IB Physics HL E.1 Structure of the atom Questions

SLSL HLHL

Practise IB Physics HL E.1 by analysing atomic spectra, nuclear structure, mass defect and binding-energy evidence in extended problems.

172 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Describe the scattering observations separately: most alphas pass undeflected, some deflect and very few scatter backwards.
For the scattering method use a thin target and detect deflection angles or positions; an angular-count graph peaks near zero deflection.
Infer a tiny, dense, positively charged nucleus containing most atomic mass; the observations do not establish neutrons or Bohr orbits.
Compare atom and nucleus length scales: atomic diameter is about 10⁵ times nuclear diameter, rather than that being a volume ratio.
Recognize atomic spectra as evidence that changed the atomic model; this short question asks for the relevant evidence, not a historical essay.
Read A as protons plus neutrons, Z as proton number and N = A − Z; balance total nucleon number and signed charge separately.
Identify a neutron by A = 1, Z = 0, a proton by A = 1, Z = 1, and an alpha particle by A = 4, Z = 2.
In beta-minus decay A is unchanged and daughter Z increases by one; include the emitted electron’s negative charge in the equation.
Alpha decay reduces A by four and Z by two; Ra with Z = 88 gives Rn with Z = 86.
Include the incoming neutron when counting fission products: ²³⁵U plus a neutron has A = 236, so Ba-144 and Kr-89 leave three neutrons.
Follow the actual branch of a decay-chain table, updating A and Z at every alpha or beta-minus step.
State the requested nucleon or proton number explicitly; an unlabelled nuclide symbol alone may not earn that mark.
Explain discrete spectral lines through photon energies equal to differences between allowed atomic levels; each link in the argument matters.
Match a line pattern using differences between energy levels, rather than assigning a spectral line to each level itself.
White light through low-pressure gas produces dark absorption lines against a coloured continuous background.
Absorption raises electron energy and emission lowers it; equal-frequency absorption and emission can connect the same pair in opposite directions.
Atomic emission is not restricted to visible light; retain the transition energy and actual spectral range.
An excited electron remains bound, whereas an ionized electron is free; higher bound levels remain distinct from the ionized state.
For bound-to-bound absorption a photon must match an allowed gap from the actual initial level; a merely larger photon energy is insufficient.
Count all accessible cascade transitions but count repeated energy gaps once when different frequencies are requested.
Order photon wavelengths inversely with downward energy gaps: smallest gap gives longest wavelength and largest gap gives highest frequency.
For a three-level model include both adjacent gaps and their sum before selecting the two smallest photon frequencies.
For the supplied visible hydrogen series ending at n = 2, higher starting levels give shorter wavelengths approaching a limit.
Use the stated final level and ΔE = hc/λ with the given Bohr formula to identify an initial level; do not assume every emission ends at ground.
For absorption, match hc/λ to an upward gap and show numerical substitution when the markscheme requires more than the formula alone.
Calculate emitted wavelength from ΔE, not one level’s binding energy; n = 3→2 in the supplied table gives about 657 nm.
Use E = hf = hc/λ with metres for wavelength and compatible units for h, converting between joules and electronvolts as needed.
When a transition energy is the sum of two others, add reciprocal wavelengths: 1/λ₁ = 1/λ₂ + 1/λ₃.
