# Photon Energy - The Energy of Light Particles

In quantum physics it has turned out that in many experiments it makes sense to consider light not as a wave but as a stream of many particles. Such a light particle is called a **photon**.

The energy of a photon is determined by the **light frequency \( f \)**. If we consider the light wave-like, then the frequency indicates *how often* the light wave oscillates *per second*. The unit of frequency is \( [f] = \frac{1}{\mathrm s}, \) though this is noted somewhat more compactly as \( \frac{1}{\mathrm s} = \mathrm{Hz} \). "Hz" stands for "Hertz" - in honor of a physicist named Heinrich Hertz.

An important insight of quantum mechanics was that the energy of a photon is connected with the so-called **Planck's constant** \( h \) - with one of the most important constants of physics! It has always the same value \( h ~=~ 6.626 \cdot 10^{-34} \, \mathrm{Js} \). As you can see, it is incredibly small and has the unit \( [h] = \mathrm{Js} \). "Js" stands for "Joule times second", where \( \mathrm{J} \) (Joule) is the unit of energy.

With the help of the Planck's constant and the light frequency you can easily answer the question about the energy of a photon:

Unit of energy \( W_{\text p} \) is Joule (J).

The formula 1

tells you only how large the energy of a *single* photon is. If you want to find out how large the energy of many photons is, you have to determine their number and multiply it by the energy of a single photon 1

. Thus, if \( n \) photons hit a capacitor plate, for example, the total energy arriving at the plate is given by:

Sometimes you know instead of the light frequency \( f \) the corresponding **wavelength** \( \lambda \) (pronunciation: "*Lamda*"). Once you know either the frequency or the wavelength, you can easily convert them into each other, because they are related by the **speed of light** \( c \):

The wavelength, as the name implies, is the *distance from one crest to another crest* of the light wave. And which unit has the wavelength? So that the unit of the (light) speed \( [c] = \frac{\text m}{\text s} \) in the equation 3

is correct, the wavelength must have the unit \( [\lambda] = \text{m} \), because the frequency has the unit \( [f] = \frac{1}{\text s} \)!

The cool thing about it is: You don't have to determine the speed of light at all. Why not? Because the speed of light is *a constant* just like the Planck's constant! In a medium (in your case it is air or vacuum) it always has the same value:

&~\approx~ 3 \cdot 10^{8} \, \frac{\mathrm m}{\mathrm s}

Using formula 3

, which gives the relation between the light frequency and light wavelength, you can express the photon energy 1

also using the wavelength:

## What color do photons have?

If you hear something like "*wavelength of green light*", then the "*green*" refers to a certain light frequency or light wavelength. Depending on the wavelength / frequency, you perceive the light in a different color.

Light wavelength | Light frequency | Energy of a photon |
---|---|---|

575 nm | 5.2 × 10^{14} Hz |
3.5 × 10^{-19} J |

546 nm | 5.5 × 10^{14} Hz |
3.6 × 10^{-19} J |

435 nm | 6.9 × 10^{14} Hz |
4.6 × 10^{-19} J |

400 nm | 7.5 × 10^{14} Hz |
5 × 10^{-19} J |

365 nm (UV light) | 8.2 × 10^{14} Hz |
5.4 × 10^{-19} J |

## What is the energy of one mole of photons?

The **energy of one mole of photons**, let us call it \(W_{\text{mol}}\), is the energy \(W_{\text p}\) of a single photon multiplied by the number of photons per mole. The **Avogadro constant** \(N_{\text A} = 6 \cdot 10^{23} \, \frac{1}{\mathrm{mol}} \) provides us with the number of photons per mole. Therefore, the photon energy per mole is given by:

Substitute equation 1

into 8

:

So if you insert a concrete wavelength \(\lambda\) into Eq. 9

, you get the energy of \( 6 \cdot 10^{23} \) photons, which just form one mole.

You can express the energy \( W_{\text p} \) of a photon with the **light frequency** \(f\). Thus, the photon energy per mole can also be calculated in the following way:

In the next lesson, we will use this knowledge about photons to explain a very important experiment in quantum physics, namely the photoelectric effect.