AQA AS Level Physics Paper 2, November 2021: Question 13

1 mark · Medium difficulty · Multiple Choice

Calculate the kinetic energy of a free electron after it collides with a lithium atom and excites it from the n = 1 to the n = 2 energy level.

Practise this question

Question

The question presents an energy level diagram for a lithium atom showing ionization at 0 J, n = 2 at -2.9 x 10^-19 J, and n = 1 at -8.6 x 10^-19 J. A free electron with kinetic energy 6.0 x 10^-19 J collides with a stationary lithium atom in its n = 1 energy level, exciting it to n = 2. Four multiple choice options are listed: A (0.3 x 10^-19 J), B (2.6 x 10^-19 J), C (3.1 x 10^-19 J), and D (5.7 x 10^-19 J).
Question text

13 Some energy levels of a lithium atom are shown below.

A free electron with kinetic energy 6.0 × 10−19 J collides with a stationary lithium atom in its

n = 1 energy level. The lithium atom is excited to the n = 2 energy level.

What is the kinetic energy of the free electron after the collision?

[1 mark]

A 0.3 × 10−19 J

B 2.6 × 10−19 J

C 3.1 × 10−19 J

D 5.7 × 10−19 J

Mark scheme

Show the mark scheme The mark scheme shows that the correct answer for question 13 is option A, which is 0.3 x 10^-19 J.

13 A 0.3 × 10−19 J

How to answer it

Electron Collision and Atomic Energy Levels

What this question tests

This question assesses your understanding of atomic energy levels, electron excitation via inelastic collisions, and the conservation of energy. You must determine the energy transferred from a free incident electron to an atom and calculate the remaining kinetic energy of the free electron.

Question 13

Determining Post-Collision Electron Kinetic Energy

✅ Correct Answer

Option A ( 0.3 × 10⁻¹⁹ J )

💡 Key Knowledge

  • Inelastic Collisions: Free electrons can transfer a portion of their kinetic energy to excite bound electrons into higher energy states.
  • Energy Conservation: Initial kinetic energy of the free electron equals the excitation energy required plus the final kinetic energy of the free electron.
  • Energy Level Notation: Energy levels are negative because they represent bound states relative to free electrons at infinity (ionization = 0 J).

🧠 Exam Technique

When dealing with negative energy levels, always find the difference between the two states to determine the exact energy packet required for excitation. Avoid sign errors by taking the absolute value of the change in energy.

❌ Common Errors

  • Forgetting to subtract the excitation energy from the incoming kinetic energy, choosing option D instead.
  • Failing to account for the negative signs correctly, leading to incorrect addition instead of subtraction.

📐 Step-by-Step Calculation

  1. Step 1: Calculate the excitation energy required.
    The atom transitions from n = 1 ( -8.6 × 10⁻¹⁹ J ) to n = 2 ( -2.9 × 10⁻¹⁹ J ).
    ΔE = E₂ - E₁ = (-2.9 × 10⁻¹⁹ J) - (-8.6 × 10⁻¹⁹ J)
    ΔE = +5.7 × 10⁻¹⁹ J (Energy absorbed by the lithium atom)
  2. Step 2: Apply conservation of energy.
    Initial KE of electron = Energy absorbed by atom + Final KE of electron
    6.0 × 10⁻¹⁹ J = 5.7 × 10⁻¹⁹ J + Final KE
  3. Step 3: Solve for final kinetic energy.
    Final KE = 6.0 × 10⁻¹⁹ J - 5.7 × 10⁻¹⁹ J = 0.3 × 10⁻¹⁹ J
Mark Scheme Allocation: 1 mark total for selecting option A.

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.