AQA AS Level Physics Paper 1, June 2018: Question 3

10 marks · Medium difficulty · Short Answer

Analyze hydrogen atom energy levels, photon emission regions, excitation mechanisms, and photoelectric effect applications including threshold frequency and maximum electron speed.

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Question

Figure 3 shows hydrogen atom energy levels from n=1 to n=4 with corresponding energies -13.6 eV, -3.4 eV, -1.5 eV, and -0.85 eV, along with three transitions labeled A, B, and C. Sub-questions 03.1 to 03.4 ask to identify the spectral regions for transitions A, B, and C, explain why photons require exact energy while free electrons need a minimum kinetic energy for excitation, determine if any transition can cause photoelectric emission given a threshold frequency of 5.1 x 10^14 Hz, and calculate the maximum speed of emitted electrons for 12.1 eV incident photons.
Question text

03 Figure 3 shows some of the energy levels for a hydrogen atom.

Figure 3

An excited hydrogen atom can emit photons of certain discrete frequencies. Three

possible transitions are shown in Figure 3.

03.1 The transitions shown in Figure 3 result in photons being emitted in the ultraviolet,

visible and infrared regions of the electromagnetic spectrum.

To which region of the spectrum do the emitted photons belong?

Tick ( ) the correct box for each transition, A, B and C.

[1 mark]

Transition Ultraviolet Visible Infrared

A

B

C 9

03.2 Two ways to excite a hydrogen atom are by collision with a free electron or by the

absorption of a photon.

Explain why, for a particular transition, the photon must have an exact amount of

energy whereas the free electron only needs a minimum amount of kinetic energy.

[3 marks]

03.3 The surface of a sample of caesium is exposed to photons emitted in each of the

three transitions shown in Figure 3.

The threshold frequency of caesium is 5.1 × 1014 Hz

Determine whether any of these transitions would produce photons that would cause

electrons to be emitted from the surface of caesium.

[3 marks]

03.4 Photons each with energy 12.1 eV are incident on the surface of the caesium sample.

Calculate the maximum speed of electrons emitted from the caesium.

[3 marks]

m s−1

maximum speed =

Mark scheme

Show the mark scheme The mark scheme provides answers for questions 03.1 through 03.4, detailing the table ticks for ultraviolet (C), visible (B), and infrared (A), explanations for photon vs electron excitation, photoelectric calculations using work function and threshold frequency, and maximum speed calculations resulting in 1.9 x 10^6 m s^-1.

Question Answers Additional Comments/Guidelines Mark

Transition Ultraviolet Visible Infrared all correct 1 mark

A

03.1 1

B

C

EITHER Any implication of photoelectric effect max 1

energy needed for electron to move to higher level/orbital Accept one energy level to another

OR

for a transition/excitation/change of levels an exact amount of energy 1

is needed 1

03.2

all the photon’s energy absorbed( in 1 to 1 interaction)

electron can transfer part of its energy (to cause a

transition/excitation)/ continues moving/ lower kinetic energy/ lower

speed

(use of = hf0)

= 6.63 × 10−34 × 5.1 × 1014 (= 3.38 × 10−19 ) If see 2.1 get these first two marks 1

= 3.38 × 10−19/1.6 × 10−19 = 2.1(1) (eV) 1

OR

= 6.63 × 10−34 × 5.1 × 1014 (= 3.38 × 10−19 ) 1

energy in J 10.2 ×1.6 × 10−19= 1.63 × 10−18

03.3

OR

energy levels in J = 10.2 ×1.6 × 10−19= 1.63 × 10−18

photons frequencies giving this energy= 2.46 × 1015

2 → 1 / C possible

last mark dependent on previous 2

03.4 (use of hf = + Ek) photoelectric equation must be used

– – –

12.1 × 1.6 × 10−19 = 2.1 × 1.6 × 10−19 + E ecf for third mark their calculated kinetic energy 1

k

E = 1.6 × 10−18(J) having used photoelectric equation even if not 1

k

v = (2 × 1.6 × 10−18/9.11 × 10−31) (= 1.9 × 106 m s−1) converted eV to J or frequency to J 1

correct answer gets(1.9 × 106 m s−1 ) full marks

Total 10

How to answer it

Hydrogen Energy Levels & Photoelectric Effect

What this question tests

This question assesses your understanding of atomic energy levels, photon emission, the electromagnetic spectrum regions (UV, Visible, IR), and how excitation mechanisms differ between photons and free electrons. It also tests application of the photoelectric equation ( hf = φ + Ek ), threshold frequency calculations, and determining maximum electron speeds.

