IB Chemistry 1.3 1.3.1
1.3.1

Emission Spectra

Light provides the evidence for the structure of the atom. We analyze it using two wave equations.

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

+ n=1 n=2 n=3 e⁻ Photon (hν) 1. e⁻ falls to lower level 2. ∆E is released 3. Released as light (photon) ∆E = hν = hc/λ

Continuous vs. Line Spectra

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

The existence of sharp, discrete lines proves that electrons can only exist at specific energy levels. If electrons could be anywhere, we would see a continuous rainbow.

Calculate Frequency from Wavelength

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


1. Convert Units: λ must be in meters.

\( 700 \text{ nm} = 700 \times 10^{-9} \text{ m} \)

2. Rearrange Formula: \( c = \nu \lambda \rightarrow \nu = c / \lambda \)

3. Solve:

\( \nu = \frac{3.00 \times 10^8}{700 \times 10^{-9}} \)

\( \nu = 4.29 \times 10^{14} \text{ Hz} \)

← Back to 1.3 1.3.2 Hydrogen Spectrum →