IB Chemistry 1.3 1.3.1
1.3.1
Structure 1.3 SL & HL ⏱️ ~4 min revision

Emission Spectra

Continuous vs line emission spectra, electron transitions, and photon quantisation.

When atoms absorb energy, their electrons jump to higher energy levels (excited state). As they fall back to lower, more stable energy levels, they emit energy in the form of photons of light. This emitted light creates an emission spectrum.

1. Wave Equation

\( c = \nu \lambda \)

c = Speed (\(3.00 \times 10^8\) m/s)

ν = Frequency (Hz or s⁻¹)

λ = Wavelength (m)

2. Photon Energy

\( E = h \nu \)

E = Energy (Joules)

h = Planck (\(6.63 \times 10^{-34}\))

ν = Frequency (Hz)

How Emission Works Diagram How Emission Works + n=1 n=2 n=3 e⁻ excited e⁻ ground Photon (hv) What Happens 1. e⁻ falls to a lower level 2. Energy difference (ΔE) released 3. Emitted as a photon of light ΔE = hv = hc/λ

Continuous vs. Line Spectra

Continuous (Rainbow) White Light
Line Spectrum (Discrete) Excited Hydrogen
Key Evidence

Quantised Energy Levels

The existence of sharp, discrete lines (not a continuous rainbow) proves that electron transitions occur only between fixed, discrete energy levels within atoms.

Worked Example

Calculate Frequency from Wavelength

Problem: Red light has a wavelength of 700 nm. Calculate its frequency.


1. Convert Units: \(\lambda\) must be in metres: \(700\text{ nm} = 700 \times 10^{-9}\text{ m} = 7.00 \times 10^{-7}\text{ m}\).

2. Rearrange Wave Equation: \(\nu = \dfrac{c}{\lambda}\)

3. Substitute Values: \(\nu = \dfrac{3.00 \times 10^8\text{ m s}^{-1}}{7.00 \times 10^{-7}\text{ m}} = \mathbf{4.29 \times 10^{14}\text{ Hz}}\)

AQA GCSE & IB Chemistry

Study this topic on the go

Get active recall flashcards, notes, and topic quizzes in ChemEasy, or build your revision schedule with ChemPlan IB.

See our apps