These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: h = 6.63 × 10⁻³⁴ J·s = 4.14 × 10⁻¹⁵ eV·s; c = 3.00 × 10⁸ m/s; 1 eV = 1.60 × 10⁻¹⁹ J; hc ≈ 1.24 × 10³ eV·nm, so λ (nm) ≈ 1240 ÷ E (eV). Visible light runs from about 400 nm to 700 nm. All atoms and gases in these questions are invented for practice.
Atom M (used in Questions 1, 3, 4 and 5) is a single-electron atom with these energy levels: n = 1: −8.0 eV; n = 2: −3.0 eV; n = 3: −1.5 eV; n = 4: −0.5 eV. Zero energy means the electron is free and at rest.
Question 1 (multiple choice · foundation)
An atom M is in its ground state. Which photon can it absorb?
- (A) 2.5 eV
- (B) 4.5 eV
- (C) 5.0 eV
- (D) 6.0 eV
Answer and explanation
Answer: (C). −8.0 eV + 5.0 eV = −3.0 eV, which is level n = 2. The photon energy matches the 1 → 2 gap exactly.
- (A) 2.5 eV is the gap between n = 2 and n = 4. An atom in the ground state cannot use it: −8.0 + 2.5 = −5.5 eV is not a level.
- (B) −8.0 + 4.5 = −3.5 eV is not a level. “Close to” a gap is not enough.
- (D) −8.0 + 6.0 = −2.0 eV lies between n = 2 and n = 3. Having more energy than the 1 → 2 gap does not help, and 6.0 eV is below the 8.0 eV binding energy, so it cannot ionize the atom.
Question 2 (multiple choice · core)
White light with a continuous spectrum passes through a tube of cool gas, then through a diffraction grating. Which statement best describes the spectrum observed?
- (A) Bright lines on a dark background, at every wavelength the gas emits when hot
- (B) A continuous spectrum with dark lines, at wavelengths that also appear in the hot gas’s emission spectrum
- (C) A continuous spectrum with dark lines, at wavelengths that never appear in the gas’s emission spectrum
- (D) A continuous spectrum with no lines, because the atoms re-emit every photon they absorb
Answer and explanation
Answer: (B). Atoms in the cool gas absorb photons whose energies match their gaps, so those wavelengths are weakened and appear dark against the continuous spectrum. The same gaps produce emission lines when the gas is hot, so each dark line matches an emission line (a cool gas usually shows fewer lines than the hot gas emits).
- (A) describes an emission spectrum from a hot, low-density gas, not light passing through a cool gas.
- (C) gets the dark lines right but the link wrong: absorption and emission use the same energy gaps.
- (D) The atoms do re-emit, but in all directions, so little of that light continues along the beam.
Question 3 (multiple choice · core)
What is the wavelength of the photon emitted when atom M drops from n = 4 to n = 2?
- (A) 165 nm
- (B) 496 nm
- (C) 827 nm
- (D) 2480 nm
Answer and explanation
Answer: (B). E = −0.5 − (−3.0) = 2.5 eV, so λ ≈ 1240 ÷ 2.5 = 496 nm (visible).
- (A) is the 4 → 1 line: 7.5 eV gives 165 nm.
- (C) is the 3 → 2 line: 1.5 eV gives 827 nm.
- (D) uses only the energy of level 4 (0.5 eV) instead of the difference between two levels.
Question 4 (multiple choice · core)
An atom M is in the n = 2 state. What is the minimum photon energy that can ionize it?
- (A) 3.0 eV
- (B) 5.0 eV
- (C) 8.0 eV
- (D) 11.0 eV
Answer and explanation
Answer: (A). The binding energy from n = 2 is 0 − (−3.0 eV) = 3.0 eV. Any photon of at least 3.0 eV can free the electron.
- (B) is the 1 → 2 gap, the energy needed to reach n = 2, not to leave the atom.
- (C) is the binding energy from the ground state. The atom is already in n = 2.
- (D) adds 8.0 eV and 3.0 eV, treating both level energies as amounts that must be supplied.
