates about atomic energy.

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   An emission spectrum is the set of discrete wavelengths (bright coloured lines on a dark background) emitted by an excited low-pressure gas as its electrons make transitions from higher to lower energy levels, each line corresponding to one transition with photon energy hf = E_higher − E_lower. Exam tip: the required conclusion is that the spectrum is DISCRETE, so the atom can only emit certain photon energies, so the electron energies inside the atom are quantised — this is the evidence for discrete energy levels. Every element has its own unique line pattern, so emission spectra are used to identify elements in stars and in flame tests.

7. DefinitionSL & HL

   Define an absorption spectrum and explain how it is produced.

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   An absorption spectrum is a continuous spectrum crossed by dark lines, produced when white light passes through a cool gas: photons whose energies exactly match a difference between two energy levels of the atoms are absorbed, exciting electrons, and are then re-emitted in all directions rather than in the original beam. Exam tip: the dark lines occur at exactly the same wavelengths as the bright lines of that element's emission spectrum — stating this is a standard mark. Do not say the photons are "destroyed" or that the gas absorbs "all" the light; only the matching energies are removed, and the re-emission is isotropic, which is why the beam appears depleted.

8. DefinitionSL & HL

   Distinguish between a continuous spectrum and a line spectrum.

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   A continuous spectrum contains all wavelengths over a range with no gaps and is emitted by hot dense solids, liquids or high-pressure gases (a black body); a line spectrum contains only certain discrete wavelengths and is emitted by a low-pressure gas of isolated atoms. Exam tip: the physical reason is the key mark — in a dense body atoms interact strongly so the energy levels smear into continuous bands, while isolated atoms retain sharp discrete levels. A star gives a continuous spectrum from its photosphere with absorption lines from its cooler outer atmosphere, which is how stellar composition is deduced.

9. DefinitionSL & HL

   Define an atomic energy level and state what is meant by quantisation of energy.

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   An energy level is one of the discrete allowed values of total energy that an electron bound in an atom may have; the electron cannot possess any energy between these values. Quantisation means the energy can take only certain separated values rather than a continuous range. Levels are quoted in electronvolts and are NEGATIVE for bound electrons, with the zero taken at infinite separation. Exam tip: the experimental evidence demanded by the mark scheme is the existence of discrete line spectra (and, at HL, the Franck–Hertz style argument). Saying "electrons orbit in shells" is a model, not the definition — define the level as an allowed energy.

10. DefinitionSL & HL

    Define the ground state and an excited state of an atom.

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    The ground state is the lowest energy level available to the electron, the state the atom occupies normally; for hydrogen it is n = 1 with E₁ = −13.6 eV. An excited state is any allowed level of higher energy (n ≥ 2), reached by absorbing a photon of exactly the right energy or by collision with a free electron. Exam tip: excited states are unstable, with typical lifetimes ~10⁻⁸ s, after which the electron falls back and emits one or more photons — the multi-step cascade is why a single excitation energy can produce several spectral lines. Energies are less negative for higher n, so "higher" means closer to zero.

11. DefinitionSL & HL

    Define ionisation and the ionisation energy of an atom.

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    Ionisation is the removal of an electron from an atom, leaving a positive ion; the ionisation energy is the minimum energy needed to remove the electron completely from a given level to infinity, where its total energy is zero and it is free with zero kinetic energy. For hydrogen in the ground state it is 13.6 eV = 2.18 × 10⁻¹⁸ J. Exam tip: it equals the MAGNITUDE of the level energy, E_ionisation = |E_n| = 13.6/n² eV, so from n = 2 only 3.40 eV is required. Any photon of energy greater than the ionisation energy can ionise, with the excess appearing as kinetic energy of the freed electron.

12. DefinitionSL & HL

    Define the electronvolt and state its relation to the joule.

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    One electronvolt is the kinetic energy gained by an electron (or by any particle of charge magnitude e) when it is accelerated through a potential difference of one volt: 1 eV = e × 1 V = 1.60 × 10⁻¹⁹ J. It is a unit of ENERGY, not of potential difference or of charge. Exam tip: multiply by 1.60 × 10⁻¹⁹ to convert eV → J and divide to convert J → eV; the commonest lost mark in E.1 is substituting a level energy in eV directly into E = hf while h is in J s. Useful multiples: 1 keV = 10³ eV, 1 MeV = 10⁶ eV.

13. DefinitionSL & HL

    Define a photon.

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    A photon is a quantum (discrete packet) of electromagnetic radiation, of energy E = hf, that carries zero rest mass and travels at c = 3.00 × 10⁸ m s⁻¹ in a vacuum. Its energy is fixed entirely by the frequency of the radiation, not by its intensity. Exam tip: in an atomic transition exactly ONE photon is emitted per electron jump, and its energy exactly equals the level difference — this one-to-one correspondence is the mark-earning statement. Increasing the brightness of a lamp increases the NUMBER of photons per second, not the energy of each; this distinction is examined repeatedly in both E.1 and E.2.

14. EquationSL & HLData booklet: Yes

    State the equation for the energy of a photon in terms of frequency and define every symbol.

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    E = hf. E is photon energy in joules (J); h is the Planck constant, 6.63 × 10⁻³⁴ J s; f is the frequency of the radiation in hertz (Hz = s⁻¹). Data-booklet status: printed. Valid for electromagnetic radiation of any frequency in any medium (f is unchanged on refraction, unlike λ). Common misuse: substituting a wavelength for f, or leaving E in eV while h is in J s. Sanity check: green light of f = 6.0 × 10¹⁴ Hz gives E = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 4.0 × 10⁻¹⁹ J ≈ 2.5 eV, the right order for visible photons.

15. EquationSL & HLData booklet: Yes

    State the equation for photon energy in terms of wavelength and show how it follows from E = hf.

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    E = hc/λ, obtained by substituting f = c/λ into E = hf. E is in joules (J); h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹ is the speed of light in a vacuum; λ is the vacuum wavelength in metres (m). Data-booklet status: printed (together with c = fλ). Rearranged forms you must know: λ = hc/E and f = E/h. Common misuse: entering λ in nanometres — 500 nm must become 5.00 × 10⁻⁷ m. Sanity check: λ = 500 nm gives E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(5.00 × 10⁻⁷) = 3.98 × 10⁻¹⁹ J = 2.49 eV.

16. EquationSL & HLData booklet: Yes

    State the equation linking an energy-level transition to the emitted photon and define the symbols.

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    ΔE = hf = hc/λ, where ΔE = E_initial − E_final is the (positive) difference between the two energy levels in joules, h = 6.63 × 10⁻³⁴ J s, f is the emitted frequency in Hz and λ the emitted wavelength in m. Data-booklet status: E = hf and E = hc/λ are printed; the identification of ΔE with the level difference is memorise. For absorption the same equation applies with ΔE = E_final − E_initial. Common misuse: using the level energy E_n itself instead of the DIFFERENCE — a transition from −3.40 eV to −13.6 eV emits 10.2 eV, not 13.6 eV. Always take the modulus so f is positive.

17. EquationSL & HLData booklet: Yes

    State the equation for the energy levels of atomic hydrogen and define its symbols.

