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Photon Energy - The Energy of Light Particles

Important Formula

Formula: Photon

$$W_{\text p} ~=~ h \, \class{violet}{f}$$ $$W_{\text p} ~=~ h \, \class{violet}{f}$$ $$\class{violet}{f} ~=~ \frac{ W_{\text p} }{ h }$$ $$h ~=~ \frac{ W_{\text p} }{ \class{violet}{f} }$$

Rearrange formula

What do the formula symbols mean?

Photon energy

$$ W_{\text p} $$ Unit $$ \mathrm{J} $$

Photon energy is the energy of a single photon (light particle). You can calculate it with the help of the light frequency $ f $. For example, if the light has the frequency $ f = 10^{15} \, \mathrm{Hz} $, then the energy of a photon is: \begin{align} W_{\text p} &= 6.6 \cdot 10^{-34} \, \mathrm{Js} ~\cdot~ 10^{15} \, \mathrm{Hz} \\\\ &= 6.6 \cdot 10^{-19} \, \mathrm{J} \end{align}

Frequency

$$ f $$ Unit $$ \mathrm{Hz} $$

Frequency of electromagnetic radiation. For example, the frequency of green light is in the range $ f = 5.5 \cdot 10^{14} \, \mathrm{Hz}$.

The frequency can be written with the wavelength $\lambda$ and the speed of light $c$ as follows: $$ f = \frac{c}{\lambda} $$

Planck's Constant

$$ h $$ Unit $$ \mathrm{Js} $$

Planck's constant is a physical constant from quantum mechanics and has the value: $$ h ~=~ 6.626 \, 070 \, 15 \,\cdot\, 10^{-34} \, \mathrm{Js} $$

Table of contents

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.

Four photons, which have different energy & color depending on frequency.

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

$$ \begin{align} W_{\text p} ~=~ h \, \class{violet}{f} \end{align} $$

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

Remember!

The larger the light frequency $ f $, the larger is the energy of a photon.

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:

$$ \begin{align} W ~=~ n\, h \, f \end{align} $$

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 $:

$$ \begin{align} c ~=~ f \, \lambda \end{align} $$

Two important wave properties: Wavelength and amplitude.

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:

$$ \begin{align} c &~=~ 299 \, 792 \, 458 \, \frac{\mathrm m}{\mathrm s} \\\\
&~\approx~ 3 \cdot 10^{8} \, \frac{\mathrm m}{\mathrm s} \end{align} $$

Example: Convert wavelength and frequency into each other

From the instruction manual of your laser pointer you read that the light emitted by the laser pointer has a wavelength $ \class{green}{\lambda} ~=~ \class{green}{500 \, \mathrm{nm}} $ ($ \mathrm{nm} $ stands for "nanometer", so $10^{-9} \, \mathrm{m}$).

You have just learned that you can easily determine the frequency $ \class{green}{f} $ from the wavelength. So you use the formula 3 rearranged for frequency:

$$ \begin{align} \class{green}{f} &~=~ \frac{c}{\class{green}{\lambda}} ~=~ \frac{3 \cdot 10^8 \, \frac{\text m}{\text s}}{ \class{green}{500 \cdot 10^{-9} \, \text{m}} } \\\\
&~=~ \class{green}{ 6 \cdot 10^{14} \, \text{Hz} } \end{align} $$

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:

$$ \begin{align} W_{\text p} ~=~ h \, \frac{c}{\lambda} \end{align} $$

Remember!

The larger the light wavelength $ \lambda $, the smaller is the energy of a photon.

Four photons, which have different energy and color depending on 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.

Table : Examples of photon energy, as well as the colors of light.
Light wavelength	Light frequency	Energy of a photon
575 nm	5.2 × 10¹⁴ Hz	3.5 × 10^-19 J
546 nm	5.5 × 10¹⁴ Hz	3.6 × 10^-19 J
435 nm	6.9 × 10¹⁴ Hz	4.6 × 10^-19 J
400 nm	7.5 × 10¹⁴ Hz	5 × 10^-19 J
365 nm (UV light)	8.2 × 10¹⁴ Hz	5.4 × 10^-19 J

Example: Photon energy of yellow light

Take an interference filter, which filters out all colors except the yellow color. The yellow color has the wavelength $ \lambda = 575 \, \mathrm{nm} $. Written out the wavelength of the yellow light is: $ \lambda = 575 \cdot 10^{-9} \, \mathrm{m} $.

To determine the photon energy having wavelength $ \lambda = 575 \, \mathrm{nm} $, you use the light quantum hypothesis 1.

Since the wavelength of the yellow light is given instead of the light frequency, you use the rewritten formula 4 instead of 1. Thus, the photon energy of the yellow light is:

$$ \begin{align} W_{\text p} &~=~ 6.6 \cdot 10^{-34} \, \mathrm{Js} ~\cdot~ \frac{3\cdot10^{8} \, \frac{\mathrm m}{\mathrm s}}{575 \cdot 10^{-9} \, \mathrm{m}} \\\\
&~=~ 3.4\cdot10^{-19} \, \mathrm{J} \end{align} $$

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:

$$ \begin{align} W_{\text{mol}} ~=~ N_{\text A} \, W_{\text p} \end{align} $$

Substitute equation 1 into 8:

$$ \begin{align} W_{\text{mol}} ~=~ N_{\text A} \, h \, \frac{c}{\lambda} \end{align} $$

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.

Example: What is the energy of one mole of photons of red light?

You want to know what is the photon energy per mole $ W_{\text{mol}} $ of a red light that has wavelength $ \lambda = 780 \, \mathrm{nm} $. This corresponds to $ \lambda = 780 \cdot 10^{-9} \, \mathrm{m} $. Insert the wavelength into Eq. 9:

$$ \begin{align} W_{\text{mol}} ~&=~ 6 \cdot 10^{23} \, \frac{1}{\mathrm{mol}} ~\cdot~ 6.6 \,\cdot\, 10^{-34} \, \mathrm{Js} ~\cdot~ \frac{3 \cdot 10^8 \, \frac{\mathrm m}{\mathrm s}}{780 \cdot 10^{-9} \, \mathrm{m}} \\\\
~&=~ 152 \, \frac{\mathrm{kJ}}{\mathrm{mol}} \end{align} $$

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:

$$ \begin{align} W_{\text{mol}} ~=~ N_{\text A} \, h \, f \end{align} $$

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