The method and the results are on the ionisation energy lab. Do the analysis yourself first, then check the answers below.
Quick answer: An atomic orbital is a region around a nucleus that holds at most two electrons, and orbitals come as s, p and d types. These answers go with the Chemistry 11-12 lab on ionisation energy: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.
The model answers use h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s and Avogadro’s constant 6.022 × 10²³ mol⁻¹. Small differences in the last figure come from rounding.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.1.1, IB S1.3
Lithium: E = 1.9878 × 10⁻²⁵ ÷ 6.70 × 10⁻⁷ = 2.97 × 10⁻¹⁹ J, which is 179 kJ mol⁻¹. Potassium: E = 1.9878 × 10⁻²⁵ ÷ 7.67 × 10⁻⁷ = 2.59 × 10⁻¹⁹ J, which is 156 kJ mol⁻¹.
Question 4 Origin of the colours
Heat gives electrons in the ion enough energy to jump to a higher energy level (excitation). They fall back to lower levels and emit the energy difference as a photon. Each ion has its own set of energy gaps, so it emits its own wavelengths and its own colour.
Question 5 Comparison with ionisation
The photon value (203 kJ mol⁻¹) is much smaller than 496 kJ mol⁻¹. The electron only moves between two levels of the atom (3p to 3s), whereas ionisation removes it completely, which needs far more energy. The ratio is 2.4.
Use a narrower slit or a calibrated spectrometer, check the scale against a known line such as the sodium yellow line, and repeat each reading several times to take a mean.