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    E_n = −13.6/n² eV, where E_n is the total energy of the electron in level n, measured in electronvolts, and n = 1, 2, 3, … is the principal quantum number. Data-booklet status: printed. The minus sign shows the electron is BOUND; the zero of energy is the electron at rest at infinity. Values: E₁ = −13.6 eV, E₂ = −3.40 eV, E₃ = −1.51 eV, E₄ = −0.850 eV, E_∞ = 0. Valid only for hydrogen (and hydrogen-like one-electron ions, where it becomes −13.6Z²/n² eV). Common misuse: dropping the minus sign, which reverses every transition. Levels crowd together as n grows, so the series converges to a limit.

18. EquationSL & HLData booklet: No – memorise

    State the conversion between electronvolts and joules and show a worked conversion both ways.

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    E(J) = E(eV) × 1.60 × 10⁻¹⁹ and E(eV) = E(J)/(1.60 × 10⁻¹⁹). The conversion factor is the elementary charge e = 1.60 × 10⁻¹⁹ C, since energy gained = qV. Data-booklet status: e is given in the constants page; the conversion itself is memorise. Example: 10.2 eV = 10.2 × 1.60 × 10⁻¹⁹ = 1.63 × 10⁻¹⁸ J; conversely 3.98 × 10⁻¹⁹ J = 3.98 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 2.49 eV. Common misuse: dividing when you should multiply — check the size, since a joule is a huge unit on the atomic scale so the number in eV is always vastly larger.

19. EquationSL & HLData booklet: No – derive

    State the shortcut relating photon energy in eV to wavelength in nm, and justify it.

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    E(eV) ≈ 1240/λ(nm), equivalently hc = 1.24 × 10⁻⁶ eV m = 1.99 × 10⁻²⁵ J m. It follows from E = hc/λ with hc = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ = 1.99 × 10⁻²⁵ J m, divided by 1.60 × 10⁻¹⁹ J eV⁻¹. Data-booklet status: not printed — derive or memorise. Use it to check answers quickly: the red hydrogen line at 656 nm corresponds to 1240/656 = 1.89 eV, which matches E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV. Common misuse: applying it with λ in metres. Visible light spans roughly 1.8 eV (700 nm) to 3.1 eV (400 nm).

20. EquationSL & HLData booklet: No – memorise

    State the relation between nucleon, proton and neutron numbers and how it is used to balance nuclear equations.

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    A = Z + N, where A is the nucleon number, Z the proton number and N the neutron number, all dimensionless integers. Data-booklet status: not printed — memorise. In any nuclear reaction both the total nucleon number and the total proton number (charge) are separately conserved: sum of A on the left = sum of A on the right, and likewise for Z. Common misuse: forgetting that a beta-minus particle is written ⁰₍₋₁₎e, so its Z of −1 must be included in the balance. Sanity check on ²³⁵₍₉₂₎U: N = 235 − 92 = 143 neutrons, and 143 > 92 as expected for a heavy stable-ish nuclide.

21. EquationSL & HLData booklet: Yes

    State the equation c = fλ as applied to photons and give the rearranged forms needed in E.1.

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    c = fλ, where c = 3.00 × 10⁸ m s⁻¹ is the speed of electromagnetic radiation in a vacuum, f is frequency in hertz (Hz) and λ is wavelength in metres (m). Data-booklet status: printed. Rearrangements used constantly here: f = c/λ and λ = c/f, feeding straight into E = hf. Valid in a vacuum; in a medium of refractive index n the speed becomes c/n and λ shortens to λ/n while f is unchanged. Common misuse: using the wavelength measured in a medium when the photon energy is wanted — always use the vacuum wavelength. Sanity check: f = 3.00 × 10⁸/6.56 × 10⁻⁷ = 4.57 × 10¹⁴ Hz for the red hydrogen line.

22. EquationSL & HLData booklet: No – derive

    State how to calculate the ionisation energy of hydrogen from any level, with symbols and units.

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    Ionisation energy from level n is E_ion = E_∞ − E_n = 0 − (−13.6/n²) = +13.6/n² eV, in electronvolts; multiply by 1.60 × 10⁻¹⁹ for joules. Symbols: n is the principal quantum number (dimensionless), 13.6 eV is the ground-state ionisation energy of hydrogen. Data-booklet status: follows from the printed E_n = −13.6/n² eV. Common misuse: quoting 13.6 eV for every level — from n = 3 only 1.51 eV is needed. The threshold wavelength for ionising ground-state hydrogen is λ = 1240/13.6 = 91.2 nm, in the far ultraviolet, so visible light cannot ionise hydrogen from its ground state.

23. Graph/diagramSL & HL

    Describe the appearance of a hydrogen emission spectrum as seen through a diffraction grating and how to extract energy-level information from it.

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    Presentation: a horizontal wavelength axis, λ/nm increasing to the right (typically 400–700 nm for the visible Balmer series), showing sharp bright lines on a black background at 410 nm (violet), 434 nm (blue), 486 nm (blue-green) and 656 nm (red). Shape: discrete lines, not a band; the lines crowd closer together towards shorter wavelengths, converging on the series limit at 365 nm. Extracting a quantity: measure λ for a line, compute E = hc/λ, and equate it to E_n − E₂. Changing a parameter: a different element gives a completely different line pattern; higher gas pressure broadens the lines and can merge them towards a continuum.

24. Graph/diagramSL & HL

    Describe the appearance of an absorption spectrum and compare it with the emission spectrum of the same gas.

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    Presentation: wavelength λ/nm on the horizontal axis, showing a full continuous rainbow background crossed by narrow DARK lines. Shape: the dark lines lie at exactly the same wavelengths as the bright lines of that element's emission spectrum, because the same level differences are involved. Extracting a quantity: read λ of a dark line and use ΔE = hc/λ to find the level difference; matching the pattern identifies the element, which is how the composition of stellar atmospheres and of the Sun (Fraunhofer lines) is determined. Changing a parameter: a cooler absorbing gas populates mainly the ground state, so lines from low-lying levels dominate; a hotter gas shows lines from excited states as well.

25. Graph/diagramSL & HL

    Describe how to draw and use an energy-level diagram for hydrogen.

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    Presentation: a vertical energy axis in eV, with 0 eV at the TOP (the ionisation limit) and levels drawn as horizontal lines at negative values: −13.6 (n = 1), −3.40 (n = 2), −1.51 (n = 3), −0.850 (n = 4), converging on 0 as n → ∞. Shape: unequally spaced lines that bunch together near the top. Extracting a quantity: a downward arrow between two levels represents emission, an upward arrow absorption, and the photon energy is the vertical length of the arrow, ΔE = hc/λ. Changing a parameter: for a hydrogen-like ion of charge Z the whole ladder scales by Z², so He⁺ levels are four times deeper and its lines lie at one quarter the wavelength.

26. Graph/diagramSL & HL

    Describe the graph of the number of scattered alpha particles against scattering angle in the Geiger–Marsden experiment.

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    Presentation: number of alpha particles detected per unit time (usually on a logarithmic vertical scale) against scattering angle θ/degrees, from near 0° to 180°. Shape: an extremely steep fall — the count is enormous at small angles and drops by many orders of magnitude by 90°, with a small but non-zero tail all the way to 180°. Extracting a quantity: Rutherford's model predicts N ∝ 1/sin⁴(θ/2), so a plot of log N against log[sin(θ/2)] gives a straight line of gradient −4, confirming an inverse-square Coulomb force from a point-like nucleus. Changing a parameter: a thicker foil or a higher-Z target raises the whole curve; faster alphas reduce large-angle scattering.

27. Graph/diagramSL & HL

    Compare the intensity–wavelength graphs of a continuous (black-body) source and of a low-pressure gas discharge.

