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Photons with $E = hf = hc/\lambda$ and $p = h/\lambda$, $hc = 1240$ eV·nm, photon counts from beam power, and the energies across the spectrum.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to convert among a photon's wavelength, frequency, energy and momentum, and count the photons in a beam.
From the waves and electromagnetism lessons you know that light is an electromagnetic wave with $c = f\lambda$, carrying energy and momentum. From the last unit you know that a massless particle has $E = pc$. This lesson adds the idea that changed physics in the twentieth century: light's energy comes in packets whose size is set by its frequency.
| Term | What it means |
|---|---|
| Photon | A quantum of light, carrying energy $hf$ and momentum $h/\lambda$. |
| Planck's constant | $h = 6.63 \times 10^{-34}$ J·s, the scale of quantum effects. |
| Electron volt | $1$ eV $= 1.60 \times 10^{-19}$ J, a convenient energy unit for photons and atoms. |
| $hc$ | $1240$ eV·nm: divide by a wavelength in nm to get a photon's energy in eV. |
| Quantum | The smallest amount of something that can be exchanged. |
| Photon flux | The number of photons crossing a surface per second. |
| Band gap | The smallest energy that frees an electron to conduct in a semiconductor. |
Light behaves as a wave when it spreads and interferes, but it is emitted and absorbed in discrete packets, photons. Each carries energy and momentum
$$E = hf = \frac{hc}{\lambda}, \qquad p = \frac{h}{\lambda},$$
with Planck's constant $h = 6.63 \times 10^{-34}$ J·s. Because $hc = 1240$ eV·nm, a photon's energy in electron volts is simply $1240$ divided by its wavelength in nanometers: red light at $620$ nm carries $2.0$ eV, violet at $400$ nm $3.1$ eV, an X-ray at $0.1$ nm about $12{,}000$ eV.
A beam's power $P$ is its number of photons per second times the energy of each, so $N = P/hf$. Brighter light has more photons, not more energetic ones; bluer light has more energetic photons.
Another way: picture
Picture light as a stream of raindrops rather than a smooth flow of water. A steady beam delivers so many drops each second that it seems continuous, but each drop lands whole. Their size depends on the color: violet drops are larger than red ones. A brighter beam is a heavier shower of the same-sized drops.
Another way: steps
In 1900 Max Planck was trying to explain the spectrum of light from hot objects. Classical physics predicted that a hot body should radiate infinitely much ultraviolet light, which is absurd. Planck found he could match the measurements if the walls of the oven exchanged energy with light only in steps of $hf$.
Planck regarded this as a mathematical trick. Einstein in 1905 took it literally: light itself consists of quanta of energy $hf$. The next lesson shows the evidence that forced the idea on physicists, and lesson 8 returns to Planck's hot oven.
Photon energies in joules are awkward numbers, around $10^{-19}$ J for visible light. The electron volt, the energy an electron gains across one volt, is the natural unit: visible photons carry $1.6$ to $3.3$ eV, comparable to the energies of chemical bonds and electrons in atoms.
Combining the constants, $hc = 6.63 \times 10^{-34} \times 3.00 \times 10^8 = 1.99 \times 10^{-25}$ J·m, which in eV and nm is $1240$ eV·nm. That single number turns every photon calculation into a division. It is worth memorizing, just as chemists memorize the gas constant.
Radio photons, with wavelengths of meters, carry about a millionth of an eV, far too little to break any chemical bond; that is why radio waves are harmless at ordinary intensities. Visible photons carry a few eV, enough to drive the chemistry of vision and photosynthesis. Ultraviolet photons, above about $4$ eV, can break the bonds in DNA, which is why they cause sunburn and skin cancer.
X-rays, thousands of eV, knock electrons out of atoms, and gamma rays from nuclei carry millions. The danger of radiation is set by the energy per photon, not by the total energy delivered: a warm bath delivers far more energy than a medical X-ray, but in photons too weak to break molecules.
A photon has no mass but carries momentum $p = h/\lambda = E/c$. The momentum of a single visible photon is tiny, about $10^{-27}$ kg·m/s, but a beam delivers many photons each second. Absorbed, the beam pushes with force $P/c$; reflected, with $2P/c$, as the solar-sail problem of the last course showed.
Photon momentum matters at the scale of atoms: an atom that absorbs a photon recoils. Laser cooling uses billions of such kicks, aimed against an atom's motion, to slow atoms to within a millionth of a degree of absolute zero, work that won Steven Chu, later U.S. Secretary of Energy, a share of the 1997 Nobel Prize.
