AQA AS Level Physics Paper 1, June 2024: Question 2

8 marks · Medium difficulty · Short Answer

Deduce the potential difference V for electron excitation of mercury atoms, calculate the wavelength of the emitted photon from a 6.7 eV energy change, and state and explain the energy transitions for an atom receiving 18.4 eV of energy.

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Question text

02 A tube contains a vapour of mercury atoms at low pressure. In an experiment, the

vapour is bombarded by a beam of electrons.

An electron in the beam gains 6.7 eV of kinetic energy by moving through a potential

difference V.

02.1 Deduce V.

[1 mark]

V = V

The electron collides with a mercury atom. The atom subsequently emits a photon of

ultraviolet radiation with an energy of 6.7 eV.

02.2 Calculate the wavelength of the emitted photon of this ultraviolet radiation.

[3 marks]

7 wavelength = m

02.3 The experiment is repeated with a different gas.

Figure 2 shows the three lowest energy levels for an atom of the gas.

Figure 2

When an electron in the beam collides with the gas atom, 18.4 eV of energy is

transferred to the atom.

The atom subsequently emits a photon of visible light.

State and explain the energy transitions that are involved.

Support your answer with appropriate calculations.

[4 marks]

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidance Mark AO

02.1 (V =) 6.7 (V) 1 AO1

02.2 Max 2 from 2nd bullet point, Use of: 3 2 × AO1

• Converts 6.7 eV to 1.07(2) × 10−18 (J) by correct rearrangement to make λ 1 × AO2

OR (E =) 6.7 × 1.6 × 10-19 seen Or

substitution of all terms with maximum of one

ℎ𝑐𝑐 POT error.

• Use of E = 𝜆𝜆

3rd bullet point:

Use of E = hf and c = f λ Allow their E = 4.1875 × 1019 without

supporting working seen

Using 6.7 as their E yields an answer =

6.63 ×10−34 × 3×108

• 𝜆𝜆 = 2.97 × 10-26 (m)

𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐸𝐸

𝑂𝑂𝑂𝑂

6.63 ×10−34 × 3×108

𝜆𝜆 =

6.7

Calculator displays: 1.8554104480 × 10-7 (m)

(λ =) 1.9 × 10−7 (m)

02.3 Idea that a transfer of energy to atomic electron causes Accept an arrow drawn from ground state to B. 4 2 × AO2

transition from ground state to B 2 × AO3

Accept an arrow drawn from B to A and

Idea of atomic electron moves from B to A and (visible) photon statement that this is the transition where

is emitted visible photon is emitted or only arrow drawn

for relaxation.

Reason for ground state to B, 1 from Reason for B to ground:

• 18.4 eV is equal to energy difference between the

Do not credit use of 18.4 eV:

levels.

• −21.56 + 18.4(0) = −3.16 seen • leads to 67.5 nm (not visible)

• 21.56 − 3.16 = 18.4(0) seen. • more energy than 6.7 eV ( uv)

• −3.16 − −21.56 = 18.4(0) seen

• 21.56 − 18.4(0) =3.16 seen

Where no other mark is scored

Reason for B to A, 1 from

Max 1 compensation mark for:

• energy difference (between B and A) is 1.8 eV and this

Atom in its ground state absorbs energy and

is less than 6.7 eV (and therefore will emit a longer

electron moves up energy level.

wavelength photon)

• other transitions (B to ground or A to ground) are too OR

big, and wavelength is too small for visible light. Atom de-excites and electron moves down

• calculates wavelength for 1.8 eV energy level an emits a photon

6.63 ×10−34 ×3 ×108

𝜆𝜆 = ( ) = 6.9(1) × 10−7(𝑚𝑚) and states

2.88×10−19

this is in visible range

Total 8

How to answer it

Energy Levels, Accelerating Potential, and Photons

AQA AS Level Physics • Quantum Phenomena

What this question tests

This question assesses your understanding of electron gun acceleration, atomic energy levels, excitation via inelastic collisions, and photon emission using E = hf = hc/λ . You must be comfortable converting between electron-volts (eV) and joules (J), and logically deducing atomic transitions from numerical energy data.

Part 02.1: Deducing Accelerating Potential V

✅ Correct Answer

6.7 V

Awarded 1 mark (AO1)

💡 Key Knowledge

When an electron is accelerated through a potential difference V , its kinetic energy gained is given by E_k = eV .
If an electron gains 6.7 eV of kinetic energy, the accelerating voltage must numerically be 6.7 V .

Part 02.2: Wavelength of Emitted Photon

✅ Correct Answer

1.9 × 10⁻⁷ m (or 1.86 × 10⁻⁷ m )

Awarded 3 marks (2 × AO1, 1 × AO2)

📐 Step-by-Step Calculation

Convert energy from eV to Joules:
E = 6.7 × 1.60 × 10⁻¹⁹ = 1.072 × 10⁻¹⁸ J
Recall wave equation / photon energy formula:
E = hc / λ therefore λ = hc / E
Substitute values:
λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (1.072 × 10⁻¹⁸)
Final evaluation & rounding:
λ = 1.855 × 10⁻⁷ m rounds to 1.9 × 10⁻⁷ m (2 s.f.)

❌ Common Errors & Traps

Unit omission: Forgetting to multiply by the elementary charge ( 1.60 × 10⁻¹⁹ ) to convert electron-volts to joules.
Power of 10 errors: Incorrectly inputting Planck’s constant or the speed of light into the calculator.

🧠 Exam Technique

Always state your energy conversion explicitly so examiners can award method marks even if your final calculator input slips up.
Match your final significant figures to the data given in the question stem (2 significant figures).

Part 02.3: Energy Transitions and Calculations

✅ Correct Answer

Transitions: Electron excites from ground state ( -21.56 eV ) to energy level B ( -3.16 eV ), then de-excites from B to A ( -4.96 eV ), emitting a visible photon.

Awarded up to 4 marks (2 × AO2, 2 × AO3)

💡 Mark Scheme Breakdown

Mark 1: Identifies energy transfer causes transition from ground state to level B .
Mark 2: Identifies subsequent move from B to A releasing a visible photon.
Mark 3: Shows calculation proving ground state to B energy difference equals 18.4 eV ( -3.16 - (-21.56) = 18.4 eV ).
Mark 4: Explains why B to A works (energy difference is 1.8 eV , yielding a visible wavelength) or why other transitions are invalid.

🧠 Exam Technique & Examiner Guidance

Showing subtraction clearly ( -3.16 - (-21.56) = 18.4 eV ) is vital to secure the calculation marks. Double-check your signs when dealing with negative energy levels!
Top-level responses explicitly calculated the wavelength for the B to A transition ( 6.9 × 10⁻⁷ m ) and confirmed it falls correctly inside the visible light spectrum ( 400 nm to 700 nm ).

❌ Common Errors

Assuming the entire 18.4 eV is emitted directly as a single photon back to the ground state (this would produce ultraviolet radiation, not visible light, and does not match the excitation path).

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.