AQA AS Level Physics Paper 1, June 2025: Question 3

12 marks · Medium difficulty · Extended Answer

Calculate the de Broglie wavelength of an electron after inelastic collision with a mercury atom, determine the photon wavelength and emission rate, and explain the emission of white light by a fluorescent tube coating.

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Question

Question 3 presents an energy level diagram for a mercury atom showing levels n = 1 (-16.64 x 10^-19 J), n = 2 (-8.80 x 10^-19 J), n = 3 (-5.92 x 10^-19 J), and n = 4 (-2.56 x 10^-19 J). Question 03.1 asks to calculate the de Broglie wavelength of a colliding electron with initial kinetic energy 13.10 x 10^-19 J that excites the atom to n = 3 (4 marks). Question 03.2 asks to calculate the wavelength in nm of a 7.83 x 10^-19 J ultraviolet photon (2 marks). Question 03.3 asks to calculate the number of UV photons emitted per second given an 18 W tube operating at 82% efficiency (2 marks). Question 03.4 asks for an explanation of how the fluorescent tube coating produces white light and why more visible photons are emitted than ultraviolet photons (4 marks).
Question text

03 A fluorescent tube contains a low-pressure gas consisting of mercury atoms.

Figure 3 shows the four lowest energy levels of a mercury atom.

Figure 3

A mercury atom inside the fluorescent tube is in its ground state n = 1

An electron is travelling inside the tube. The electron has a kinetic energy of

13.10 × 10−19 J when it collides with the mercury atom.

As a result of the collision, the mercury atom is excited to its n = 3 energy level.

Immediately after the collision, the electron is moving with a speed that is smaller than

before the collision.

Assume that the kinetic energy of the mercury atom does not change.

03.1 Calculate the de Broglie wavelength of this electron immediately after the collision.

[4 marks]

de Broglie wavelength = m

The excited mercury atoms inside the fluorescent tube can emit photons of

ultraviolet light, each of energy 7.83 × 10−19 J.

03.2 Calculate, in nm, the wavelength of a photon with this energy.

*07* [2 marks]

wavelength = nm

03.3 The fluorescent tube has an input power of 18 W.

82% of the input power is converted into ultraviolet photons, each of

energy 7.83 × 10−19 J.

Calculate the number of these photons emitted in one second.

[2 marks]

9number of photons =

03.4 Explain how the coating of the fluorescent tube emits white light.

In your answer you should:

• describe the processes that take place in the coating

• suggest why there are many more visible photons than ultraviolet photons emitted

by the tube.

[4 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 03 detailing marks for each subsection: 03.1 awards 4 marks for finding energy absorbed (10.72 x 10^-19 J), remaining kinetic energy (2.38 x 10^-19 J), calculating momentum or velocity, and finding wavelength (1.01 x 10^-9 m). 03.2 awards 2 marks for using E = hc/λ and converting to nanometres to get 254 nm. 03.3 awards 2 marks for finding 82% of 18 W (14.76 W) and dividing by photon energy to obtain 1.9 x 10^19 photons per second. 03.4 awards 4 marks for explaining photon absorption, excitation and de-excitation of coating electrons, multiple energy transitions producing white light, smaller energy differences leading to longer wavelengths, and one UV photon producing multiple visible photons.

Question Answers Additional Comments/Guidance Mark AO

03.1 Any three from 4 2 × AO2

• (energy absorbed by atom =) 16.64 – 5.92 In 2nd bullet point: condone 13.1 – their energy 2 × AO1

−19 absorbed by the atom

= 10.72 (× 10 ) (1)

−19 Expect to see speed of electron immediately

• (Ek of scattered electron =) 13.1 – 10.72 = 2.38 (× 10 ) (2) 5 -1

after collision = 7.23 × 10 ms

• Use of Ek = m v by substitution with an Ek OR Expect to see momentum of electron

2 immediately after collision = 6.6 × 10-25 kg ms-1

rearrangement where v is subject

OR their Ek and speed

p2

Ek = by substitution with an Ek OR rearrangement

2m

where p is subject (3)

h h

• Use of λ = OR λ = with their v or their p (4)

mv p

Accept a correct answer given to at least

(de Broglie wavelength =) 1.01 × 10−9 (m) 2 significant figures

Calculator value 1.00681775 × 10−9

03.2 Any one from 2 1 × AO1

hc Allow use of h = 6.6 × 10−34 J s where answer is

(1) 1 × AO2

• use of E = seen as 253 nm if use already penalised in

λ

03.1.

• use of E = h f and c = f λ (2) −7

Expect to see λ = 2.54 × 10

• conversion of their wavelength in metres to

nanometres seen (3)

Accept a correct answer given to at least

(wavelength = ) 254 (nm) 2 significant figures.

