3 Radiation & Atomic Theory+

**3.3 Energy of a Photon**

3.4 Matter Interactions with EM Radiation

3.5 Atomic Theory for those in a Hurry

3.6 Quantum Numbers

3.8 Nomenclature

3.9 Ionic Bonding

3.11 Covalent Bonding

❮ previous chapter next chapter ❯

__external links__

Have you heard? Light not only behaves like a wave, but also like a particle. Yeah - a flyin' through the air chunk of energy. Just pure energy and no mass zipping around at the speed of light. The name of that chunk of energy/light is a **photon**. Yep - a photon is a "particle of light". Photons still have their alter-ego, wavelength and frequency - but they also are quantized into chunks of energy. So how small is that chunk of light? Use the Planck Equation!

E = hν

Pretty simple eh? The E is the total energy in joules of a single photon with a frequency of ν (nu). Planck's constant is h and is defined as such:

*h* = 6.626 × 10^{-34} J·s

Also notice that frequency ν is *directly proportional* to the energy. They scale with each other: if you double the frequency, you will double the energy.

Your eyes receive color information via the interaction of photons with the rods and cones in your eye. When you see white light you are really seeing the mixing of all the colors of the spectrum (aka rainbow). So a blast of white light is really a blast of a multitude of photons with varying energies across the visible spectrum. Below is my simplified diagram of a blast of white light. Notice there are distinct colored photons in the grouping. Technically speaking, I should show a photon for pretty much every wavelength in the spectrum. I've chosen just 7 total colors to get the point across.

Also note how ALL the photons are traveling at the same speed. It doesn't matter about the energy of the photon because that varying greatly across the total electromagnetic spectrum - ALL photons travel at the speed of light which is 3.00 × 10^{8} m/s.

What IS different for the different colors is the energy per photon in joules. That is where the Planck equation comes in. If you know the frequency, use the previous equation. If you have the wavelength, you can convert it to frequency or use the wavelength version of the equation:

\[E_{\rm photon} = {h c \over \lambda} \]

Make sure your wavelength is in meters (m) though.

(a) What is the energy (joules per photon, J/photon) of a photon that has a wavelength of 700 nm? That would be a *red* photon. (b) What is the total energy of a mole of photons that all have wavelengths of 700 nm? Answer in kJ/mol.

**Solution (a):** Since the question gave us wavelength (\(\lambda \)) we will use the wavelength version of the Planck equation:

\[E_{\rm photon} = {h c\over \lambda}\]

\[\require{cancel} \newcommand\ccancel[2][black]{\color{#1}{\bcancel{\color{black}{#2}}}} E_{\rm photon} = {(6.626\times 10^{-34}\;{\rm J\cdot \ccancel[red]{s}})(3.00\times 10^8 \,{\rm \ccancel[green]{m}\cdot \ccancel[red]{s^{-1}}})\over 700\times 10^{-9}\,\ccancel[green]{\rm m}} \]

\[E_{\rm photon} = 2.84\times 10^{-19}\;{\rm J/photon}\]

**Solution (b):** All we need to do is take the answer from part (a) for one photon and scale it up by Avogadro's number (a mole):

\[E_{\rm mole\;of\;photons} = E_{\rm photon}\cdot N_{\rm A}\]

\[E = \left(2.84\times 10^{-19}\;{\rm J\over photon}\right)\left(6.022\times 10^{23}\;{\rm mol^{-1}}\right)\]

\[E = 1.71\times 10^5\;{\rm J/mol}\]

\[E = 171\;{\rm kJ/mol}\]