AQA A-Level Physics Paper 1, June 2025: Question 33
1 mark · Medium difficulty · Multiple Choice
Determine how many different photon frequencies are emitted as hydrogen atoms excited by 12.74 eV photons return to their ground state.
Practise this questionQuestion
Question text
33 The diagram shows some of the energy levels for a hydrogen atom.
Photons of energy 12.74 eV excite hydrogen atoms that are all initially in the
ground state (n = 1).
How many different photon frequencies are emitted as the atoms return to their
ground state?
[1 mark]
A 3
B 5
C 6
D 9
Mark scheme
Show the mark scheme
33 C 6 AO1
How to answer it
Atomic Energy Levels & Photon Emission
What this question tests
This question assesses your ability to determine the excitation of an electron via photon absorption and to systematically count all possible radiative de-excitation transitions (cascade emission) from an excited state down to the ground state.
Determining Number of Emitted Photon Frequencies
✅ Correct Answer
C — 6
The absorbed photon excites the electron from n = 1 to n = 4 . From n = 4 down to n = 1 , there are 6 distinct transitions possible.
💡 Key Knowledge
- Excitation by Photons: A photon is only absorbed if its energy precisely matches the difference between two discrete levels: ΔE = Efinal − Einitial .
- Emission: When returning to the ground state, an electron can de-excite directly or cascade through any intermediate levels.
- Combinatorics Rule: For an upper state n returning to ground state ( n = 1 ), the number of possible downward transitions is given by:
Total = n(n − 1) / 2
🔧 Step-by-Step Calculation
- Find the excited state energy:
Starting at ground level ( n = 1 ): E₁ = −13.59 eV
Excitation energy absorbed: ΔE = +12.74 eV
New level: E = −13.59 + 12.74 = −0.85 eV - Identify the quantum level:
Looking at the energy level diagram, −0.85 eV corresponds exactly to n = 4 . - List every unique downward transition:
• From n = 4 : 4 → 3 (0.66 eV), 4 → 2 (2.55 eV), 4 → 1 (12.74 eV) — 3 paths
• From n = 3 : 3 → 2 (1.89 eV), 3 → 1 (12.08 eV) — 2 paths
• From n = 2 : 2 → 1 (10.19 eV) — 1 path - Calculate total unique transitions:
3 + 2 + 1 = 6 distinct energy drops.
Because each energy drop has a unique ΔE , each produces a unique frequency according to f = ΔE / h .
❌ Common Traps & Errors
- Only counting direct steps (Answer A = 3): Thinking the electron can only cascade 4 → 3 → 2 → 1 (3 steps) or only drop directly back to ground. Remember, in a gas of many atoms, all possible routes occur simultaneously across the sample.
- Misreading negative signs: Adding 12.74 eV incorrectly to −13.59 eV , leading to an incorrect level such as n = 3 or n = 5 .
- Confusing photon excitation with electron collision: An incoming colliding electron only needs kinetic energy ≥ ΔE , but a photon must equal ΔE exactly.
🧠 Exam Technique & Shortcut
- Quick combination formula: For transitions between n levels, the number of distinct lines is simply:
N = ⁿC₂ = n(n − 1) / 2
For n = 4 : 4 × 3 / 2 = 6 .
This saves over a minute in multiple-choice exams! - Diagram Check: Quickly sketch lines connecting the 4 levels: 3 arrows from top level, 2 from next, 1 from the second-lowest. Sum: 3 + 2 + 1 = 6 .
Topics
Physics · 3.2 Particles and radiation
Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.