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. Elements, bonds and dyes labelled with a letter are fictional, and their data are invented for practice. Data for every question: h = 6.626 × 10⁻³⁴ J s; c = 2.998 × 10⁸ m s⁻¹; N_A = 6.022 × 10²³ mol⁻¹; 1 cm = 10⁻² m; 1 nm = 10⁻⁹ m.
Question 1 (multiple choice · foundation)
A gas absorbs microwave radiation of wavelength 1.25 cm. What is the frequency of this radiation?
- (A) 4.17 × 10⁻¹¹ s⁻¹
- (B) 3.75 × 10⁶ s⁻¹
- (C) 2.40 × 10⁸ s⁻¹
- (D) 2.40 × 10¹⁰ s⁻¹
Answer and explanation
Answer: (D). Convert first: 1.25 cm = 1.25 × 10⁻² m. Then ν = c / λ = (2.998 × 10⁸ m s⁻¹) ÷ (1.25 × 10⁻² m) = 2.40 × 10¹⁰ s⁻¹.
- (A) calculates λ / c, which is upside down; its unit would be s, not s⁻¹.
- (B) multiplies c by λ; the unit would be m² s⁻¹.
- (C) uses 1.25 m instead of 1.25 cm, so it is 100 times too small.
Question 2 (multiple choice · foundation)
Lamp X emits photons of wavelength 300 nm at low brightness. Lamp Y emits photons of wavelength 600 nm at high brightness. Which statement is correct?
- (A) Each photon from X has twice the energy of each photon from Y.
- (B) Each photon from Y has twice the energy of each photon from X.
- (C) The photons have equal energy because Y’s greater brightness makes up for its longer wavelength.
- (D) Photons from X have half the frequency of photons from Y.
Answer and explanation
Answer: (A). E = hc/λ, so energy is inversely proportional to wavelength. Halving the wavelength (600 nm → 300 nm) doubles the energy per photon.
- (B) reverses the relationship: a longer wavelength means less energy per photon.
- (C) confuses brightness (number of photons) with the energy of each photon.
- (D) reverses c = λν: the shorter wavelength has the higher frequency (twice Y’s).
Question 3 (multiple choice · core)
What is the energy of one mole of photons of wavelength 380 nm?
- (A) 5.23 × 10⁻¹⁹ kJ mol⁻¹
- (B) 315 kJ mol⁻¹
- (C) 3.15 × 10⁵ kJ mol⁻¹
- (D) 3.15 × 10⁻⁷ kJ mol⁻¹
Answer and explanation
Answer: (B). Per photon: E = hc/λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ (3.80 × 10⁻⁷ m) = 5.228 × 10⁻¹⁹ J. Per mole: 5.228 × 10⁻¹⁹ J × 6.022 × 10²³ mol⁻¹ = 3.148 × 10⁵ J mol⁻¹ = 315 kJ mol⁻¹.
- (A) is the energy of one photon (in J), given the wrong unit; it was never multiplied by N_A.
- (C) is the answer in J mol⁻¹ labelled as kJ mol⁻¹; it misses the ÷ 1000.
- (D) uses λ = 380 m (no nm → m conversion), giving an answer 10⁹ times too small.
Question 4 (calculation · core)
The bond in a fictional diatomic molecule, A–B, has a bond energy of 310 kJ mol⁻¹. One photon can break one bond.
(a) Calculate the minimum energy, in joules, needed to break one A–B bond. (b) Calculate the longest wavelength of light that can break the bond, in nm. (c) State the region of the spectrum this wavelength is in.
Worked solution
- (a) Per bond: 310 kJ mol⁻¹ × 1000 J kJ⁻¹ ÷ 6.022 × 10²³ mol⁻¹ = 5.15 × 10⁻¹⁹ J.
- (b) λ = hc / E = (6.626 × 10⁻³⁴ J s × 2.998 × 10⁸ m s⁻¹) ÷ 5.148 × 10⁻¹⁹ J = 3.859 × 10⁻⁷ m = 386 nm.
- (c) 386 nm is just shorter than 400 nm, so it is in the ultraviolet (near the violet edge of the visible).
Why “longest”? A longer wavelength means a less energetic photon. Any wavelength longer than 386 nm gives photons with less than 5.15 × 10⁻¹⁹ J, which cannot break the bond on their own.
Suggested mark points (3): 1 for dividing by N_A (with kJ → J) to get the energy per bond; 1 for 386 nm from E = hc/λ with λ converted to nm; 1 for ultraviolet. Carry forward an error from (a) into (b).
Common error: using 310 000 J in E = hc/λ without dividing by N_A gives λ ≈ 6.41 × 10⁻³¹ m, a meaningless wavelength far smaller than an atomic nucleus.
Question 5 (constructed response · core)
Atoms of a fictional element, R, have a ground level (level 1) and two excited levels (levels 2 and 3). In the emission spectrum of R:
- the drop from level 3 to level 1 gives a line at 300. nm;
- the drop from level 3 to level 2 gives a line at 655 nm.
(a) Calculate the energy of level 3 above the ground level, in J. (b) Calculate the energy of level 2 above the ground level, in J. (c) Calculate the wavelength of the line for the drop from level 2 to level 1, and state its region. (d) Ground-state R atoms are lit with white light. State one wavelength they absorb, and explain why they do not absorb 655 nm light.
Worked solution and suggested Marlbridge rubric
(a) E(3 → 1) = hc/λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ 3.00 × 10⁻⁷ m = 6.62 × 10⁻¹⁹ J above ground.
