AQA A-Level Physics Paper 1, June 2022: Question 15

1 mark · Easy difficulty · Multiple Choice

Identify the correct description of the function of the coating in a fluorescent tube containing a gas.

Practise this question

Question

Multiple choice question 15 stating: A fluorescent tube contains a gas. The coating of the tube [1 mark]. Options are A: becomes ionised by the gas and emits photons of ultraviolet light. B: absorbs photons of ultraviolet light from the gas and emits visible light. C: absorbs photons of ultraviolet light from the gas and emits photoelectrons. D: absorbs several photons of visible light from the gas and then emits one photon of ultraviolet light.
Question text

15 A fluorescent tube contains a gas.

The coating of the tube

[1 mark]

A becomes ionised by the gas and emits photons of ultraviolet light.

B absorbs photons of ultraviolet light from the gas and emits visible light.

C absorbs photons of ultraviolet light from the gas and emits photoelectrons.

D absorbs several photons of visible light from the gas and then emits one

photon of ultraviolet light.

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is B (AO1), corresponding to the statement that the coating absorbs photons of ultraviolet light from the gas and emits visible light.

15 B (AO1) absorbs photons of ultraviolet light from the gas and emits visible light.

How to answer it

The Physics of Fluorescent Tubes

What this question tests

This question assesses your core knowledge of the mechanism behind fluorescent lighting (AO1). It tests your understanding of atomic energy levels, photon absorption and emission, and the crucial role played by the phosphor coating inside a fluorescent tube in converting invisible ultraviolet radiation into visible light.

Question 1.5

The Role of the Tube Coating

✅ Correct Answer: Option B

absorbs photons of ultraviolet light from the gas and emits visible light.

Mark: 1 / 1 (AO1)

💡 Key Knowledge

  • Free electrons collide with mercury atoms inside the tube, exciting mercury electrons to higher energy levels.
  • When these mercury electrons drop back down, they release energy as ultraviolet (UV) photons.
  • The coating (phosphor) on the inside of the tube absorbs these high-energy UV photons.
  • Through a process of de-excitation across smaller energy steps, the coating re-emits the energy as lower-energy visible light photons.

🧠 Exam Technique

  • Energy Conservation Check: UV photons have higher frequency and energy than visible light. Since energy cannot be created from nothing, a high-energy UV photon must be absorbed to produce a lower-energy visible light photon. This immediately eliminates options involving energy creation in reverse (like option D).
  • Process Elimination: Fluorescent tubes rely on a phosphor coating to convert invisible UV light into visible light. They do not emit photoelectrons to create light (eliminating C).

❌ Common Errors

  • Confusing the direction of energy conversion, incorrectly thinking visible light is absorbed to create UV light (Option D).
  • Assuming the tube coating undergoes photoelectric emission (Option C), confusing fluorescence with the photoelectric effect.
  • Mixing up which component generates the UV light (the gas/mercury vapor) versus which component absorbs it and emits visible light (the phosphor coating).

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.