For two lines ending at the same level, the difference between their photon energies equals the difference between their initial electron energies.
Find incident photon count from total incident energy divided by hf; use intensity, area and exposure time before any detector efficiency.
Intensity equals photon arrival rate per area times photon energy; doubling wavelength and photon arrival rate leaves intensity unchanged.
For electrons accelerated from rest through V, photon energy cannot exceed eV, so λmin = hc/(eV); use the spectrum’s cutoff rather than a peak.
Identify an element by matching stellar absorption or emission line wavelengths with a known reference pattern specific to that element.
Explain stellar comparison as specific gas absorption, observation of the spectrum on Earth, then matching known chemical spectra.
Obtain a laboratory hydrogen spectrum using a discharge tube or hot gas, a prism or grating, and a screen or telescope.
Use R = R₀A^(1/3) with nucleon number, and R₂/R₁ = (A₂/A₁)^(1/3); convert fm and do not substitute Z for A.
In reverse radius problems use A = (R/R₀)³, halving any stated diameter first; a diameter ratio of three implies a nucleon ratio of 27.
Prove approximately constant nuclear density by showing both mass ∝ A and volume ∝ R³ ∝ A, then cancelling A in mass divided by volume.
An eightfold A doubles nuclear radius but leaves density approximately unchanged; density-versus-A is horizontal in this model.
Add nuclear radii for the stated contact estimate of centre separation; obtain volume from 4πR³/3 using one consistent accepted radius route.
At fixed electron wavelength, a larger nucleus gives a smaller diffraction angle while nuclear density remains approximately the same.
High-energy alphas can probe finite nuclear structure and enter strong-force range, causing departures from a Coulomb-only scattering model.
High-energy scattering deviations give evidence for the strong interaction and allow a nuclear-radius estimate, rather than measuring alpha size.
Increasing incident alpha energy and decreasing target proton number permit closer approach and greater deviations.
Compare closest approach with nuclear radius: a much larger approach distance leaves the alpha outside strong-force range, so no deviation is expected.
For head-on closest approach set initial kinetic energy equal to kqQ/r, using alpha q = 2e, target Q = Ze and centre separation r.
If acceleration through voltage supplies K = qV, q cancels in r = kqQ/K; same-Z isotopes then give equal distance and force in the fixed-target model.
At the deviation threshold identify nuclear radius approximately with closest approach, then explain E₀ = kQq/r; show the required calculation precision.
Apply Eₙ = −13.6/n² eV, keeping negative bound-state energies; n = 2 has energy −3.40 eV.
Recognize orbital radius, quantized energy and quantized angular momentum mvr = nh/(2π) as Bohr features; radioactive decay is a separate model.
Read the actual n-arrows: absorption energy ratios use gap ratios, while emission wavelength ratios use inverse gap ratios.
For allowed states L ∝ n; if n doubles and radius quadruples, speed halves rather than angular momentum remaining constant.
Bohr postulated non-radiating allowed orbits to address classical radiating-orbit collapse; the supplied ground-radius expression gives 5.3 × 10⁻¹¹ m.
Check the one-electron condition: removing one electron from neutral helium leaves a hydrogen-like ion for the qualitative Bohr-model test.
In the stated nuclear-size probe comparison, electrons avoid the strong nuclear interaction; do not claim they have shorter wavelength at equal energy.
Compare probe scale qualitatively: lower-energy electrons probe the whole nucleus, while higher-energy electrons can resolve nucleons within it.