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    Presentation: relative intensity on the vertical axis against wavelength λ/nm on the horizontal axis. The black body gives a single smooth continuous curve with a broad peak whose position obeys λ_max T = 2.9 × 10⁻³ m K; the discharge tube gives a set of narrow spikes at fixed wavelengths with essentially zero intensity between them. Extracting a quantity: from the black-body curve read λ_max to find surface temperature; from the line spectrum read λ of a spike to find a level difference. Changing a parameter: raising the temperature of the black body raises and shifts its peak to shorter λ, but the positions of the discharge lines do not move — only their brightness changes.

28. Graph/diagramSL & HL

    Describe the linearised graph used to determine the Planck constant from a set of light-emitting diodes.

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    Presentation: plot the LED activation (threshold) potential difference V/V on the vertical axis against 1/λ on the horizontal axis, with λ the peak emission wavelength in metres, so the horizontal unit is m⁻¹. Shape: a straight line through (or very near) the origin, because eV = hc/λ gives V = (hc/e)(1/λ). Extracting a quantity: gradient = hc/e, so h = (gradient × e)/c; with e = 1.60 × 10⁻¹⁹ C and c = 3.00 × 10⁸ m s⁻¹ a gradient of about 1.24 × 10⁻⁶ V m returns h ≈ 6.6 × 10⁻³⁴ J s. Changing a parameter: a systematic error in judging the turn-on point shifts the line vertically, producing a false intercept but leaving the gradient usable.

29. Concept/explainSL & HL

    Explain how the results of the Geiger–Marsden–Rutherford alpha-scattering experiment provide evidence for the nuclear model of the atom.

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    • A collimated beam of alpha particles from a radioactive source was fired at a very thin gold foil in an evacuated chamber and detected by scintillations on a rotatable zinc sulfide screen
    • The vast majority passed through with little or no deflection, showing the atom is mostly empty space
    • A very small fraction (about 1 in 8000) was deflected through more than 90°, some almost straight back
    • Such large deflections require a very large repulsive Coulomb force, so the positive charge and almost all the mass must be concentrated in a tiny, dense core — the nucleus
    • The rarity of back-scattering shows the nucleus is about 10⁻¹⁵ m across compared with 10⁻¹⁰ m for the atom. Exam tip: students state the observations but omit the inference; each observation must be paired with the conclusion it supports.
30. Concept/explainSL & HL

    Explain why the alpha-scattering results could not be accounted for by the "plum pudding" model of the atom.

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    • In the plum pudding model the positive charge is spread uniformly through the whole volume of the atom
    • The electric field inside such a diffuse sphere is small, so the Coulomb force on an alpha particle is weak everywhere
    • Electrons are about 7000 times less massive than an alpha particle and can produce only negligible deflection
    • The model therefore predicts that every alpha particle is deflected by at most a fraction of a degree, with no possibility of back-scattering
    • The observation of rare but very large-angle deflections falsified the model, and Rutherford replaced it with a concentrated nucleus. Exam tip: this is a Nature of Science point — a single reproducible observation that contradicts a prediction is enough to reject a model, however well established.
31. Concept/explainSL & HL

    Explain why the alpha-scattering experiment must be carried out in a vacuum and with an extremely thin foil.

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    • Alpha particles are strongly ionising and have a range of only a few centimetres in air, so residual gas would absorb them and also scatter them randomly before they reached the foil
    • A vacuum ensures that any deflection observed is due to the foil alone
    • The foil must be only a few hundred atoms thick so that each alpha particle undergoes at most one significant scattering event
    • Multiple scattering would blur the angular distribution and prevent comparison with the single-nucleus Coulomb prediction
    • Gold is used because it is highly malleable and can be beaten into a foil about 10⁻⁷ m thick, and its large Z gives strong deflections. Exam tip: "so the alphas can get through" alone is not enough; the single-scattering requirement is the marking point.
32. Concept/explainSL & HLData booklet: Yes

    Explain how the existence of line emission spectra provides evidence that the energy of an atomic electron is quantised.

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    • A hot low-pressure gas emits light only at a set of sharply defined discrete wavelengths, seen as bright lines on a dark background
    • Each photon carries energy E = hf = hc/λ, so a discrete set of wavelengths means a discrete set of photon energies
    • Photons are emitted when an electron makes a transition from a higher to a lower energy level, and energy conservation gives hf = E_higher − E_lower
    • Only discrete differences are possible, so the electron energies themselves must be restricted to particular allowed values, i.e. quantised
    • A continuous range of allowed energies would produce a continuous spectrum, which is not observed. Exam tip: the mark is for linking discrete wavelengths → discrete photon energies → discrete energy differences → discrete levels, not merely asserting quantisation.
33. Concept/explainSL & HL

    Distinguish between an emission spectrum and an absorption spectrum, and explain why the lines of the two occur at the same wavelengths for a given element.

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    • An emission spectrum consists of bright coloured lines on a dark background, produced by excited atoms of a low-pressure gas de-exciting and emitting photons
    • An absorption spectrum consists of dark lines on a continuous coloured background, produced when white light passes through a cooler gas of the same element
    • In absorption a photon is absorbed only if its energy exactly equals a difference between two levels, so the same energy differences are involved
    • The absorbed photons are re-emitted in all directions and after cascades, so the intensity in the original direction is greatly reduced, giving dark lines
    • Hence both spectra map the same energy-level structure and act as a fingerprint of the element. Exam tip: the dark lines are not "missing energy" — the light is scattered out of the beam, and answers must say re-emitted in all directions.
34. Concept/explainSL & HL

    Outline how line spectra are used to determine the chemical composition of a distant star.

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    • Every element has a unique set of energy levels and therefore a unique pattern of spectral line wavelengths
    • The hot dense photosphere emits an approximately continuous black-body spectrum
    • Light passing outward through the cooler stellar atmosphere is absorbed at wavelengths matching transitions in the atoms present, producing dark absorption lines
    • The observed pattern is compared with laboratory spectra of known elements to identify which elements are present, and the relative line strengths indicate abundance
    • Any overall Doppler shift of the whole pattern is allowed for before matching. Exam tip: this is the standard Nature of Science link — a laboratory result applied to objects that can never be sampled directly; students should mention comparison with laboratory reference spectra explicitly.
35. Concept/explainSL & HL

    Explain the meaning of each symbol in the nuclear notation ᴬ_Z X and describe how two isotopes of the same element differ.

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    • X is the chemical symbol of the element
    • Z is the proton number (atomic number), the number of protons in the nucleus, which fixes the element and the chemical behaviour
    • A is the nucleon number (mass number), the total number of protons plus neutrons
    • The number of neutrons is therefore N = A − Z
    • Isotopes of an element have the same Z but different A, so they have the same number of protons and electrons but different numbers of neutrons
    • They are chemically identical but differ in mass, in density and in nuclear stability. Exam tip: A is a count of nucleons, not a mass in kg or u; writing "atomic mass" for A is penalised, and "different number of nucleons" must be traced to neutrons.
36. Concept/explainSL & HLData booklet: Yes

    Explain why the energies of the levels in the hydrogen atom are quoted as negative values and why the levels become closer together as n increases.

