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)
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
Paper 2 StyleRed 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:
\( \nu = \frac{3.00 \times 10^8}{700 \times 10^{-9}} \)
\( \nu = 4.29 \times 10^{14} \text{ Hz} \)
Student Practice Set
1.
Which relationship is correct?