Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

1.3.5Additional lessons

Ionisation Energy from an Emission Spectrum (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 12 of 172

📚 IB Chemistry HL Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
0%read

Part 1 of 5Convergence and Ionisation

The lines in a series come closer together because the energy levels converge at high n.

Convergence limit
  • Convergence limit The highest frequency in the Lyman series, where the lines merge, corresponds to a transition from n = ∞ to n = 1.
Meaning
  • Meaning At n = ∞ the electron is completely removed from the atom, so this is ionisation.
Link
  • Link The energy at the convergence limit therefore equals the first ionisation energy of hydrogen.
Relationship
  • Relationship E = hν gives the energy per atom, which is multiplied by Avogadro's constant for one mole.

Part 2 of 5Calculation Steps

Frequency at the limit is converted into energy per mole.

1
Read the frequency, ν, at which the lines converge from the spectrum or the data given.
2
If a wavelength is given instead, find the frequency using ν = c ÷ λ.
3
Calculate the energy per atom with E = hν in joules.
4
Multiply by Avogadro's constant, 6.02 × 10²³ mol⁻¹, to get energy per mole.
5
Divide by 1000 to give kJ mol⁻¹, the usual unit for ionisation energy.

Part 3 of 5Interpreting Results

Interpreting Results
Comparison
  • Comparison The calculated value for hydrogen agrees well with the accepted first ionisation energy of about 1312 kJ mol⁻¹.
Only hydrogen
  • Only hydrogen The method is simple for hydrogen because it has one electron and a single series of lines.
Other series
  • Other series Convergence in the Balmer series gives the energy needed to ionise from n = 2.
Check units
  • Check units The result in joules per atom is very small, around 10⁻¹⁸ J, until multiplied by NA.

Part 4 of 5Worked Example

Question

  • The Lyman series of hydrogen converges at 3.28 × 10¹⁵ Hz.
  • Calculate the first ionisation energy in kJ mol⁻¹.

  1. E = hν = 6.63 × 10⁻³⁴ × 3.28 × 10¹⁵ = 2.175 × 10⁻¹⁸ J per atom.
  2. Per mole2.175 × 10⁻¹⁸ × 6.02 × 10²³ = 1.309 × 10⁶ J mol⁻¹.
  3. Convert1309 kJ mol⁻¹.

Answer First ionisation energy ≈ 1.31 × 10³ kJ mol⁻¹

Part 5 of 5Exam Checklist & Tips

1.3.5 – Key Points for the Exam

  • Convergence limit The frequency at which the lines of a series merge.
  • First ionisation energy Energy to remove one mole of electrons from one mole of gaseous atoms.
  • Avogadro constant 6.02 × 10²³ mol⁻¹.
  • Quantised Having only certain fixed values.

Common Mistakes

  • Forgetting to multiply by Avogadro's constant to convert from atom to mole.
  • Using the frequency of the first line in the series instead of the convergence limit.
  • Leaving the answer in J mol⁻¹ when kJ mol⁻¹ is needed.
  • Using the Balmer limit when the question concerns the first ionisation energy from n = 1.

Last updated:

\n