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    • The zero of potential energy is defined as the electron being at rest at infinite separation from the proton
    • The Coulomb force is attractive, so work must be done on the electron to remove it; any bound state therefore has energy less than zero, i.e. negative
    • E_n = −13.6 eV/n² gives E₁ = −13.6 eV, E₂ = −3.40 eV, E₃ = −1.51 eV, so the spacing falls off as the levels crowd towards zero
    • The magnitude of E_n measures the energy needed to ionise the atom from that level
    • As n → ∞ the energy tends to zero, which is the ionisation limit, and beyond it the electron is free with a continuous range of kinetic energies. Exam tip: students lose marks by treating the sign as arbitrary — the negative sign is the direct consequence of the chosen zero at infinity.
37. Concept/explainSL & HL

    Explain the origin of a continuous spectrum and contrast it with the line spectrum of a low-pressure gas.

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    • A continuous spectrum contains all wavelengths over a broad range with no gaps, and is emitted by a hot solid, a hot liquid or a hot high-pressure gas
    • In such dense matter the atoms are close together and interact strongly, so the sharp atomic levels are broadened and merged into effectively continuous bands
    • The emitted spectrum then approximates a black-body curve whose peak wavelength depends only on temperature
    • In a low-pressure gas the atoms are far apart and effectively isolated, so their energy levels remain sharp and discrete and only certain photon energies are emitted
    • The result is a set of narrow bright lines characteristic of the element. Exam tip: the distinction is one of atomic separation and interaction, not of temperature alone.
38. Concept/explainSL & HL

    Explain what happens to an atom when it absorbs a photon whose energy is greater than its ionisation energy, and how this differs from absorption of a photon matching a level difference.

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    • If the photon energy exactly matches a difference between two bound levels, it is absorbed and the electron is excited to the higher level; a photon of any other energy passes straight through
    • Once the photon energy exceeds the ionisation energy the electron can be removed completely, so the transition is to an unbound state
    • Unbound electrons may have any kinetic energy, so there is no matching condition and photons of a continuous range of energies can be absorbed
    • Energy conservation gives E_photon = ionisation energy + E_k of the freed electron
    • This produces a continuous absorption region above the series limit rather than discrete lines. Exam tip: many answers claim all photons are absorbed; below the ionisation limit only exactly matching energies are.
39. Worked problemSL & HLData booklet: Yes

    A photon has a wavelength of 486 nm. Calculate its energy in joules and in electronvolts.

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    Principle: E = hf and c = fλ, so E = hc/λ. Substitution: E = (6.63 × 10⁻³⁴ J s × 3.00 × 10⁸ m s⁻¹)/(486 × 10⁻⁹ m). Numerator: hc = 1.989 × 10⁻²⁵ J m. Division: E = 1.989 × 10⁻²⁵/4.86 × 10⁻⁷ = 4.09 × 10⁻¹⁹ J. Convert: E = 4.09 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 2.56 eV. Answers: 4.09 × 10⁻¹⁹ J (3 s.f.) and 2.56 eV (3 s.f.). Check/Trap: convert nm to m before dividing — using 486 gives an answer 10⁹ times too small. A useful memory check is that visible photons are always between about 1.8 eV (700 nm) and 3.1 eV (400 nm), so 2.56 eV for blue-green light is sensible.

40. Worked problemSL & HLData booklet: Yes

    An electron in a hydrogen atom makes a transition from the n = 3 level to the n = 2 level. Determine the wavelength of the emitted photon.

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    Principle: E_n = −13.6 eV/n², and hf = E_higher − E_lower. Levels: E₃ = −13.6/9 = −1.51 eV; E₂ = −13.6/4 = −3.40 eV. Photon energy: ΔE = −1.51 − (−3.40) = 1.89 eV = 1.89 × 1.60 × 10⁻¹⁹ = 3.02 × 10⁻¹⁹ J. Wavelength: λ = hc/ΔE = 1.989 × 10⁻²⁵/3.02 × 10⁻¹⁹ = 6.58 × 10⁻⁷ m ≈ 658 nm. Answer: 6.58 × 10⁻⁷ m, i.e. red light, the Balmer H-α line (accepted value 656 nm). Check/Trap: subtract the levels including their negative signs and take the magnitude; writing ΔE = 1.51 + 3.40 = 4.91 eV is the commonest error. Also convert eV to J before using λ = hc/E.

41. Worked problemSL & HLData booklet: Yes

    A helium–neon laser emits a continuous beam of power 3.0 mW at a wavelength of 633 nm. Estimate the number of photons emitted each second.

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    Principle: power = (photons per second) × (energy per photon). Photon energy: E = hc/λ = 1.989 × 10⁻²⁵/(6.33 × 10⁻⁷) = 3.14 × 10⁻¹⁹ J (equivalently 1.96 eV). Rate: n = P/E = 3.0 × 10⁻³/3.14 × 10⁻¹⁹ = 9.6 × 10¹⁵ photons s⁻¹. Answer: about 9.6 × 10¹⁵ s⁻¹ (2 s.f., matching the 2 s.f. data). Check/Trap: the answer is a rate, so the unit is s⁻¹ and not J or W; a common slip is to divide energy by power. The huge number explains why laser light appears continuous rather than granular, and why photon effects are invisible in everyday optics.

42. Worked problemSL & HLData booklet: Yes

    Calculate the longest wavelength of electromagnetic radiation that can ionise a hydrogen atom in its ground state.

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    Principle: ionisation from n = 1 requires the electron to be raised from E₁ = −13.6 eV to E = 0, so the minimum photon energy is 13.6 eV; minimum energy corresponds to maximum wavelength because E = hc/λ. Convert: E = 13.6 × 1.60 × 10⁻¹⁹ = 2.18 × 10⁻¹⁸ J. Wavelength: λ = hc/E = 1.989 × 10⁻²⁵/2.18 × 10⁻¹⁸ = 9.13 × 10⁻⁸ m ≈ 91 nm. Answer: 9.1 × 10⁻⁸ m, in the ultraviolet. Check/Trap: "longest wavelength" means "smallest photon energy", which is the exact ionisation energy — shorter wavelengths also ionise, leaving the electron with surplus kinetic energy. Do not use the n = 1 to n = 2 difference of 10.2 eV.

43. Worked problemSL & HLData booklet: Yes

    A hydrogen atom in its ground state is bombarded in turn by photons of energy 10.2 eV, 11.0 eV and 13.9 eV. Deduce the outcome in each case.

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    Levels: E₁ = −13.6 eV, E₂ = −3.40 eV, E₃ = −1.51 eV, E₄ = −0.85 eV; ionisation limit 0 eV. Required excitation energies from n = 1: to n = 2, 10.2 eV; to n = 3, 12.1 eV; to n = 4, 12.75 eV; to ionise, 13.6 eV. 10.2 eV: matches the n = 1 → n = 2 gap exactly, so the photon is absorbed and the atom is excited. 11.0 eV: matches no level difference and is below the ionisation energy, so the photon is not absorbed and passes through. 13.9 eV: exceeds 13.6 eV, so the atom is ionised and the freed electron carries E_k = 13.9 − 13.6 = 0.3 eV. Check/Trap: partial absorption is impossible — a photon is absorbed whole or not at all.

44. Worked problemSL & HLData booklet: No – memorise

    An energy-level diagram for an atom shows four levels. Determine the number of different spectral lines that can appear in its emission spectrum, and identify which transition gives the shortest wavelength.

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    Principle: each downward transition between a distinct pair of levels gives one line. Counting: the number of pairs from 4 levels is 4C2 = (4 × 3)/2 = 6, i.e. 4→3, 4→2, 4→1, 3→2, 3→1, 2→1. Shortest wavelength: λ = hc/ΔE, so shortest λ corresponds to the largest ΔE, which is the transition from the highest level directly to the ground state, 4→1. Answers: 6 lines; shortest wavelength from n = 4 → n = 1. Check/Trap: cascades such as 4→3→2→1 are already counted as their individual steps, so do not add extra lines. For n levels the general result is n(n − 1)/2, worth memorising for Paper 1.

