# Photons

Particles are the parts that make up everything. Some examples are: electrons, protons, atoms, molecules, dust, and sand. Particles and waves act differently. A wave spreads out in all possible directions at once and can combine constructively and destructively with other waves.

Physicists were surprised to discover wave-like behavior for known particles, like electrons. They were equally surprised to discover particle-like behavior for known waves, like light.

A central idea in quantum mechanics is wave-particle duality. This duality is best explained with math. It is poorly summarized by this statement:

Particles act like waves until they interact with other particles.

When isolated, a particle's probable position spreads out as a wave capable of diffraction and interference. When measured, the particle's position has a precise value.

The wave behavior of particles is common at the subatomic scale. At larger scales wave behavior is less noticeable because of the difficulty isolating larger objects. This is why humans, or cats, don't act like waves.

## Photon Energy

In 1899, after investigating the thermal radiation spectrum, Maxwell Planck reluctantly hypothesized that the energy of light is only released in small quantities. This idea was the beginning of quantum mechanics.

This small quantity of light is called a photon. A photon acts like a wave. It has a frequency, a wavelength, and a constant speed. Yet, when detected, a photon acts like a particle with its own energy and position.

Each photon has an energy determined by its frequency or wavelength. Photons have energy but zero mass.

# $$E=hf \quad \quad E=\frac{hc}{\lambda}$$

$$E$$ = energy of a photon [J, Joules, kg m²/s²]
$$h$$ = 6.626 × 10 -34 = Planck's constant [J s]
$$f$$ = frequency [Hz, 1/s]
$$\lambda$$ = wavelength [m]
$$c$$ = speed of light = 3 × 10 8 [m/s]
Question: Which has more energy: a red photon or a blue photon?
solution
Red photons have a lower frequency and a lower energy.
Blue photons have a higher frequency and a higher energy.
Example: The frequency of a photon is 3.60 × 10 15 Hz. What's the photon's energy?
solution $$E=hf$$ $$E = (6.626 \times 10^{-34})(3.602 \times 10^{15})$$ $$E = 2.385 \times 10 ^{-18}J$$
Example: Find the energy for a photon that has a wavelength of 0.3m.
solution $$c = \lambda f$$ $$f = \frac{c}{\lambda}$$ $$f = \frac{ 3 \times 10^{8}}{ 0.3 }$$ $$f = 10^{9} Hz$$ $$E = hf$$ $$E = (6.626 \times 10^{-34})(10^{9})$$ $$E = 6.626 \times 10^{-25} J$$
Example: Find the wavelength of a photon that has 2.65 × 10 -19 J of energy.
solution $$E=\frac{hc}{\lambda}$$ $$\lambda=\frac{hc}{E}$$ $$\lambda=\frac{(6.626 \times 10^{-34})(3 \times 10^8)}{2.65 \times 10^{-19}}$$ $$\lambda = 7.5 \times 10 ^{-7} m$$ $$\lambda = 750nm$$
Example: A cheap helium neon laser outputs 5mW of optical power. Lookup the wavelength of the light and then calculate the number of photons produced every second.
hint
Wikipedia says the wavelength of the photons is 632.8 nm.
Convert the wavelength into energy.

The number of photons is the energy output divided by the energy per photon.
The energy output is 0.005J every second.

solution $$\lambda = 632.8 nm$$ $$E=\frac{hc}{\lambda}$$ $$E=\frac{(6.626 \times 10^{-34})(3 \times 10^8)}{632.8 \times 10^{-9}}$$ $$E = 3.14 \times 10^{-19}J$$
$$\frac{0.005W}{3.14 \times 10^{-19}J} = 1.59 \times 10^{16} \tfrac{1}{s}$$
15,900,000,000,000,000 photons/second from a low power laser beam!

## The Photoelectric Effect

The photoelectric effect occurs when metals dislodge electrons after being hit by light. Bright red light does not produce the effect, but dim UV light does.

In 1905, Albert Einstein published an explanation of the photoelectric effect. He suggested that light is made up of many small packets of energy, and each packet interacts with a single electron. Only high frequency light triggers the effect because it has more energy per packet.

The photoelectric effect can be understood with conservation of energy. The kinetic energy of a released electron is equal to the difference between the energy of an incoming photon and the energy needed to dislodge the electron.

# $$K_{max}=hf - \Phi$$

$$K_{max}$$ = maximum kinetic energy of released electron [J, Joules]
If the kinetic energy is below 0 an electron is not released.
$$h$$ = 6.626 × 10 -34 = Planck's constant [J s]
$$f$$ = frequency of incoming light [Hz, 1/s]
$$\Phi$$ = Work function, the minimum energy to dislodge an electron [J]
The work function depends on the material. It is lowest for metals.

Energy at the atomic level is often calculated in electron volts (eV) instead of Joules. We can convert between eV and J by multiplying or dividing by the charge of an electron.