(b) E(3 → 2) = hc ÷ 6.55 × 10⁻⁷ m = 3.033 × 10⁻¹⁹ J. Level 2 lies this far below level 3, so its energy above ground is 6.622 × 10⁻¹⁹ − 3.033 × 10⁻¹⁹ = 3.59 × 10⁻¹⁹ J.
(c) λ = hc / E = 1.9865 × 10⁻²⁵ J m ÷ 3.589 × 10⁻¹⁹ J = 5.535 × 10⁻⁷ m = 554 nm, in the visible region (green-yellow).
(d) From level 1 they can absorb 554 nm (to level 2) or 300 nm (to level 3, though 300 nm is ultraviolet, so white light from an ordinary visible lamp may contain little of it). They do not absorb 655 nm, because 3.03 × 10⁻¹⁹ J is the gap between levels 2 and 3, and ground-state atoms are not in level 2. No gap from level 1 equals that energy.
| Point | What earns it |
|---|---|
| 1 | 6.62 × 10⁻¹⁹ J for level 3, with nm → m conversion |
| 1 | Level 2 found by subtracting the 3 → 2 photon energy from level 3’s energy |
| 1 | 554 nm (accept 553–554 nm from rounded intermediate values) and visible |
| 1 | 554 nm (or 300 nm) absorbed, and 655 nm explained: it matches the 2 → 3 gap, but the atoms start in level 1 |
Alternative method for (c): 1/λ(2 → 1) = 1/λ(3 → 1) − 1/λ(3 → 2), because energy is proportional to 1/λ. This gives the same 554 nm.
Question 6 (calculation · core)
In a photochemistry experiment, a lamp delivers 1.50 J of ultraviolet light, all of wavelength 365 nm, to a solution of a fictional light-sensitive compound, S.
(a) Calculate the energy of one photon of this light. (b) Calculate the number of photons delivered, and the amount in moles of photons. (c) Only 40.0% of the photons are absorbed, and each absorbed photon converts one molecule of S into a new compound. Calculate the amount, in mol, of S converted.
Worked solution and suggested Marlbridge rubric
(a) E = hc/λ = (6.626 × 10⁻³⁴ J s × 2.998 × 10⁸ m s⁻¹) ÷ 3.65 × 10⁻⁷ m = 5.44 × 10⁻¹⁹ J per photon.
(b) Number of photons = 1.50 J ÷ 5.442 × 10⁻¹⁹ J per photon = 2.76 × 10¹⁸ photons. Moles of photons = 2.756 × 10¹⁸ ÷ 6.022 × 10²³ mol⁻¹ = 4.58 × 10⁻⁶ mol.
(c) Absorbed photons = 0.400 × 4.577 × 10⁻⁶ mol = 1.83 × 10⁻⁶ mol, so 1.83 × 10⁻⁶ mol of S is converted (one molecule per photon).
| Point | What earns it |
|---|---|
| 1 | Photon energy 5.44 × 10⁻¹⁹ J with nm → m conversion |
| 1 | Number of photons = total energy ÷ energy per photon |
| 1 | Moles of photons by dividing by N_A |
| 1 | 1.83 × 10⁻⁶ mol, using the 40.0% and the 1 : 1 photon-to-molecule ratio |
Dimensional analysis check for (b): J ÷ (J per photon) = photons; photons ÷ (photons mol⁻¹) = mol.
Question 7 (explanation · stretch)
A fictional dye, D, has an electronic transition with an energy gap of 3.60 × 10⁻¹⁹ J. A student has two lamps:
- a dim lamp emitting light of wavelength 552 nm;
- a very bright lamp emitting light of wavelength 690 nm.
The student claims: “The bright lamp delivers far more energy each second, so it will cause more D molecules to make this transition than the dim lamp.”
Evaluate the claim. Support your answer with a calculation.
Model answer and suggested Marlbridge rubric
Calculation.
- 552 nm photon: E = hc/λ = 1.9865 × 10⁻²⁵ J m ÷ 5.52 × 10⁻⁷ m = 3.60 × 10⁻¹⁹ J. This matches the gap.
- 690 nm photon: E = 1.9865 × 10⁻²⁵ ÷ 6.90 × 10⁻⁷ = 2.88 × 10⁻¹⁹ J. This is less than the gap, and matches no transition here.
Evaluation. The claim is incorrect. A molecule absorbs one photon at a time, and the photon’s energy must equal the gap. Each 690 nm photon carries too little energy, however many arrive each second, so the bright lamp causes none of these transitions. Every 552 nm photon has exactly the right energy, so the dim lamp causes some transitions. Brightness only changes the number of photons per second, which affects how many molecules make the transition, not whether the transition is possible.
| Point | What earns it |
|---|---|
| 1 | Correct photon energies for both lamps (3.60 × 10⁻¹⁹ J and 2.88 × 10⁻¹⁹ J) |
| 1 | Photon energy must match the gap; one photon per transition |
| 1 | Claim rejected: brightness changes the number of photons, not the energy of each photon |
Do not award the second point for “the 690 nm light has less energy” alone: the bright lamp may deliver more total energy; the point is the energy of each photon.
How did you do?
- Q1 or Q3 wrong: check your unit conversions (cm and nm to m) and the per photon → per mole step in the study guide, Worked example 1.
- Q2 or Q7 wrong: re-read “Brightness is not photon energy” and the proportions in the revision notes.
- Q4 or Q6 incomplete: write units on every line so that J, photons and mol cancel correctly.
- Q5 incomplete: practise subtracting level energies with Figure 1 and Worked example 2.
Then tick off the topic checklist.