Photons

Light as a Particle

Light acts like a wave. It has a frequency, a wavelength, and a constant speed, yet light waves can also collapse into a particle with a single value for energy, but no mass. We call a particle of light a photon.

After investigating blackbody radiation, Maxwell Planck hypothesized that the energy of light is equal to a even multiple of a constant times the frequency of the wave. This lead to the idea of a photon as a particle of light.

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

\(E\) = energy [J, Joules, kg m² s⁻²]
\(h\) = 6.626 x 10-34 = Planck's constant [J s]
\(f\) = frequency [Hz, s⁻¹]
Example: If the frequency of a photon is 3.60 × 1015 Hz, what’s its 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 $$c = \lambda f $$ $$f = \frac{c}{\lambda} $$ $$f = \frac{ 3 \times 10^{8}}{ 0.3 } $$ $$f = 10^{9} Hz$$ $$ E = hf $$ $$ E = (6.626 \times 10^{-34})(10^{9}) $$ $$ E = 6.626 \times 10^{-25} J$$

The Photoelectric Effect

We can use conservation of energy to find how much maximum kinetic energy an electron will have after being hit by a photon.

The energy of an incoming photon minus the energy needed to dislodge an electron equals the maximum possible kinetic energy of the electron.

$$KE_{max}=hf - \Phi$$

\(KE_{max}\) = maximum kinetic energy of an electron [J, Joules, kg m² s⁻²]
\(h\) = 6.626 x 10-34 = Planck's constant [J s]
\(f\) = frequency [Hz, s⁻¹]
\(\Phi\) = Work function, the minimum energy to dislodge an electron [J]
The work function depends on the material, and it is lowest for metals.

The photoelectric effect was described in one of Albert Einstein's early papers in 1905. It guided research towards technology like solar panels and digital cameras.

Example: After being hit by light with 7.0eV 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 $$KE_{max} = hf - \Phi$$ $$\Phi = hf - KE_{max}$$ $$\Phi = 7.0eV - 4.0eV$$ $$\Phi = 3.0eV$$
Example: Find the max velocity of a magnesium electron after being hit by a photon with a frequency of 6.00 x 1014 Hz. Look up the work function for Magnesium.
solution $$\Phi = (3.66)(1.6 \times 10^{-19})$$ $$\Phi = 5.856 \times 10^{-19} J$$
$$hf = (6.626 \times 10^{-34})(6.00 \times 10^{14})$$ $$hf = 3.9756 \times 10^{-19} J$$
$$KE_{max}=hf - \Phi$$ $$KE_{max}= (3.9756 \times 10^{-19}) - (5.856 \times 10^{-19}) $$ $$KE_{max}= -1.88 \times 10^{-19} J $$

A negative kinetic energy means the electron will not escape.

Line Emission and Absorption

When an electron in an atom collides with a particle, 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.

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Line emission is single frequency light emitted when electrons transition to lower energy levels. Line emission produces photons with an energy equal to the energy difference between energy levels.

Hydrogen has a very simple emission spectrum as you can see below.

Iron has more electrons and more possible energy levels in the spectrum below




Line absorption is when 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 absorption lines 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 lines in the infrared.

You can see the hydrogen absorption spectrum below. Light is absorbed at the same energies as hydrogen's emission spectrum.




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.

Line emission is the working principle behind fluorescent lights. To make light a tube is filled with various gases. The gases are ionized to bring 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.

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

image/svg+xml 250 500 750 1000 1250 1500 1750 2000 2250 2500 Wavelength (nm) Spectrum of Solar Radiation (Earth) 0 0.5 1 1.5 2 2.5 Irradiance (W/m²/nm) 2 H O Atmospheric absorption bands H O 2 H O 2 H O 2 H O 2 CO 2 O 2 O 3 UV Visible Infrared Sunlight without atmospheric absorption 5778K blackbody Sunlight at sea level 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.

strong>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.

103 nm n = 1 n = 2 n = 3 n = 4 n = 6 n = 5 434 nm 122 nm Lyman series Balmer series Paschen series 94 nm 410 nm 486 nm 656 nm 1875 nm 1282 nm 1094 nm 97 nm 95 nm 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 $$\lambda = 486nm = 4.86 \times 10^{-7}m$$
$$c = f\lambda $$ $$f = \frac{c}{\lambda}$$
$$E = hf$$ $$E = \frac{hc}{\lambda}$$ $$E = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{4.86 \times 10^{-7}}$$ $$E = 4.09 \times 10^{-19} J $$ $$E = 4.09 \times 10^{-19} J / (1.6 \times 10^{-19}) = 2.56 eV$$
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