45. Worked problemSL & HL

    A nuclide is written as ²³⁵_92 U. State the number of protons, neutrons and nucleons, and compare it with ²³⁸_92 U.

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    Reading the notation: Z = 92 is the proton number, so there are 92 protons; A = 235 is the nucleon number, so there are 235 nucleons. Neutrons: N = A − Z = 235 − 92 = 143. Neutral atom: 92 electrons. Comparison with ²³⁸_92 U: same Z = 92, so the same 92 protons and the same chemistry; A = 238 gives N = 238 − 92 = 146 neutrons, three more than in U-235. The two are therefore isotopes of uranium, differing only in neutron number, hence in nuclear mass and stability. Check/Trap: subtract Z from A, never the other way round, and remember the difference between the isotopes is purely nuclear — no spectroscopic chemical difference is expected.

46. Worked problemSL & HLData booklet: Yes

    A spectral line in a hydrogen discharge tube is measured at 1.03 × 10⁻⁷ m. Deduce the transition responsible.

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    Principle: photon energy E = hc/λ, then match to E_n = −13.6 eV/n². Energy: E = 1.989 × 10⁻²⁵/1.03 × 10⁻⁷ = 1.93 × 10⁻¹⁸ J = 1.93 × 10⁻¹⁸/1.60 × 10⁻¹⁹ = 12.1 eV. Matching: transitions ending on n = 1 have energies 10.2 eV (2→1), 12.1 eV (3→1), 12.75 eV (4→1). Since E₃ − E₁ = −1.51 − (−13.6) = 12.1 eV, the line arises from n = 3 → n = 1. Answer: the n = 3 to n = 1 transition (Lyman series, ultraviolet). Check/Trap: an ultraviolet wavelength immediately signals a Lyman line ending on the ground state; visible lines (400–700 nm) end on n = 2 and infrared lines on n = 3 or higher.

47. Exam technique/trapSL & HL

    Explain the common error made when students are asked to state the evidence from alpha scattering for a small dense nucleus, and give the wording that earns full marks.

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    • The trap: writing only the observations ("most went straight through, some bounced back") without the deductions
    • Why students fall for it: the observations are memorable and the mark scheme looks obvious, but "state the evidence for X" requires observation plus what it shows
    • Correct approach: pair each result with an inference — most undeflected → atom mostly empty space; a few deflected by more than 90° → a large repulsive force from a concentrated positive charge; very few such events → the nucleus occupies a tiny fraction of the atomic volume; back-scattering of a massive alpha → the nucleus is much more massive than the alpha
    • Command-term guidance: "outline" needs a brief account, "explain" demands the reason each observation implies the conclusion. Exam tip: quantify where you can — about 1 in 8000, nucleus ~10⁻¹⁵ m versus atom ~10⁻¹⁰ m.
48. Exam technique/trapSL & HLData booklet: Yes

    Identify the errors students most often make with the eV, keV and MeV units in atomic and nuclear calculations.

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    • The trap: using energies in eV directly in E = hf or λ = hc/E, which gives answers a factor of 1.60 × 10⁻¹⁹ out
    • Why: the data booklet quotes E_n in eV while h and c are in SI units, so the mixture looks harmless
    • Correct approach: multiply by 1.60 × 10⁻¹⁹ to convert eV → J before any SI formula, and divide to go back; 1 keV = 10³ eV, 1 MeV = 10⁶ eV
    • A second trap is quoting an electronvolt as a unit of potential difference — it is an energy, the work done moving charge e through 1 V
    • Exam technique: state the conversion line explicitly so the marker can award method marks even if arithmetic slips. Exam tip: 1 eV = 1.60 × 10⁻¹⁹ J is in the data booklet, so there is no excuse for guessing it.
49. Exam technique/trapSL & HLData booklet: Yes

    Explain the sign and direction traps in energy-level transition questions and how to avoid them.

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    • The trap: adding the magnitudes of two negative level energies, e.g. taking the 3→2 gap as 1.51 + 3.40 = 4.91 eV instead of 1.89 eV
    • Why: students ignore that both energies are negative and subtract carelessly
    • Correct approach: always compute ΔE = E_final − E_initial with signs, then use the magnitude for the photon energy
    • Second trap: claiming a photon is emitted when the electron moves to a higher level — absorption raises the electron, emission lowers it
    • Third trap: describing the electron as "gaining energy and falling down"
    • Command terms: "deduce" requires you to show the numerical matching, not just name the transition
    • Sketching the level diagram with values marked prevents nearly all of these errors. Exam tip: an upward arrow means absorption, a downward arrow emission — label arrows on any diagram you draw.
50. Exam technique/trapSL & HL

    Describe an experiment to observe and measure the wavelengths of the visible lines in the hydrogen emission spectrum, including variables, limitations and improvements.

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    • Apparatus: hydrogen discharge tube with EHT supply, single narrow slit, diffraction grating of known spacing d on a spectrometer table with a vernier angular scale, telescope
    • Method: level and focus the spectrometer, locate the zero-order image, then measure the angle θ of each coloured first-order line on both sides and halve the difference; use nλ = d sin θ
    • Independent variable: order n or the line observed; dependent variable: diffraction angle θ; controlled: grating spacing, slit width, tube current, room darkened
    • Limitations: faint violet lines are hard to see, slit width broadens lines, backlash in the angular scale
    • Improvements: measure left and right and average to remove zero error, use second-order lines for greater angular dispersion, use a photodiode or spectrometer app to locate line centres. Exam tip: always quote θ to the nearest minute of arc and propagate that uncertainty into λ.
51. Exam technique/trapSL & HLData booklet: No – derive

    Describe how a Geiger–Marsden type scattering investigation is analysed, including how counting uncertainties are handled.

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    • Apparatus: sealed alpha source in a collimator, thin metal foil, evacuated chamber, detector (scintillation screen and microscope, or a solid-state detector) mounted on an arm rotatable about the foil
    • Independent variable: scattering angle θ; dependent variable: count N in a fixed time interval; controlled: source activity, foil material and thickness, counting time, source–foil–detector geometry
    • Analysis: plot count rate against θ; the rate falls very steeply with angle, so a log scale or a plot against 1/sin⁴(θ/2) is used to test the Coulomb prediction, which should give a straight line through the origin
    • Uncertainty: radioactive counting is random, so the absolute uncertainty in N is √N and the fractional uncertainty is 1/√N — at large angles N is small so the error bars are large
    • Improvement: count for far longer at large angles. Exam tip: subtract the background count before analysis.
52. Exam technique/trapSL & HL

    Outline how to handle significant figures and uncertainties when a photon wavelength is calculated from a measured energy.

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    • Rule: the final answer carries the same number of significant figures as the least precise datum, usually 2 or 3 s.f. in IB questions
    • Because λ = hc/E, wavelength and energy are inversely proportional, so the fractional uncertainty transfers directly: Δλ/λ = ΔE/E
    • Fractional (percentage) uncertainties add for products and quotients, absolute uncertainties add for sums and differences — so a transition energy found as a difference of two level energies uses added absolute uncertainties first
    • Do not round intermediate values; carry extra digits and round once at the end
    • Quote the uncertainty to 1 s.f. and match the decimal place of the value, e.g. λ = (658 ± 4) nm
    • Precision refers to spread of repeats, accuracy to closeness to the accepted value. Exam tip: a stray "3 × 10⁸" written as "3" costs the s.f. mark.
53. DefinitionHL only

    Define the distance of closest approach in alpha-particle scattering and state what it tells us.