Calculator value 2.540229885 × 10−7

03.3 Any one from 2 1 × AO1

• calculates 82% of 18 W Expect to see 14.76 (W) 1 × AO2

18 x 0.82 is insufficient on its own.

−19 Accept any answer that rounds to two

• divides their energy per second by 7.83 × 10 significant figure answer of 1.9 × 1019

(Number of photons per second = ) 1.9 × 1019

03.4 Any two from 4 2 × AO1

• idea of photon being absorbed (by atoms in the coating) (1) 2 × AO3

• idea that (atomic) electron (in the coating) moves to a higher

energy level and returns to lower energy level (emitting a

photon)

OR Condone:

Idea of excitation (of atom in coating) followed by de- Electrons (in the coating) excite then de-excite

excitation (emitting a photon) (emitting a photon)

(2)

• idea that there are many different energy transitions that

16 lead to the production of white light (3)

Accept lower frequency in place of longer

Energy difference between energy levels (in the coating) is wavelength

smaller (than the difference in energy levels in mercury atom)

therefore emits a radiation with a longer wavelength (4)

Idea that one uv photon being absorbed can produce several

lower energy photons when energy is re-emitted (5)

Total 12

How to answer it

Quantum Phenomena: Collisions, de Broglie Wavelength & Fluorescent Tubes

AQA AS Level Physics • Paper 1 / Section 3 • Total: 12 Marks

📋 What this question tests

  • Atomic excitation by electron collision: Energy conservation during inelastic collisions between free electrons and bound atomic electrons.
  • Wave-particle duality: Calculating momentum and the de Broglie wavelength ( λ = h / p ) of moving particles.
  • Photon energy & electromagnetic spectrum: Converting photon energy to wavelength ( E = hc / λ ) and working with nanometre ( nm ) conversions.
  • Power, efficiency, and photon flux: Calculating the rate of photon emission given percentage efficiency and energy per photon.
  • Mechanism of a fluorescent tube: Step-by-step description of UV absorption, excitation, and downward cascade transitions in the phosphor coating producing visible light.
Part 03.1 • 4 Marks

de Broglie Wavelength of the Scattered Electron

Calculate the de Broglie wavelength of the electron immediately after exciting the mercury atom.

📐 Step-by-Step Calculation

1 Energy transferred to the atom:
Excitation from n = 1 to n = 3 :
ΔE = (-5.92 × 10⁻¹⁹) - (-16.64 × 10⁻¹⁹) = 1.072 × 10⁻¹⁸ J (or 10.72 × 10⁻¹⁹ J )

2 Remaining kinetic energy of the electron:
Ek = Einitial - ΔE = 13.10 × 10⁻¹⁹ - 10.72 × 10⁻¹⁹ = 2.38 × 10⁻¹⁹ J

3 Find speed ( v ) or momentum ( p ):
Using Ek = ½ m v² (where electron mass m = 9.11 × 10⁻³¹ kg ):
v = √[ 2 × 2.38 × 10⁻¹⁹ / (9.11 × 10⁻³¹) ] = 7.23 × 10⁵ m s⁻¹
Alternatively, p = √(2 m Ek) = 6.586 × 10⁻²⁵ kg m s⁻¹

4 Calculate de Broglie wavelength:
λ = h / (m v) = (6.63 × 10⁻³⁴) / (9.11 × 10⁻³¹ × 7.23 × 10⁵) = 1.01 × 10⁻⁹ m

✅ Final Answer

de Broglie wavelength: 1.01 × 10⁻⁹ m (accept 1.0 × 10⁻⁹ m to 2 s.f., exact calculator value 1.007 × 10⁻⁹ m ).

[4 marks total] Awarded for:
  • Calculation of energy absorbed ( 10.72 × 10⁻¹⁹ J ) [1]
  • Calculation of scattered Ek ( 2.38 × 10⁻¹⁹ J ) [1]
  • Use of Ek = ½mv² or p = √(2mEk) [1]
  • Use of λ = h/p leading to 1.01 × 10⁻⁹ m [1]

🧠 Exam Technique

Always keep full precision in your calculator memory between intermediate steps. Rounding speed prematurely to 7.2 × 10⁵ m s⁻¹ can cause rounding discrepancies in the final decimal place.

❌ Common Errors

  • Using the initial kinetic energy ( 13.10 × 10⁻¹⁹ J ) to calculate wavelength instead of the energy remaining after collision.
  • Confusing de Broglie wavelength of a particle ( λ = h / p ) with photon wavelength ( λ = hc / E ). Never use speed of light c for an electron!
  • Incorrectly reading energy level: transferring to n = 2 or n = 4 instead of n = 3 .
Part 03.2 • 2 Marks

Wavelength of Emitted UV Photon

Calculate, in nm, the wavelength of a photon with energy 7.83 × 10⁻¹⁹ J.