Part 03.1

Identifying Photon Regions in the EM Spectrum

✅ Correct Answer

  • Transition A (n = 4 to n = 3): Infrared
  • Transition B (n = 3 to n = 2): Visible
  • Transition C (n = 2 to n = 1): Ultraviolet

💡 Key Knowledge

Calculate energy changes ( ΔE ) for each transition to judge wavelength/frequency:

  • Small energy drops (A) yield low frequency / long wavelength (Infrared).
  • Large energy drops (C) yield high frequency / short wavelength (Ultraviolet).
Mark: 1 mark for all three transitions correctly identified.
Part 03.2

Photon vs. Free Electron Excitation

💡 Key Knowledge

Explain the fundamental difference between absorption and inelastic collisions:

  • Photons: Must be completely absorbed in a 1-to-1 interaction. Therefore, photon energy ( hf ) must exactly equal the energy difference ( ΔE ) between levels.
  • Free Electrons: Can transfer part of their kinetic energy to excite the atom, as long as their initial kinetic energy is greater than or equal to the required threshold amount. The electron simply carries away the remaining kinetic energy.

❌ Common Errors

Students frequently lose marks by confusing the photoelectric effect rules with atomic excitation, or by stating free electrons are absorbed by the atom.

Mark: 3 marks total (1 for exact energy needed for photon, 1 for all-or-nothing photon absorption, 1 for electron transferring only part of its kinetic energy).
Part 03.3

Threshold Frequency and Caesium Emission

📐 Step-by-Step Calculation

  1. Calculate work function (φ) or energy of transitions:
    Threshold frequency f0 = 5.1 × 1014 Hz
    φ = hf0 = (6.63 × 10-34) × (5.1 × 1014) = 3.38 × 10-19 J
    Convert to eV: 3.38 × 10-19 / 1.6 × 10-19 = 2.11 eV
  2. Evaluate Transition Energies from Fig. 3:
    Transition C ( n = 2 → 1 ): -3.4 eV - (-13.6 eV) = 10.2 eV
  3. Compare & Conclude:
    Since 10.2 eV > 2.11 eV (or photon energy exceeds work function), Transition C produces photons capable of causing photoelectric emission.

🧠 Exam Technique

Always convert work function and photon energies into the same unit (either both in Joules or both in eV) before making your comparison statement to avoid unit mismatch errors.

Mark: 3 marks total (1 for calculating work function/threshold energy, 1 for identifying photon energies from transitions, 1 for correctly concluding transition C is capable).
Part 03.4

Calculating Maximum Speed of Emitted Electrons

📐 Step-by-Step Calculation

  1. Recall Photoelectric Equation:
    hf = φ + Ek(max)
  2. Substitute values (converting eV to Joules):
    Incident photon energy = 12.1 eV
    Work function (φ) = 2.1 eV (from part 03.3)
    Ek = 12.1 eV - 2.1 eV = 10.0 eV
    Convert 10.0 eV to Joules:
    10.0 × 1.6 × 10-19 = 1.6 × 10-18 J
  3. Rearrange for velocity ( v ):
    Ek = 0.5 × m × v2
    v = √(2Ek / m)
    v = √(2 × 1.6 × 10-18 / 9.11 × 10-31)
    v = 1.87 × 106 m s-1 (rounds to 1.9 × 106 m s-1 to 2 s.f.)

❌ Common Calculation Traps

  • Forgetting to multiply by the elementary charge ( 1.6 × 10-19 ) when converting eV to Joules before calculating kinetic energy and speed.
  • Using the mass of a proton or nucleon instead of the electron mass ( 9.11 × 10-31 kg ).
Mark: 3 marks total (1 for correct application of photoelectric equation / substitution, 1 for correct kinetic energy in Joules, 1 for correct final velocity with proper unit and significant figures).

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.