WJEC A-Level Chemistry AS Unit 1, June 2025: Question 3

1 mark · Easy difficulty · Short Answer

Compare the frequency and transition energy of spectral lines in the visible emission spectrum of hydrogen using the terms higher or lower.

Practise this question

Question

Question 3 states: 'The visible emission spectrum of hydrogen shows prominent lines at wavelengths of 434 nm, 486 nm and 656 nm. Complete the sentences by using the words higher or lower.' Two fill-in-the-blank sentences follow: 'The frequency of the line at 656 nm is [blank] than the frequency of the line at 434 nm.' and 'The line at 656 nm is caused by an electronic transition of [blank] energy than the line at 486 nm.' It is worth 1 mark.

Mark scheme

Show the mark scheme Mark scheme for Question 3 shows the required answers: 'lower ... lower', with 1 mark allocated.

How to answer it

Hydrogen Emission Spectrum: Wavelength, Frequency & Energy

📋 What This Question Tests

This question assesses your fundamental understanding of the electromagnetic spectrum in the context of atomic emission:

  • The inverse relationship between wavelength (λ) and frequency (ν): c = νλ
  • The direct relationship between frequency and photon energy: ΔE = hν
  • The inverse relationship between wavelength and transition energy: ΔE = hc / λ
  • Reading and comparing spectral line data in nanometres (nm).

Question 3 Walkthrough

Completing the sentences using "higher" or "lower" [1 Mark]

✅ Correct Answer

The sentences completed correctly:

  • "The frequency of the line at 656 nm is lower than the frequency of the line at 434 nm."
  • "The line at 656 nm is caused by an electronic transition of lower energy than the line at 486 nm."
Mark Scheme: lower ... lower [1 mark for both correct].

💡 Key Knowledge

  • Wave equation: c = νλ (where c is the speed of light). Since c is constant, frequency is inversely proportional to wavelength (ν ∝ 1/λ).
  • Planck's equation: ΔE = hν = hc / λ . Energy is directly proportional to frequency, but inversely proportional to wavelength.
  • Hydrogen Balmer Series (Visible):
    • 656 nm: Longest λ → lowest frequency → lowest energy (n = 3 → n = 2, red).
    • 486 nm: Intermediate λ (n = 4 → n = 2, cyan).
    • 434 nm: Shortest λ → highest frequency → highest energy (n = 5 → n = 2, blue-violet).

📐 Step-by-Step Deduction

  1. Compare 656 nm with 434 nm (Frequency):
    656 nm is a longer wavelength than 434 nm.
    Because wavelength and frequency are inversely related, a longer wavelength must mean a lower frequency.
  2. Compare 656 nm with 486 nm (Energy):
    656 nm is a longer wavelength than 486 nm.
    Because photon energy is inversely proportional to wavelength ( ΔE ∝ 1/λ ), a longer wavelength corresponds to a lower energy transition.

❌ Common Errors & Misconceptions

  • Assuming larger number = higher energy: Students see 656 nm > 434 nm and incorrectly deduce "higher" frequency or energy, forgetting the inverse relationship.
  • All-or-nothing marking: Both blanks must be correct to earn the 1 mark. Getting one right and one wrong yields 0 marks.
  • Ignoring the prompt terms: The question specifies using the words "higher" or "lower". Do not use words like "less", "smaller", "greater", or "decreased".

🧠 Exam Technique & Quick Check Method

Always write down the two quick arrows on your exam paper as soon as you see spectrum questions:

λ ↑ (increases)  ⟹  ν ↓ (decreases)  ⟹  ΔE ↓ (decreases)

Since 656 nm is the largest wavelength in the given data, anything compared directly to it will result in lower frequency and lower energy.

Topics

Physical Chemistry · 1.2 Basic ideas about atoms

Question and mark scheme from the WJEC A-Level Chemistry examination, AS Unit 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.