📐 Step-by-Step Calculation

1 Photon energy relationship:
E = h f = hc / λ ⇒ λ = hc / E

2 Substitute fundamental constants:
λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (7.83 × 10⁻¹⁹)
λ = 2.540 × 10⁻⁷ m

3 Convert metres to nanometres ( nm ):
Since 1 nm = 10⁻⁹ m :
λ = 2.540 × 10⁻⁷ / 10⁻⁹ = 254 nm

✅ Final Answer

Wavelength: 254 nm (or 250 nm to 2 s.f.).

[2 marks total]
• Mark 1: Correct use of E = hc / λ (or E = hf and c = fλ ).
• Mark 2: Correct answer in nm ( 254 nm ).

❌ Common Errors

  • Leaving the answer in metres ( 2.54 × 10⁻⁷ m ) and forgetting the unit request specified in the question stem ( nm ).
  • Multiplying by 10⁻⁹ instead of dividing when converting from m to nm .

🧠 Sanity Check

Ultraviolet radiation lies between 10 nm and 400 nm . Visible light is 400–700 nm . An answer of 254 nm fits squarely in the UV-C band, confirming your result is physically sensible.

Part 03.3 • 2 Marks

Photon Emission Rate

Calculate the number of UV photons emitted in one second for an 18 W tube operating at 82% efficiency.

📐 Step-by-Step Calculation

1 Useful UV power output:
PUV = 0.82 × 18 W = 14.76 W
(This means 14.76 J of UV energy emitted each second).

2 Number of photons per second:
Number = Total Energy / Energy per photon
N = 14.76 / (7.83 × 10⁻¹⁹) = 1.885 × 10¹⁹ s⁻¹
N = 1.9 × 10¹⁹ (to 2 s.f.)

✅ Final Answer

Number of photons: 1.9 × 10¹⁹ (accept values that round to 1.9 × 10¹⁹ ).

[2 marks total]
• Mark 1: Calculates 82% of 18 W ( 14.76 W ). Note: Writing only "18 × 0.82" without working out the value is insufficient if the final answer is wrong.
• Mark 2: Divides their useful power by 7.83 × 10⁻¹⁹ J .

💡 Key Formulae

Power (P) = Etotal / t . For t = 1 s , Etotal = P × 1 .

Total energy emitted by N identical photons: Etotal = N × Ephoton .

❌ Common Errors

Dividing the full 18 W by photon energy and forgetting the 82% efficiency factor, yielding an incorrect answer of 2.3 × 10¹⁹ .

Part 03.4 • 4 Marks

Mechanism of the Fluorescent Coating

Explain how the coating of the fluorescent tube emits white light.

💡 Core Physics Process

  1. Absorption: Ultraviolet photons emitted by de-exciting mercury atoms are absorbed by the atoms/electrons in the fluorescent phosphor coating.
  2. Excitation: Electrons in the coating atoms are raised to higher atomic energy states.
  3. De-excitation / Cascade: Electrons de-excite in multiple smaller downward steps, emitting photons of lower energy (longer wavelengths in the visible spectrum).
  4. White Light Generation: Different transitions release a range of visible wavelengths; their mixture produces white light.

✅ Mark Scheme Breakdown

Award any four points from the following:

  • [1] Absorption: UV photons are absorbed by atoms in the coating.
  • [2] Excitation & De-excitation: Atomic electrons in the coating are excited to higher energy levels and then return/de-excite to lower levels (emitting photons).
  • [3] Many transitions: There are many different energy level transitions in the coating, producing a spectrum of visible frequencies that combine to form white light.
  • [4] Lower energy / longer wavelength: The energy differences (ΔE) between levels in the coating are smaller than in the mercury atom, so emitted photons have lower frequencies / longer wavelengths.
  • [5] Photon multiplication: One absorbed high-energy UV photon causes a cascade of transitions, emitting several lower-energy visible photons.

🧠 How to Structure a 4-Mark Explanation

Address both bullet points given in the question prompt systematically:

Bullet 1 (Processes in coating): State absorption of UV → excitation of electrons → de-excitation via intermediate states emitting visible light.

Bullet 2 (Why more visible photons than UV): State that visible photons have lower energy than UV photons ( E = hf ). By energy conservation, one high-energy UV photon can produce multiple lower-energy visible photons as electrons fall down through stepped intermediate levels.

❌ Common Errors

  • Confusing the gas mechanism (collisional excitation by free beam electrons) with the coating mechanism (photo-excitation by absorbed UV photons).
  • Stating that "electrons emit light when going up" instead of during downward de-excitation.
  • Vaguely saying "it reflects visible light" rather than explaining absorption, excitation, and re-emission.

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.