Question 5 (calculation · core)
A discharge tube contains a low-density gas of atom M. Collisions excite some atoms as high as n = 4.
(a) How many different emission lines can the gas produce? List the transitions. (b) Calculate the wavelength of the line with the longest wavelength, and state the part of the electromagnetic spectrum it is in. (c) Show that only one line is visible, and find its frequency.
Worked solution
- (a) Four levels give 4 × 3 ÷ 2 = 6 lines: 4 → 3, 4 → 2, 4 → 1, 3 → 2, 3 → 1, 2 → 1.
- (b) The longest wavelength comes from the smallest gap. Gaps: 4 → 3 = 1.0 eV, 3 → 2 = 1.5 eV, 4 → 2 = 2.5 eV, 2 → 1 = 5.0 eV, 3 → 1 = 6.5 eV, 4 → 1 = 7.5 eV. The smallest is 1.0 eV, so λ ≈ 1240 ÷ 1.0 = 1240 nm (1.24 × 10⁻⁶ m), in the infrared.
- (c) Wavelengths of the others: 3 → 2: 827 nm (infrared); 4 → 2: 496 nm; 2 → 1: 248 nm; 3 → 1: 191 nm; 4 → 1: 165 nm (all ultraviolet). Only 496 nm lies between 400 nm and 700 nm.
- Frequency: f = c/λ = (3.00 × 10⁸ m/s) ÷ (496 × 10⁻⁹ m) = 6.05 × 10¹⁴ Hz. (Or f = E/h = 2.5 eV ÷ 4.14 × 10⁻¹⁵ eV·s = 6.04 × 10¹⁴ Hz.)
Suggested mark points (4): 1 for six lines with the transitions listed; 1 for picking the smallest gap (4 → 3) for the longest wavelength; 1 for 1240 nm, infrared; 1 for identifying 496 nm as the only visible line with f ≈ 6.0 × 10¹⁴ Hz.
Common error: choosing the line from the top level to the ground state (4 → 1) as “longest” because it is the longest arrow on the diagram. A long arrow is a large energy and so a short wavelength.
Question 6 (constructed response · core)
A student has three lamps, each filled with one invented gas. She records the visible emission lines of each:
| Gas | Emission lines (nm) |
|---|---|
| A | 420, 515, 610 |
| B | 455, 540, 670 |
| C | 470, 590 |
She then shines white light through a cool sample of an unknown gas mixture. The spectrum shows dark lines at 420 nm, 470 nm, 515 nm and 590 nm, and no others in the visible range.
(a) Which gas do the data show is in the mixture? Justify. (b) Is there any evidence that gas B is in the mixture? Explain what you can and cannot conclude. (c) Another student says gas A cannot be present, because there is no dark line at 610 nm. Evaluate this claim. (d) Explain why the lines look dark even though the gas atoms soon re-emit the energy they absorb.
Worked solution and suggested Marlbridge rubric
(a) Gas C. Both of its lines, 470 nm and 590 nm, appear as dark lines, and its pattern is fully matched.
(b) No. None of gas B’s lines (455, 540, 670 nm) appears, so there is no evidence for it. This does not fully rule B out: if all three of its visible lines came from drops between excited states, a cool sample would absorb none of them (see part (c)). The absorption spectrum of a pure cool sample of B would settle it.
(c) The claim is not justified. A cool gas has almost all its atoms in the ground state, so it absorbs only transitions that start at the ground state. An emission line such as 610 nm may come from a drop between two excited states. Cool atoms are not in that lower excited state, so they cannot absorb at 610 nm. The 420 nm and 515 nm lines of gas A are both present, which supports gas A being in the mixture. (To be sure, the student could compare with the absorption spectrum of a pure cool sample of gas A.)
