The convergence limit is the point where an atom’s emission lines run together and stop. Run that transition backwards and you have the energy needed to tear the electron off completely — which means a spectrum can be used to measure ionisation energy directly.
📚 What you need to know
The first ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms.
The convergence limit of the Lyman series corresponds to the n = ∞ to n = 1 transition.
Reversed, that transition is exactly the ionisation of a ground-state atom.
Use c = fλ to get the frequency, then E = hf to get the energy per atom.
Multiply by Avogadro’s constant to get energy per mole, then divide by 1000 for kJ mol−1.
Both equations and all three constants are in the data booklet.
Keep full precision in the calculator and round only at the end.
What ionisation energy means
First ionisation energy
X(g) → X+(g) + e−
Notice the state symbols. Ionisation energies are defined for gaseous atoms, because otherwise you would also be paying the energy cost of separating the atoms from each other, and the number would no longer be a property of the atom alone.
Only the Lyman convergence limit gives the first ionisation energy, because only the Lyman series ends at n = 1, the ground state.
The calculation route
Every question of this type follows the same four steps. Learn the route rather than individual examples and none of them can surprise you.
A useful sense check: first ionisation energies for real elements land between about 400 and 2400 kJ mol−1. Anything wildly outside that range means a step was skipped.
The reason step three exists is a definition, not a piece of physics. E = hf gives you the energy to ionise one atom, but ionisation energy is always quoted per mole. Multiplying by 6.02 × 1023 is simply converting from one atom to a mole of them. Miss it and your answer is out by twenty-three orders of magnitude, which is at least easy to spot.
WORKED EXAMPLE
The convergence limit in the emission spectrum of lithium occurs at a wavelength of 230 nm. Calculate the first ionisation energy of lithium in kJ mol−1.
Step 1: convert and find the frequencyλ = 230 × 10−9 = 2.30 × 10−7 mf = c ÷ λ = 3.00 × 108 ÷ 2.30 × 10−7 = 1.30 × 1015 s−1Step 2: energy for one atomE = hf = 6.63 × 10−34 × 1.30 × 1015 = 8.65 × 10−19 JStep 3: scale up to one mole8.65 × 10−19 × 6.02 × 1023 = 5.21 × 105 J mol−1Step 4: convert to kJ mol−15.21 × 105 ÷ 1000 = 521IE1(Li) = 521 kJ mol−1The data booklet gives 520 kJ mol−1, so the method checks out.
WORKED EXAMPLE
The convergence limit for potassium occurs at a frequency of 1.05 × 1015 s−1. Calculate its first ionisation energy in kJ mol−1.
Frequency is given, so skip straight to E = hfE = 6.63 × 10−34 × 1.05 × 1015 = 6.96 × 10−19 JThere is no wavelength to convert, so there is no c = fλ step at all.Per mole6.96 × 10−19 × 6.02 × 1023 = 4.19 × 105 J mol−1Into kJIE1(K) = 419 kJ mol−1Sense check against lithiumPotassium is below lithium in Group 1, so its outer electron is further from the nucleus and better shielded. A lower ionisation energy than lithium’s 521 is exactly what you would expect.
WORKED EXAMPLE
The first ionisation energy of magnesium is 738 kJ mol−1. Calculate the wavelength of its convergence limit, in nm.
Run the route backwards, starting per atom738 × 1000 = 7.38 × 105 J mol−1E = 7.38 × 105 ÷ 6.02 × 1023 = 1.23 × 10−18 J per atomRearrange E = hf for frequencyf = E ÷ h = 1.23 × 10−18 ÷ 6.63 × 10−34 = 1.85 × 1015 s−1Then c = fλ for wavelengthλ = c ÷ f = 3.00 × 108 ÷ 1.85 × 1015 = 1.62 × 10−7 mλ = 162 nmDoes that make sense?162 nm is well into the ultraviolet, which is right: convergence limits for ionisation always fall in the UV, since ionisation is a large energy jump.
💡 Exam tip
Write out the four steps before substituting anything, so no stage gets skipped.
Convert nm to m at the very start by multiplying by 10−9.
Always multiply by Avogadro’s constant — ionisation energy is per mole.
Divide by 1000 at the end to reach kJ mol−1.
Keep full precision in the calculator and round only the final answer.
Sense-check against the range 400 to 2400 kJ mol−1.
⚠️ Common mix-up
Forgetting Avogadro’s constant, leaving an answer around 10−19.
Leaving the wavelength in nanometres before using c = fλ.
Giving the answer in J mol−1 when kJ mol−1 was asked for.
Using a convergence limit from the Balmer series, which ends at n = 2 and is not ionisation.
Rounding at every step, which drifts the final answer.
Omitting (g) from the ionisation equation, when the gaseous state is part of the definition.
Up next: Successive Ionisation Energies — you have removed one electron. The next page keeps going, and shows how the pattern of what happens next reveals an element’s group without being told what it is.
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