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    The distance of closest approach d is the least separation between the centres of an alpha particle and a target nucleus in a head-on (180° backscattering) collision, reached at the instant when all the alpha's initial kinetic energy has been converted into electric potential energy. It is measured in metres, typically a few times 10⁻¹⁴ m. Exam tip: d is an UPPER BOUND on the nuclear radius, not the radius itself, because the alpha is turned around by the Coulomb force before it touches the nucleus; saying "this gives the radius of the nucleus" loses the mark. Using higher-energy alphas gives a smaller d and hence a tighter limit.

54. DefinitionHL only

    Define the nuclear radius and state the meaning and value of R₀.

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    The nuclear radius R is the radius of the approximately spherical volume within which the nucleons of a nucleus are contained, given by R = R₀A^(1/3). R₀ is the Fermi radius, the constant of proportionality, equal to 1.2 × 10⁻¹⁵ m (1.2 fm); it is numerically the radius a nucleus with A = 1 would have. Units: metres, conveniently the femtometre (fermi), 1 fm = 10⁻¹⁵ m. Exam tip: R is not sharply defined because the nuclear surface is diffuse; it is conventionally taken where the nucleon density falls to half its central value. Typical values: about 3 fm for ¹⁶O, about 7 fm for ¹⁹⁷Au.

55. EquationHL onlyData booklet: Yes

    State the Bohr quantisation condition, define every symbol, and show what it implies for orbital radii.

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    mvr = nh/(2π). m is the electron mass, 9.11 × 10⁻³¹ kg; v its orbital speed in m s⁻¹; r the orbit radius in m; n = 1, 2, 3, … the principal quantum number (dimensionless); h = 6.63 × 10⁻³⁴ J s. Angular momentum has units kg m² s⁻¹ (= J s). Data-booklet status: printed. Combining it with kZe²/r² = mv²/r eliminates v and gives r ∝ n²/Z, so r_n = n² × 5.3 × 10⁻¹¹ m for hydrogen. Common misuse: forgetting the 2π, or using it for a non-circular path. Sanity check: for n = 1, mvr = 6.63 × 10⁻³⁴/(2π) = 1.06 × 10⁻³⁴ J s.

56. EquationHL onlyData booklet: No – derive

    Outline how the Bohr model produces E_n = −13.6/n² eV and state which parts are booklet and which are derived.

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    Set the Coulomb force equal to the centripetal force, ke²/r² = mv²/r, giving E_k = ½mv² = ke²/(2r); the potential energy is E_p = −ke²/r, so the total is E = E_k + E_p = −ke²/(2r). Substituting r_n = n²h²/(4π²kme²) yields E_n = −(2π²k²e⁴m)/(h²n²) = −13.6/n² eV. Symbols: k = 8.99 × 10⁹ N m² C⁻², e = 1.60 × 10⁻¹⁹ C, m = 9.11 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s. Data-booklet status: E_n = −13.6/n² eV and mvr = nh/(2π) printed; the derivation itself must be reproduced. Note E_p = 2E and E_k = −E, a standard checkable relation.

57. EquationHL onlyData booklet: No – derive

    State the energy-conservation equation for the distance of closest approach and define the symbols.

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    ½mv² = kQq/d, so d = 2kQq/(mv²) = kQq/E_k. E_k = ½mv² is the initial kinetic energy of the alpha particle in joules; m is its mass (6.64 × 10⁻²⁷ kg) and v its initial speed in m s⁻¹; k = 8.99 × 10⁹ N m² C⁻²; Q = Ze is the nuclear charge and q = 2e the alpha charge, both in coulombs; d is in metres. Data-booklet status: derive from Coulomb's law (printed) plus conservation of energy. Valid for a head-on approach to a nucleus assumed stationary and point-like. Common misuse: leaving E_k in MeV. Sanity check: a 5.0 MeV alpha on gold (Z = 79) gives d = 4.5 × 10⁻¹⁴ m.

58. EquationHL onlyData booklet: Yes

    State the nuclear radius equation, define its symbols and give a worked value.

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    R = R₀A^(1/3). R is the nuclear radius in metres; R₀ = 1.2 × 10⁻¹⁵ m is the Fermi radius; A is the nucleon number (dimensionless). Data-booklet status: printed. Rearranged forms: A = (R/R₀)³ and R₂/R₁ = (A₂/A₁)^(1/3). Valid for stable, roughly spherical nuclei with A greater than about 20; light and strongly deformed nuclei deviate. Common misuse: cubing instead of taking the cube root, or expecting R to double when A doubles — it rises only by a factor 2^(1/3) = 1.26. Sanity check: for ¹⁹⁷Au, A^(1/3) = 5.82 so R = 1.2 × 10⁻¹⁵ × 5.82 = 7.0 × 10⁻¹⁵ m, comfortably smaller than the 4.5 × 10⁻¹⁴ m closest-approach limit.

59. EquationHL onlyData booklet: No – derive

    Derive an expression for nuclear density and evaluate it.

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    ρ = m/V with m ≈ Au and V = (4/3)πR³ = (4/3)πR₀³A, so ρ = Au/[(4/3)πR₀³A] = 3u/(4πR₀³). Symbols: u = 1.66 × 10⁻²⁷ kg is the unified atomic mass unit, R₀ = 1.2 × 10⁻¹⁵ m, ρ is in kg m⁻³; A cancels, so ρ is independent of the nuclide. Data-booklet status: derive from the printed R = R₀A^(1/3). Evaluation: ρ = (3 × 1.66 × 10⁻²⁷)/(4π × (1.2 × 10⁻¹⁵)³) = 4.98 × 10⁻²⁷/2.17 × 10⁻⁴⁴ = 2.3 × 10¹⁷ kg m⁻³. Common misuse: forgetting to cube R₀, which changes the answer by 30 orders of magnitude.

60. Graph/diagramHL only

    Describe the linearised graph used to test R = R₀A^(1/3) and to find R₀.

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    Presentation: nuclear radius R/10⁻¹⁵ m on the vertical axis against A^(1/3) (dimensionless) on the horizontal axis, using radii obtained from electron-diffraction measurements on a range of nuclides. Shape: a straight line passing through the origin, confirming the cube-root dependence. Extracting a quantity: the gradient is R₀, expected to be about 1.2 × 10⁻¹⁵ m; the zero intercept is itself a test of the model. Alternatively plot ln R against ln A, which should be a straight line of gradient 1/3 and intercept ln R₀. Changing a parameter: light nuclei (A < 20) and deformed nuclei fall slightly off the line, showing the model is an approximation.

61. Graph/diagramHL only

    Describe how the alpha-scattering graph deviates from the Rutherford prediction at high alpha energy and what this reveals.

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    Presentation: scattered intensity (log scale) against scattering angle θ, with the Rutherford 1/sin⁴(θ/2) prediction drawn as a reference curve. Shape: at low alpha energies the data lie exactly on the predicted curve; as the energy is raised the measured counts at large angles fall BELOW the Rutherford curve. Interpretation: at higher energy the alpha's distance of closest approach shrinks until it reaches the nuclear surface, where the attractive short-range strong nuclear force acts in addition to the Coulomb repulsion, so the pure inverse-square analysis fails. Extracting a quantity: the energy at which the departure begins gives d ≈ R, an estimate of the nuclear radius, of order 10⁻¹⁵ m.

