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**** =**** h****ν*** * , where

*E*is the energy (SI units of J),

*h*is Planck's constant (

*h*= 6.626 x 10

^{–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 $\lambda =c/\nu $ , where

*c*, the speed of light, is 2.998 x 10

^{8}m s

^{–1}. Another quantity that you will often see is wavenumber, $\sigma =1/\lambda $, which is commonly reported in units of cm

^{–1}.

The energy of a single photon that has the wavelength *λ* is given by:

$$E\text{}=\text{}\frac{hc}{\lambda}=\text{}\frac{1.986\times \text{}{10}^{-16}{\text{Jnmphoton}}^{-1}}{\lambda}$$

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:

$${E}_{m}\text{}=\text{}\frac{hc{N}_{A}}{\lambda}=\text{}\frac{1.196\text{}\times \text{}{10}^{8}{\text{Jnmmol}}^{-1}}{\lambda}$$

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.

#### Check Your Understanding

Consider the reaction O_{3} + UV → O_{2} + O*. 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]

$$\lambda =\text{}\frac{1.196\text{}\times \text{}{10}^{8}{\text{Jnmmol}}^{-1}}{{E}_{m}}\text{}=\frac{1.196\text{}\times \text{}{10}^{8}{\text{Jnmmol}}^{-1}}{386\text{}\times \text{}{10}^{3}{\text{Jmol}}^{-1}}=309\text{nm}$$