Covered syllabus points

E.1.1—Rutherford experiment E.1.2—Nuclear notation E.1.3—Atomic energy levels E.1.4—Atomic transitions E.1.5—Photon energy E.1.6—Spectra and composition E.1.7 (HL)—Nuclear radius E.1.8 (HL)—High-energy scattering deviations E.1.9 (HL)—Closest approach E.1.10 (HL)—Bohr hydrogen levels E.1.11 (HL)—Bohr angular momentum

Exam points

Describe the observations separately: most alpha particles pass undeflected, some are deflected and very few are scattered backwards.
For the scattering setup, direct particles at a thin target and detect their deflection angles or positions; these are distinct experimental steps.
Choose an angular-count graph with the greatest count near zero deflection and much smaller counts at large angles.
Describe a tiny nucleus containing most of the atom’s mass and all its positive charge, with electrons surrounding it.
Infer a small, dense, positively charged nucleus from rare large deflections; do not claim that these observations establish neutrons or Bohr orbits.
Compare atomic and nuclear length scales: the atom’s diameter is about 10⁵ times its nucleus’s diameter.

[Maximum number: 2]

Rutherford constructed a model of the atom based on the results of the alpha particle scattering experiment. Describe this model.

Exam points

Read A as the number of protons plus neutrons and Z as the proton number; obtain the neutron count from N = A − Z.
Balance total nucleon number and signed charge separately in a supplied nuclear equation; report both A and Z when both are requested.
Identify a neutron by A = 1 and Z = 0, and a proton by A = 1 and Z = 1, using the actual reactants and products.
Recognize an alpha particle as ⁴₂He: in the supplied D–T fusion equation, balancing A and Z leaves helium-4 plus a neutron.
In beta-minus decay the daughter’s A is unchanged and Z increases by one; include the emitted electron’s negative charge in equation balance.
In alpha decay the daughter’s A decreases by four and Z by two; Ra with Z = 88 therefore produces Rn with Z = 86.
Include the incoming neutron when counting fission products: ²³⁵U plus one neutron has total A = 236 before splitting.
Track A and Z through each stated decay-chain branch; beta-minus changes Z but not A, while each alpha emission removes four nucleons and two protons.
Give the particular number asked for explicitly: a nuclide symbol alone may not earn the mark for a requested nucleon or proton number.

[Maximum number: 2]

This question is in two parts. Part 1 is about kinematics and gravitation. Part 2 is about radioactivity.

Part 1 Kinematics and gravitation

A ball is released near the surface of the Moon at time t=0. The point of release is on a straight line between the centre of Earth and the centre of the Moon. The graph below shows the variation with time t of the displacement s of the ball from the point of release.

(a)

Part 2 Radioactivity

Radium-224 $\left({ }_{88}^{224} \mathrm{Ra}\right)$ is a radioactive nuclide that decays to form radon-220. Radon-220 is itself radioactive and undergoes a further decay. The table shows the series of radioactive nuclides that are formed as the decays proceed. The series ends with a stable isotope of lead.

For the final thallium nuclide, identify the

[ 2 ]

(i)

nucleon number.

[ 1 ]

(ii)

proton number.

[ 1 ]

Exam points

Explain the evidence chain: discrete spectral wavelengths imply discrete photon energies, equal to differences between discrete atomic energy levels.
Match an emission pattern using differences between allowed energy levels, rather than treating each level itself as a spectral line.
White light passing through low-pressure gas produces an absorption spectrum: dark lines appear against a coloured continuous background.
Absorption raises an electron’s energy; emission lowers it. Equal-frequency absorption and emission can connect the same pair of levels in reverse.
Atomic emission lines can lie outside the visible spectrum; do not identify every emission transition with visible light.
An excited atom still has a bound electron, while ionization leaves the electron unbound; distinguish this from merely occupying a higher bound level.
Recognize atomic spectra as evidence that drove a change in the atomic model; this question tests model-changing evidence rather than circuit laws.

[Maximum number: 3]

The diagram represents the three principal spectral lines in the visible region of the spectrum of atomic hydrogen.

The electron in the hydrogen atom can only occupy certain allowed energy levels.

Outline how the spectral lines provide evidence for the existence of these energy levels.

Exam points

An excited electron emits a photon when moving to a lower energy level; the photon’s energy is the difference between those levels.
For bound-to-bound absorption, the incident photon must match an allowed upward energy gap from the atom’s actual initial level.
Choose a downward arrow for emission before comparing gaps: the smallest gap gives the longest wavelength and the largest gap the highest frequency.
Count distinct photon energies, including cascades, rather than only direct drops; repeated energy gaps produce the same frequency.
Use photon frequency ΔE/h for every allowed downward gap; for adjacent three-level gaps also consider their sum.
Match a cascade spectrum by comparing all energy differences and ordering wavelength inversely with those differences.
For the supplied hydrogen visible series ending at n = 2, higher starting levels give shorter wavelengths approaching a limit.
Use the stated final level and ΔE = hc/λ with the given Bohr formula to identify an initial hydrogen level; do not assume every line ends at ground.
For absorption, identify an upward transition whose energy difference matches hc/λ, converting joules and electronvolts consistently.
The longest emitted wavelength comes from the smallest allowed energy gap; calculate hc/ΔE using the gap, not one level’s binding energy.

