# Basic principles of the photon model

#### Prerequisites

Since the photon model is in some sense a mental blend of the particle and wave models of light that can't be easily resolved into a since conceptual model, the most useful way to think about it is to establish some core anchor equations that express the basic ideas of the model as part of a set of foothold ideas.

## Foothold ideas for photons

Einstein started with the math of statistical mechanics, as described in the page the photon model of light. What he found was that he needed aspects of both models. Here are the key elements of the model:

• Light consists of packets of energy (photons) that can only interact with the atoms and molecules of matter by being absorbed or emitted in discrete units. You can't absorb or emit a fraction of a photon.
• Photons are NOT conserved. When an atom or molecule absorbs a photon, the atom or molecule gains the photon's energy and the photon disappears. When an atom or molecule emits a photon, it loses the photon's energy and the photon appears out of nothing. (It wasn't somehow "stored" in the atom and then emitted.)
• Photons carry momentum and energy like particles, but they also have a frequency and a wavelength like a sinusoidal wave.

This only becomes useful when we add the critical equations:

## Anchor equations for photons

Einstein found that he could make sense of Planck's black-body spectrum that describes the amount of energy in each frequency for electromagnetic radiation in thermal equilibrium if he took the momentum and energy of the photons to be given by

$$E = hf\quad \quad \ p = h/\lambda$$

In dealing with sinusoidal waves we often found it convenient to talk in terms of wave number, $k$, and angular frequency, $\omega$, instead of wavelength ($\lambda$) and frequency ($f$). These are defined by

$$\omega = 2\pi f \quad \quad k = 2\pi/\lambda.$$

We can use these if we define the constant "h-bar" equal to

$$\hbar = h/2\pi.$$

With this, our anchor equations for photons become

$$E = \hbar \omega \quad \quad p = \hbar k.$$

It's also sometimes convenient to express the energy in terms of the wavelength using

$$λf = c$$

where $c$ is the speed of light, giving the energy of a photon as

$$E = hc/λ.$$

Therefore, the amount of energy of a photon is inversely linked to its wavelength. Longer wavelength photons have lower energy than shorter wavelength photons. (This is why you sometimes see photons in chemistry specified by giving their inverse wavelength.)

## Doing the numbers

Planck's constant has the value,

$$h = 6.626 \times 10^{-34} \mathrm{\;J\;s}.$$

Since we often like to describe light in terms of its wavelength and, when we're dealing with atoms and molecules, we like to have energy in eV. As a result, eqs. (6) is particularly useful. Therefore, we combine a is the speed of light, $c = 3.0 x 10^8 \mathrm{\;m/s}$ and $h$ to get

$$hc = 1.99 \times 10^{-25} \mathrm{\;J\;}m.$$

This is not particularly convenient for atomic and molecular systems since we typically don't use Joules for an atom (though we do for moles of them). If we convert to eV and nm we get the more useful number

$$hc = 1240\; \mathrm{eV nm}.$$

For most purposes, a convenient mnemonic is "$hc = 1234 \;\mathrm{ eV nm}.$". That's off by less than 0.5% and easily allows you to remember how many places there are. This is a convenient form for doing estimations with atomic and molecular systems.

Using this information, we can determine how many photons there are in a light source such as a laser. Lasers emit a stream of photons (or a beam of light). The amount of light emitted is typically given in Watts (where 1 Watt equals 1 Joule per second).

A classroom laser pointer is often red in color, and a wavelength of 650 nm, and a power of 3 mW. The amount of energy in a single photon at this wavelength is just

$$E = (6.626 \times 10^{-34} \mathrm{\;J\;s}) \times (3.0 \times 10^8 \mathrm{\;m/s}) / (650 \times 10^{-9} \mathrm{\;m})$$

or $3 \times 10^{-19}{\mathrm{ J}}$ per photon. Since the laser has 3 mW of power, this means that there are

$$(3 \times 10^{-3}\mathrm{ J/s}) / (3 \times 10^{-19}\mathrm{ J/photon}) = 10^{16} \;\mathrm{ photons/ second}$$

coming out of that laser. Not bad for a small hand held device!

## This works: Does it make sense?

Although up to now it seems like photons are just particles, you should be troubled by the fact that your particles have wave properties. This is true of everything at the quantum level. Things are even worse for photons, since a single photon can interfere with itself (a wave property), photons can get "entangled" (a quantum property) and know about each others' states even when they are far apart, and the "number" of photons in a system might not be well defined.

While you wont need most of the quantum weirdness associated with photons for most biological applications (where all that is involved is energy conservation), the quantum properties of photons are looking extremely promising as ways to probe biological systems. Look forward to some really spectacularly interesting uses of photons in biology in the not too distant future!

Karen Carleton, Wolfgang Losert, and Joe Redish 4/17/13 and 7/10/19

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