62. Concept/explainHL onlyData booklet: Yes

    State the postulates of the Bohr model of the hydrogen atom and explain how they account for the discrete line spectrum.

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    • The electron moves in a circular orbit about the proton under the Coulomb force, but only in orbits for which the angular momentum is quantised: mvr = nh/(2π), with n = 1, 2, 3, …
    • While in such a stationary state the electron does not radiate, contrary to classical electromagnetism
    • Radiation is emitted or absorbed only when the electron jumps between allowed orbits, with hf = E_initial − E_final
    • Combining the quantisation condition with Coulomb's law and the circular-motion condition gives discrete radii r ∝ n² and discrete energies E_n = −13.6 eV/n²
    • Since only discrete energy differences exist, only discrete photon frequencies appear, matching the observed hydrogen series exactly. Exam tip: the non-radiating postulate is an assumption imposed to save the model, not a derived result — say so when asked to evaluate the model.
63. Concept/explainHL only

    Discuss the limitations of the Bohr model of the atom.

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    • It works quantitatively only for hydrogen and other one-electron systems; it fails for helium and all multi-electron atoms because electron–electron repulsion is ignored
    • It predicts the wavelengths but not the relative intensities of spectral lines, nor the lifetimes of excited states
    • It cannot explain fine structure, the splitting of lines in magnetic or electric fields (Zeeman and Stark effects), or spectral line shapes
    • The non-radiating stationary state is an arbitrary postulate inconsistent with classical electromagnetism, and the model mixes classical orbits with a quantum condition
    • It assumes the electron has a definite position and momentum simultaneously, which conflicts with the uncertainty principle; the modern description uses probability-density orbitals
    • Nature of Science: the Bohr model is retained as a useful stepping-stone model despite being superseded. Exam tip: give at least three distinct limitations, not three rewordings of "only works for hydrogen".
64. Concept/explainHL onlyData booklet: No – derive

    Explain how the distance of closest approach in alpha scattering gives an upper limit for the radius of a nucleus, and why it is only an upper limit.

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    • Consider an alpha particle fired directly at a nucleus; it decelerates as the Coulomb repulsion does negative work on it
    • At closest approach it is momentarily at rest, so all the initial kinetic energy has become electric potential energy: ½mv² = kQq/d with Q = Ze and q = 2e
    • Solving for d gives d = 2kZe²/(½mv²), a distance of order 10⁻¹⁴ m for MeV alphas
    • If the alpha were to touch the nuclear surface the radii would satisfy d = R_nucleus + R_alpha, so d overestimates the nuclear radius
    • The estimate is an upper limit because the alpha never actually reaches the surface at these energies, and because the alpha itself has finite size
    • Higher-energy alphas approach more closely and give better estimates. Exam tip: the charge of the alpha is 2e and of the nucleus Ze — omitting either factor is the standard error.
65. Concept/explainHL onlyData booklet: Yes

    Explain how deviations from Rutherford scattering at high alpha energies provide evidence for the size of the nucleus and for the strong nuclear force.

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    • The Rutherford formula assumes only the inverse-square Coulomb repulsion between two point charges
    • At low alpha energies the measured angular distribution matches this prediction exactly, confirming that the nucleus behaves as a point charge on that scale
    • As the alpha energy is raised the distance of closest approach shrinks, and above a threshold the measured scattering at large angles falls below the Coulomb prediction
    • This shows the alpha has reached a separation where an additional, attractive interaction acts — the short-range strong nuclear force
    • The energy at which deviation begins gives the distance at which the alpha touches the nuclear surface, hence an estimate of the nuclear radius, consistent with R = R₀A^(1/3)
    • Similar deviations for different targets confirmed the A^(1/3) dependence. Exam tip: say deviations occur at large angles and high energies — both conditions correspond to close approach.
66. Concept/explainHL onlyData booklet: Yes

    Explain why the density of nuclear matter is approximately the same for all nuclei, and state its order of magnitude.

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    • Experiment gives the nuclear radius as R = R₀A^(1/3) with R₀ = 1.2 × 10⁻¹⁵ m
    • The nuclear volume is therefore V = (4/3)πR³ = (4/3)πR₀³A, i.e. volume is directly proportional to nucleon number
    • The nuclear mass is approximately Au, since each nucleon has a mass close to 1 u
    • Hence ρ = m/V = Au/[(4/3)πR₀³A] = u/[(4/3)πR₀³], in which A cancels, so the density is independent of the nuclide
    • Evaluating gives ρ ≈ 2.3 × 10¹⁷ kg m⁻³, about 10¹⁴ times the density of water
    • Physically this means nucleons are packed at a fixed spacing because the strong force saturates, and nuclear matter behaves like an incompressible liquid drop. Exam tip: the mark is for showing A cancels, not merely quoting the numerical value.
67. Worked problemHL onlyData booklet: No – derive

    An alpha particle of kinetic energy 5.5 MeV is fired head-on at a stationary gold nucleus (Z = 79). Determine the distance of closest approach.

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    Principle: at closest approach all kinetic energy has become electric potential energy: ½mv² = kQq/d, so d = kQq/E_k. Charges: q_alpha = 2e = 3.20 × 10⁻¹⁹ C; Q_gold = 79e = 79 × 1.60 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷ C. Energy: E_k = 5.5 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.80 × 10⁻¹³ J. Substitution: d = (8.99 × 10⁹ × 3.20 × 10⁻¹⁹ × 1.264 × 10⁻¹⁷)/8.80 × 10⁻¹³. Numerator: 3.64 × 10⁻²⁶. Result: d = 4.1 × 10⁻¹⁴ m. Answer: 4.1 × 10⁻¹⁴ m (2 s.f.), about 40 fm. Check/Trap: this is roughly six times the actual gold nuclear radius of 7 fm, which is why the alpha never touches the nucleus and Rutherford scattering holds; forgetting the factor 2 for the alpha charge halves the answer.

68. Worked problemHL onlyData booklet: Yes

    Determine the minimum kinetic energy, in MeV, that an alpha particle must have to just reach the surface of an aluminium nucleus (Z = 13, A = 27).

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    Radius: R = R₀A^(1/3) = 1.2 × 10⁻¹⁵ × 27^(1/3) = 1.2 × 10⁻¹⁵ × 3 = 3.6 × 10⁻¹⁵ m. Principle: E_k = kQq/d with d taken as R (ignoring the size of the alpha). Charges: q = 2e = 3.20 × 10⁻¹⁹ C, Q = 13e = 2.08 × 10⁻¹⁸ C. Substitution: E_k = (8.99 × 10⁹ × 3.20 × 10⁻¹⁹ × 2.08 × 10⁻¹⁸)/3.6 × 10⁻¹⁵ = 5.99 × 10⁻²⁷/3.6 × 10⁻¹⁵ = 1.66 × 10⁻¹² J. Convert: 1.66 × 10⁻¹²/1.60 × 10⁻¹⁹ = 1.0 × 10⁷ eV = 10 MeV. Answer: about 10 MeV. Check/Trap: natural alpha sources give only 4–9 MeV, which is why Rutherford's alphas never reached the nuclear surface; a truer treatment adds the alpha radius to R, lowering the required energy.