[Maximum number: 3]

This question is in two parts.

The energies of the principal energy levels in atomic hydrogen measured in eV are given by the expression

E_{n}=-\frac{13.6}{n^{2}} \text { where } n=1,2,3

The visible lines in the spectrum correspond to electron transitions that end at n=2.

Show that the spectral line of wavelength $\lambda=485 \mathrm{~nm}$ is the result of an electron transition from n=4.

energy difference between levels $=\frac{h c}{\lambda e}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{4.85 \times 10^{-7} \times 1.6 \times 10^{-19}}$ ;
=2.55 eV;
$[3.40-2.55]=0.85=\frac{13.6}{n^{2}}$ to give $n^{2}=16$ ;
n=4;

Marking guidance:

Award [3] for reversed argument.

Exam points

Use E = hf = hc/λ, convert wavelengths to metres, and convert joules to electronvolts when requested; h’s units must match the energy units.
For emission choose a downward transition: the smallest energy gap gives the longest wavelength and the largest gap gives the highest frequency.
When one transition energy equals the sum of two others, add reciprocal wavelengths: 1/λ₁ = 1/λ₂ + 1/λ₃, rather than adding wavelengths.
For two emission lines ending at the same level, subtract their photon energies to find the difference between their initial electron energies.
For hydrogen’s first-excited-to-ground transition, subtract −13.6 eV from −3.4 eV to obtain a photon energy of 10.2 eV.
Find incident photon count from N = IAΔt/(hf) or IAΔtλ/(hc); apply exposure time and area, without detector efficiency for incident photons.
Intensity is photon arrival rate per unit area multiplied by photon energy; doubling wavelength and doubling that rate leaves intensity unchanged.
For electrons accelerated from rest through V, energy conservation limits one photon to eV, giving λmin = hc/(eV); read cutoff rather than a peak.

[Maximum number: 2]

The diagram represents the three principal spectral lines in the visible region of the spectrum of atomic hydrogen.

The electron in the hydrogen atom can only occupy certain allowed energy levels.

Determine the difference in energy between the two levels from which electron transitions give rise to the $\mathrm{H}_{\alpha}$ and $\mathrm{H}_{\gamma}$ spectral lines respectively.

difference in energy $=hc\left[\frac{1}{\lambda_{\gamma}}-\frac{1}{\lambda_{\alpha}}\right]$ ;
$=6.6 \times 10^{-34} \times 3 \times 10^{8} \times 10^{7} \times 0.0780=1.54 \times 10^{-19} \mathrm{~J}$ ;
or
$E_{\alpha}=\frac{hc}{\lambda_{\alpha}}=3.02 \times 10^{-19} \mathrm{~J}$ or $1.89 \mathrm{~eV}$ ;
$E_{\gamma}=\frac{hc}{\lambda_{\gamma}}=4.56 \times 10^{-19} \mathrm{~J}$ or $2.85 \mathrm{~eV}$ ;
difference in energy $=1.54 \times 10^{-19} \mathrm{~J}$ or $0.963 \mathrm{~eV}$ ;

Exam points

Identify elements by matching observed absorption or emission line wavelengths with known reference spectra; each element has its own line pattern.
Explain the full stellar method: gases absorb specific wavelengths, the absorption spectrum is observed on Earth, and its lines are compared with known spectra.
Confirm hydrogen or helium by matching the star’s spectral lines to that element’s lines; a fusion prediction alone is not empirical confirmation.
Obtain a laboratory hydrogen spectrum with a discharge tube or hot hydrogen gas, disperse its light using a prism or grating, then observe it on a screen.

[Maximum number: 2]

The diagram shows a simplified energy-balance model for the Earth surface-atmosphere system.

