# 6.4: How is energy related to the wavelength of radiation?

- Page ID
- 3385

We can think of radiation either as waves or as individual particles called photons. The energy associated with a single photon is given by

\[E = hν \]

where \(E\) is the energy (SI units of J), * h* is Planck's constant (

*= 6.626 x 10*

*h*^{–34}J s), and \(\nu\)

*is the frequency of the radiation (SI units of s*

^{–1}or Hertz, Hz) (see figure below). Frequency is related to wavelength by \(λ=c/ν\), where \(c\), the speed of light, is 2.998 x 10

^{8}m s

^{–1}. Another quantity that you will often see is wavenumber, \(σ=1/λ\), which is commonly reported in units of cm

^{–1}.

The energy of a single photon that has the wavelength \(λ\) is given by:

\[E=\frac{h c}{\lambda}=\frac{1.986 \times 10^{-16} \mathrm{J} \,\mathrm{nm} \text { photon }^{-1}}{\lambda}\]

Note that as the wavelength of light gets shorter, the energy of the photon gets greater. The energy of a mole of photons that have the wavelength \(λ\) is found by multiplying the above equation by Avogadro's number:

\[E_{m}=\frac{h c N_{A}}{\lambda}=\frac{1.196 \times 10^{8} \mathrm{J} \mathrm{nm} \mathrm{mol}^{-1}}{\lambda}\]

In the lesson on atmospheric composition, you saw how solar UV radiation was able to break apart molecules to initiate atmospheric chemistry. These molecules are absorbing the energy of a photon of radiation, and if that photon energy is greater than the strength of the chemical bond, the molecule may break apart.

Exercise

Consider the reaction

\[\ce{O3 + UV -> O2 + O*}. \nonumber\]

If the bond strength between O_{2} and O* (i.e., excited state oxygen atom) is 386 kJ mol^{–1}, what is the longest wavelength that a photon can have and still break this bond?

**Click for answer.**-
ANSWER: Solve for wavelength in equation [6.2b]

\(\lambda=\frac{1.196 \times 10^{8} \mathrm{J} \mathrm{nm} \mathrm{mol}^{-1}}{E_{m}}=\frac{1.196 \times 10^{8} \mathrm{J} \mathrm{nm} \mathrm{mol}^{-1}}{386 \times 10^{3} \mathrm{Jmol}^{-1}}=309 \mathrm{nm}\)