$$1 eV = 1.6 \times 10^{-19}J$$ $$32 \times 10^{-19}J \left( \frac{1eV}{1.6 \times 10^{-19}J} \right)= 20eV$$
Example: After being hit by light with 7.0 eV per photon, the rare earth metal terbium releases electrons. The electrons have a maximum kinetic energy of 4.0eV. What is the work function of terbium in electron-volts?
solution $$K_{max} = hf - \Phi$$ $$\Phi = hf - K_{max}$$ $$\Phi = 7.0eV - 4.0eV$$ $$\Phi = 3.0eV$$
Question: How could a low luminosity UV light produce the photoelectric effect, but even when much brighter red light doesn't? The red light has more total energy. Shouldn't it have a better chance to dislodge electrons?
solution
Each single photon is hitting a single electron. Low energy photons don't have enough energy to dislodge an electron. Adding more photons doesn't help because two photons can't hit the same electron at once.
Example: Find the max kinetic energy of a magnesium electron after being hit by a photon with a frequency of 6.00 × 10¹⁴ Hz. Look up the work function for Magnesium and convert it into Joules.
solution $$\Phi = (3.66eV) \left( \frac{1.6 \times 10^{-19}J}{1 eV} \right)$$ $$\Phi = 5.856 \times 10^{-19} J$$
$$hf = (6.626 \times 10^{-34})(6.00 \times 10^{14})$$ $$hf = 3.9756 \times 10^{-19} J$$
$$K_{max}=hf - \Phi$$ $$K_{max}= (3.9756 \times 10^{-19}J) - (5.856 \times 10^{-19}J)$$ $$K_{max}= -1.88 \times 10^{-19} J$$
A negative kinetic energy means the electron will not escape the atom.
Example: A 100nm photon strikes a lump of magnesium. How much kinetic energy could the released electrons have?
solution $$c = f \lambda$$ $$f = \frac{c}{\lambda}$$ $$f = \frac{(3.00 \times 10^{8})}{100 \times 10^{-9}}$$ $$f = 3.00 \times 10^{15} Hz$$
$$\Phi = 3.66 eV \left( \frac{1.6 \times 10 ^{-19}J}{1eV} \right) = 5.856 \times 10^{-19} J$$
$$K_{max} = hf - \Phi$$ $$K_{max} = (6.626 \times 10^{-34})(3.00 \times 10^{15}) - 5.856 \times 10^{-19}$$ $$K_{max} = 19.878 \times 10^{-19} - 5.856 \times 10^{-19}$$ $$K_{max} = 14.022 \times 10^{-19} J$$

How fast would the electron be moving?
solution $$m_e = 9.10 \times 10^{-31} kg$$ $$K = \tfrac{1}{2}m_ev^2$$ $$v = \sqrt{\frac{2K}{m_e}}$$ $$v = \sqrt{\frac{2(14.022 \times 10^{-19})}{9.10 \times 10^{-31}}}$$ $$v = \sqrt{3.08 \times 10^{12}}$$ $$v = 1.76 \times 10 ^6 \tfrac{m}{s}$$
Example: Use the work function table to decide which elements would only produce electrons from light of wavelengths near 427nm.
hint
The lowest energy photon that can dislodge electrons will have the same energy as the work function. Convert the photon's wavelength into electronvolts and compare it to the work functions in the table.

solution $$E=\frac{hc}{\lambda}$$ $$E=\frac{(6.626 \times 10^{-34})(3 \times 10^8)}{427 \times 10^{-9}}$$ $$E = 4.66 \times 10^{-19}J$$ $$E = 4.66 \times 10^{-19}J \left( \frac{1 eV}{1.602 \times 10^{-19}J} \right)= 2.91eV$$

Elements with a work function near 2.91eV are: Ce, Gd, Li

## Emission and Absorption Energies

Each electron in an atom has an energy. These energy levels are limited to only a few possible values because the electrons form standing waves that must be even multiples of the electron's wavelength.

Electrons tend to occupy the lowest unfilled energy level, but if an electron gains the exact energy needed to jump to a higher energy level it will transition.

When an electron in an atom collides with a photon with the right energy, the electron can enter a higher unoccupied energy level in the atom. Electrons are unstable at higher energy levels. They will quickly collapse down to an unoccupied energy level and emit a photon. The photon will have an energy equal to the difference in energy levels.

Photon emission occurs when electrons transition to lower energy levels. Each electron transition emits a photon with an energy equal to the energy difference between energy levels.

Hydrogen has a very simple emission spectrum.

Iron has more electrons and more possible energy transitions.

Photon emission is the working principle behind fluorescent lights. To make light, a tube is filled with various gases. The gases are electrically charged up which brings the electrons to a higher energy level. The electrons are unstable in the higher energy levels. They eventually fall back down to their ground state and emit light. Try looking at fluorescent light reflected off a CD to see the separate bands of color.

Absorption is the reverse of emission. Single frequency light is absorbed by electrons causing them to transition to higher energy levels. Exposing atoms to a full spectrum of light will produce an absorption spectrum that also match the energy difference between electron energy levels.

If there isn't an energy difference that matches the energy of the colliding photon, the material is transparent to that frequency. A substance may be clear in one range of the spectrum but not in others. For example: glass is mostly transparent to visible light, but it has many absorption frequencies in the infrared.

Light absorption occurs at the same energies as emission. Hydrogen's absorption frequencies are the same as its emission frequencies.

Analyzing the spectrum of emission or absorption can actually be used like a fingerprint to identify the elements or molecules being observed. This technique is used in fields like forensics and astronomy.

Sunlight is mostly thermal radiation with large sections of absorption from the molecules in earth's atmosphere.

Example: What molecules are absorbing most of the infrared rays from the sun?
solution

H₂O, CO₂ and O₂ absorb infrared light and then re-emit the light in a random direction.

High H₂O and CO₂ concentrations lead to a warmer earth. This happens because incoming sunlight is mostly visible and is not absorbed by these gases, but light leaving the earth is mostly infrared and is absorbed.

Example: What molecule is responsible for absorbing the UV rays from the sun?
solution

O₃ (Ozone) absorbs some of the UV spectrum.
solution $$\lambda = 95nm$$
solution $$\lambda = 486nm = 4.86 \times 10^{-7}m$$ $$E = \frac{hc}{\lambda}$$ $$E = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{4.86 \times 10^{-7}}$$ $$E = 4.09 \times 10^{-19} J$$ $$E = 4.09 \times 10^{-19} J / (1.6 \times 10^{-19}) = 2.56 eV$$