The following data are given:

\begin{aligned} \text { Average albedo of Earth } & =0.30 \\ \text { Average global temperature of the surface } & =288 \mathrm{~K} \\ \text { Average Earth-Sun distance } & =1.5 \times 10^{11} \mathrm{~m} \end{aligned}

The primary energy source of the Sun is the proton-proton (p-p) chain of fusion reactions. Four protons and two electrons produce a helium nucleus together with neutrinos and gamma photons. The overall reaction is:

4{ }_{1}^{1} \mathrm{p}+2{ }_{-1}^{0} \mathrm{e} \rightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{0}^{0} v_{e}+4 \gamma

Outline how the presence of helium in the Sun can be confirmed empirically.

[Maximum number: 3]

(a)

State how the density of a nucleus varies with the number of nucleons in the nucleus.

[ 1 ]

(b)

Show that the nuclear radius of phosphorus-31 $\left({ }_{15}^{31} \mathrm{P}\right)$ is about 4 fm .

[ 1 ]

$R=1.20\times10^{-15}\times31^{\frac{1}{3}}=3.8\times10^{-15}\ \mathrm{m}$

(c)

$\quad{ }_{15}^{32} \mathrm{P}$ is formed when a nucleus of deuterium $\left({ }_{1}^{2} \mathrm{H}\right)$ collides with a nucleus of ${ }_{15}^{31} \mathrm{P}$ . The radius of a deuterium nucleus is 1.5 fm .

[ 1 ]

(i)

State the maximum distance between the centres of the nuclei for which the production of ${ }_{15}^{32} \mathrm{P}$ is likely to occur.

[ 1 ]

[Maximum number: 2]

In an experiment, alpha particles of initial kinetic energy 5.9 MeV are directed at stationary nuclei of lead-207. The distance of closest approach is about:

4\times10^{-14}\ \mathrm{m}

The radius of a nucleus of ${ }_{82}^{207} \mathrm{~Pb}$ is $7.1 \times 10^{-15} \mathrm{~m}$ . Suggest why there will be no deviations from Rutherford scattering in the experiment described above.

[Maximum number: 2]

In an experiment, alpha particles of initial kinetic energy 5.9 MeV are directed at stationary nuclei of lead $\left({ }_{82}^{207} \mathrm{~Pb}\right)$ . Show that the distance of closest approach is about $4 \times 10^{-14} \mathrm{~m}$ .

2e AND 82e seen

OR

$3.2 \times 10^{-19}$ «C» AND $1.312 \times 10^{-17}$ «C» seen
$d=\frac{8.99 \times 10^{9} \times(2 e)(82 e)}{5.9 \times 10^{6} \times e}=3.998 \times 10^{-14} \approx 4 \times 10^{-14}<\mathrm{m} \gg$

Must see either clear substitutions or answer to at least 4 s.f. for MP2.

[Maximum number: 1]

This question is in two parts.

The energies of the principal energy levels in atomic hydrogen measured in eV are given by the expression

E_{n}=-\frac{13.6}{n^{2}} \text { where } n=1,2,3

The visible lines in the spectrum correspond to electron transitions that end at n=2.

Calculate the energy of the level corresponding to n=2.

[Maximum number: 4]

(a)

Bohr modified the Rutherford model by introducing the condition $m v r=n \frac{h}{2 \pi}$ .

Outline the reason for this modification.

[ 3 ]

the electrons accelerate and so radiate energy;
they would therefore spiral into the nucleus/atoms would be unstable

electrons have discrete/only certain energy levels

the only orbits where electrons do not radiate are those that satisfy the Bohr condition $mvr=n\frac{h}{2\pi}$

(b)

Calculate the electron's orbital radius in (c)(ii).

[ 1 ]

$\begin{aligned} r=\frac{(6.63\times10^{-34})^2}{4\pi^2\times8.99\times10^9\times9.11\times10^{-31}\times(1.6\times10^{-19})^2} \\ r=5.3\times10^{-11}\ \mathrm{m} \end{aligned}$

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