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.

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.

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.

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.
This reduces ionizing radiation.

Example: What wavelength of light could make an electron jump from energy level n = 1 to n = 5?
solution
$$\lambda = 95nm$$

Example: Find the energy of a photon produced as an electron drops from energy level n = 4 to n = 2?
solution
$$\lambda = 486nm = 4.86 \times 10^{-7}m$$ $$E = \frac{hc}{\lambda}$$ $$E = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{4.86
\times 10^{-7}}$$ $$E = 4.09 \times 10^{-19} J $$ $$E = 4.09 \times 10^{-19} J / (1.6 \times 10^{-19}) = 2.56 eV$$