69. Worked problemHL onlyData booklet: Yes

    Calculate the radius of a gold-197 nucleus and the ratio of its radius to that of a carbon-12 nucleus.

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    Principle: R = R₀A^(1/3) with R₀ = 1.2 × 10⁻¹⁵ m. Gold: 197^(1/3) = 5.81, so R = 1.2 × 10⁻¹⁵ × 5.81 = 6.97 × 10⁻¹⁵ ≈ 7.0 × 10⁻¹⁵ m (7.0 fm). Ratio: R_Au/R_C = (197/12)^(1/3) = (16.4)^(1/3) = 2.54. Answers: R_Au = 7.0 × 10⁻¹⁵ m; the gold nucleus is about 2.5 times the radius of the carbon nucleus. Check/Trap: the ratio depends only on A, never on Z — a very common slip is to use 79 and 6. Note that the radius grows only as the cube root, so a 16-fold increase in nucleon number gives only a 2.5-fold increase in radius, consistent with constant nuclear density.

70. Worked problemHL onlyData booklet: Yes

    Show that the density of nuclear matter is approximately 2 × 10¹⁷ kg m⁻³.

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    Principle: ρ = m/V with m ≈ Au and V = (4/3)πR³, R = R₀A^(1/3). Volume: V = (4/3)πR₀³A. Density: ρ = Au/[(4/3)πR₀³A] = u/[(4/3)πR₀³], so A cancels. Substitution: R₀³ = (1.2 × 10⁻¹⁵)³ = 1.73 × 10⁻⁴⁵ m³; (4/3)π × 1.73 × 10⁻⁴⁵ = 7.24 × 10⁻⁴⁵ m³ per nucleon. Result: ρ = 1.66 × 10⁻²⁷/7.24 × 10⁻⁴⁵ = 2.3 × 10¹⁷ kg m⁻³. Answer: ρ ≈ 2 × 10¹⁷ kg m⁻³, independent of the nuclide. Check/Trap: cube R₀ before multiplying, and remember (10⁻¹⁵)³ = 10⁻⁴⁵. A sanity check: this is about 10¹⁴ times the density of water, so 1 cm³ of nuclear matter would have a mass of about 2 × 10¹¹ kg.

71. Worked problemHL onlyData booklet: Yes

    For the n = 1 state of hydrogen the orbital radius is 5.3 × 10⁻¹¹ m. Calculate the speed of the electron and its angular momentum in the n = 3 state.

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    Principle: Bohr quantisation mvr = nh/(2π). Speed in n = 1: v = h/(2πmr) = 6.63 × 10⁻³⁴/(2π × 9.11 × 10⁻³¹ × 5.3 × 10⁻¹¹). Denominator: 2π × 4.83 × 10⁻⁴¹ = 3.03 × 10⁻⁴⁰. So v = 2.19 × 10⁶ m s⁻¹. Angular momentum for n = 3: L = 3h/(2π) = 3 × 6.63 × 10⁻³⁴/6.283 = 3.17 × 10⁻³⁴ J s (equivalently kg m² s⁻¹). Answers: v = 2.2 × 10⁶ m s⁻¹; L = 3.2 × 10⁻³⁴ J s. Check/Trap: v/c ≈ 0.007, so the non-relativistic treatment is justified — an answer exceeding c means the 2π has been dropped. Angular momentum comes in whole multiples of h/2π only.

72. Worked problemHL onlyData booklet: Yes

    Using the Bohr model, determine the radius of the n = 4 orbit of hydrogen and the energy required to excite the atom from n = 2 to n = 4.

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    Radii: the Bohr model gives r_n = n²a₀ with a₀ = 5.3 × 10⁻¹¹ m. For n = 4: r₄ = 16 × 5.3 × 10⁻¹¹ = 8.5 × 10⁻¹⁰ m. Energies: E_n = −13.6 eV/n², so E₂ = −13.6/4 = −3.40 eV and E₄ = −13.6/16 = −0.85 eV. Excitation energy: ΔE = E₄ − E₂ = −0.85 − (−3.40) = 2.55 eV = 2.55 × 1.60 × 10⁻¹⁹ = 4.08 × 10⁻¹⁹ J. Answers: r₄ = 8.5 × 10⁻¹⁰ m; ΔE = 2.55 eV = 4.1 × 10⁻¹⁹ J. Check/Trap: radius scales as n² while energy magnitude scales as 1/n², so the electron is much further out and much less tightly bound — an n = 4 hydrogen atom is already larger than a typical ground-state atom. Do not use n rather than n² in either formula.

73. Exam technique/trapHL onlyData booklet: No – derive

    Explain the traps in questions on the distance of closest approach and how to structure a full-mark answer.

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    • Trap 1: using the alpha charge as e instead of 2e, or the target charge as A rather than Z — the nucleus carries Ze
    • Trap 2: leaving the kinetic energy in MeV when substituting into d = kQq/E_k
    • Trap 3: quoting d as "the nuclear radius" — it is an upper limit, since the alpha is repelled before contact and has its own finite size
    • Trap 4: assuming a head-on collision when the question specifies a glancing approach; only the head-on case gives the minimum d
    • Correct structure: state energy conservation ½mv² = kQq/d, convert MeV to J, substitute with both charges shown, then comment that d > R
    • Command terms: "estimate" invites reasonable approximations, "determine" expects a full calculation with unit. Exam tip: the mark scheme awards a mark for the explicit statement that all kinetic energy has become electric potential energy.
74. Exam technique/trapHL onlyData booklet: Yes

    Identify the errors students make with R = R₀A^(1/3) and nuclear-density calculations, and give the correct method.

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    • Trap 1: substituting Z instead of A — the radius depends on the total number of nucleons, since neutrons contribute volume too
    • Trap 2: cubing rather than cube-rooting A, giving absurd radii
    • Trap 3: forgetting that ratios of radii need (A₁/A₂)^(1/3) while ratios of volumes need A₁/A₂ directly
    • Trap 4: in density questions, using the atomic mass in u without converting with u = 1.66 × 10⁻²⁷ kg
    • Correct approach: compute R, then V = (4/3)πR³, then ρ = Au/V; better, show A cancels to prove density is nuclide-independent
    • Command terms: "show that" requires every substitution written out and a final value quoted to more figures than the target. Exam tip: R₀ = 1.2 × 10⁻¹⁵ m is in the data booklet — do not invent 1.0 or 1.4 fm.
75. Exam technique/trapHL onlyData booklet: Yes

    Outline how the Bohr model should be evaluated in an exam answer, with guidance on the command terms used.

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    • "Outline the Bohr model" wants the three postulates briefly: quantised angular momentum mvr = nh/(2π), non-radiating stationary states, and photon emission on transitions with hf = ΔE
    • "Explain how the model accounts for the hydrogen spectrum" wants the logical chain from quantised angular momentum to r ∝ n², to E_n = −13.6 eV/n², to discrete photon energies
    • "Discuss the limitations" or "evaluate" requires balance: successes (hydrogen wavelengths, ionisation energy, the concept of quantisation) set against failures (multi-electron atoms, line intensities, fine structure, conflict with the uncertainty principle)
    • Trap: students describe electron shells from chemistry instead of the physics postulates and gain nothing
    • Nature of Science: a model can be quantitatively successful in a narrow domain yet be conceptually wrong. Exam tip: quote the booklet form of the quantisation condition exactly.

Practise this topic with exam-style questions: E.1 Structure of the atom questions (SL) · E.1 Structure of the atom questions (HL) · all flashcards

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