(d) The excited atoms re-emit photons in all directions, not only along the beam. So only a small fraction of the absorbed energy at those wavelengths travels on towards the grating, and the line is much dimmer than its surroundings.
| Point | What earns it |
|---|---|
| 1 | Gas C, because both of its lines appear |
| 1 | No evidence for gas B, because none of its lines appear as dark lines |
| 1 | States that a cool gas absorbs mainly from the ground state |
| 1 | Concludes the missing 610 nm line can be explained by a transition between excited states, so the claim is not justified (supported by the 420 and 515 nm lines) |
| 1 | Re-emission is in all directions, so less light continues along the beam at those wavelengths |
Do not award the (c) point for “gas A is present” with no reasoning about the ground state or excited states.
Question 7 (constructed response · stretch)
A different invented single-electron atom has its ground state at −9.0 eV and only two excited states below zero. When the gas is hot, its complete emission spectrum has exactly three lines: 207 nm, 155 nm and 620 nm. The two shorter wavelengths are transitions to the ground state.
(a) Calculate the photon energy of each line in eV. (b) Find the energies of the two excited states and draw the energy level diagram, with each line shown as an arrow. (c) Use your diagram to justify that the 620 nm line is the transition between the two excited states. (d) A photon of 7.0 eV and a photon of 10.0 eV each hit an atom in the ground state. State what happens in each case. For any electron that is freed, give its kinetic energy in eV and in joules.
Worked solution and suggested Marlbridge rubric
(a) E ≈ 1240 ÷ λ: 207 nm → 5.99 eV (≈ 6.0 eV); 155 nm → 8.00 eV; 620 nm → 2.00 eV.
(b) Transitions to the ground state: E₂ = −9.0 + 6.0 = −3.0 eV; E₃ = −9.0 + 8.0 = −1.0 eV. The diagram has lines at −9.0, −3.0, −1.0 and 0 eV, with downward arrows 2 → 1 (207 nm), 3 → 1 (155 nm) and 3 → 2 (620 nm).
(c) E₃ − E₂ = −1.0 − (−3.0) = 2.0 eV, which gives λ ≈ 620 nm. This matches the third line. Also, the energies add up: 6.0 eV + 2.0 eV = 8.0 eV, so the route 3 → 2 → 1 gives out the same total energy as the direct 3 → 1 drop, as energy conservation requires.
(d) 7.0 eV: −9.0 + 7.0 = −2.0 eV is not a level, and 7.0 eV is less than the 9.0 eV binding energy. The photon is not absorbed. 10.0 eV: this exceeds the 9.0 eV binding energy, so the atom is ionized. The electron’s kinetic energy is 10.0 − 9.0 = 1.0 eV = 1.6 × 10⁻¹⁹ J.
| Point | What earns it |
|---|---|
| 1 | All three photon energies correct (6.0, 8.0, 2.0 eV) |
| 1 | E₂ = −3.0 eV and E₃ = −1.0 eV from the transitions to the ground state |
| 1 | Diagram with correctly ordered levels and three downward arrows, each labelled with its wavelength |
| 1 | Justifies 620 nm by the 2.0 eV gap between −1.0 eV and −3.0 eV (or by energy addition) |
| 1 | 7.0 eV photon not absorbed, with a reason (no level at −2.0 eV and below the binding energy) |
| 1 | 10.0 eV photon ionizes the atom; K = 1.0 eV = 1.6 × 10⁻¹⁹ J |
Accept energies found with E = hc/λ in joules then converted (for example 207 nm → 9.6 × 10⁻¹⁹ J → 6.0 eV). Small rounding differences are fine: using 5.99 eV gives E₂ = −3.01 eV and a predicted 3 → 2 line near 617 nm, which agrees with 620 nm within the precision of the data.
How did you do?
- Q1 or Q4 wrong: re-read “Photons in, photons out” and “Binding energy and ionization” in the study guide, then redo Worked example 2.
- Q3 or Q5 wrong: practise E = |ΔE| and λ ≈ 1240 ÷ E with Worked example 1.
- Q2 or Q6 wrong: revisit “Line spectra and what they tell you” and Figure 2.
- Q7 incomplete: draw the diagram first, then read each line as a gap.
Then tick off the topic checklist.