AP Chemistry / Unit 3: Properties of Substances and Mixtures
3.12 Properties of Photons
Notes
$$E=h\nu,\qquad c=\lambda\nu$$
- $h=6.626\times10^{-34}$ J·s; $c=3.00\times10^8$ m/s.
- Higher frequency = shorter wavelength = higher energy.
Practice
As wavelength increases, photon energy:
$\displaystyle E=\frac{hc}{\lambda}$.
Frequency of light with $\lambda=3\times10^{-7}$ m:
$\displaystyle \nu=\frac{c}{\lambda}=\frac{3\times10^8}{3\times10^{-7}}$.
Emission spectra show lines because electrons:
Quantized energies.
Ready for more? Take the Unit 3 test.