A $1$ mW red laser pointer emits about $3 \times 10^{15}$ photons each second. The eye can detect a flash of a few photons in the dark, and modern detectors can count them one at a time. At very low light, photographs build up grain by grain, each grain the arrival of a single photon, as in the astronomical cameras that count photons from galaxies at the edge of the universe.
The graininess of light sets a limit on measurement called shot noise. A camera collecting $N$ photons in a pixel has a random uncertainty of about $\sqrt{N}$, which is why dim photographs look speckled, and why astronomers use long exposures.
Checking an answer. Visible photons must come out between about $1.6$ and $3.3$ eV. Bluer light must give larger energies. $E = pc$ must hold. And photon counts from everyday sources must be enormous, of order $10^{15}$ per milliwatt or more.
The relation $E = hf$ is an experimental law, confirmed by the photoelectric effect, by the spectra of atoms, and by photon-counting experiments. Planck's constant has been measured to parts per billion and, since 2019, is fixed exactly in the definition of the kilogram.
Using $E = pc$ for photons follows from the relativistic energy–momentum relation with $m = 0$. It agrees with Maxwell's classical result that light carries momentum $E/c$; the quantum picture adds that this momentum, too, comes in packets of $h/\lambda$.
Light shows both faces. It diffracts and interferes like a wave, which needs a wavelength; it is absorbed and emitted in whole packets, which needs particles. The resolution, developed in the quantum mechanics of the 1920s, is that the wave gives the probability of finding a photon, and each detection finds a whole photon at one place.
Send light through two slits one photon at a time and each lands at a single point on the screen. After thousands have arrived, their points form the interference pattern of a wave. Unit 3 shows the same happens with electrons, which are particles that also behave as waves.
Every camera sensor counts photons. In a phone camera's pixels, each visible photon with more than the silicon band gap, $1.12$ eV, can free one electron, and the charge collected is read out as brightness. Photons below the gap, in the infrared beyond about $1100$ nm, pass through unseen, which is why silicon cameras are blind to thermal infrared.
Night-vision goggles use photocathodes with small work functions that release electrons even for near-infrared photons, then multiply each electron thousands of times. Infrared cameras for heat use materials with much smaller gaps, cooled so that their own warmth does not swamp the signal. The choice of material in each case comes down to the energy per photon.
Plants run on photon energies. Chlorophyll absorbs red photons near $680$ nm, about $1.8$ eV, and blue ones near $430$ nm, about $2.9$ eV, and reflects the green in between, which is why leaves look green. Each absorbed photon lifts an electron in the chlorophyll to a higher energy, and a chain of molecules passes that electron along, storing its energy in chemical bonds.
Splitting one water molecule and fixing its hydrogen into sugar takes the energy of several photons, delivered one at a time. Because the process works photon by photon, a plant under dim red light and one under bright red light run the same reactions; the bright one simply runs them more often. Agricultural researchers growing lettuce under LEDs in indoor farms in New Jersey and Ohio tune the lamps' colors to the chlorophyll peaks, delivering the most useful photons per watt of electricity.
First Solar, headquartered in Tempe, Arizona, makes solar panels from cadmium telluride, with a band gap of $1.5$ eV; most other panels use silicon, with $1.12$ eV. A photon can free an electron only if its energy reaches the gap, so cadmium telluride can use light out to about $830$ nm and silicon to about $1100$ nm.
Photons with more energy than the gap free an electron too, but the excess is lost as heat. That trade-off limits a single-gap cell to about $33$ percent efficiency. Research cells at the National Renewable Energy Laboratory in Golden, Colorado, stack several materials with different gaps, each harvesting its own slice of the spectrum, and have exceeded $47$ percent.
The UV index reported in American weather forecasts measures the intensity of sunlight's ultraviolet, weighted by how harmful each wavelength is to skin. The weighting follows photon energy: UV-B, $280$ to $315$ nm, with photons of $4$ to $4.4$ eV, is hundreds of times more damaging per watt than UV-A at $315$ to $400$ nm.
Sunscreens absorb those energetic photons with molecules or mineral particles such as zinc oxide, whose band gap of $3.3$ eV lets it absorb wavelengths below about $375$ nm while passing visible light. Its whiteness comes from scattering, not absorption; the absorption is invisible because it lies in the ultraviolet.
A bright red light and a dim blue light: which has more energetic photons? The blue, always. Brightness is the number of photons per second; the energy of each depends only on the frequency. Turning up a red lamp delivers more photons of the same small energy, which is why no amount of red light can do what a single ultraviolet photon does to a molecule.
A related error is to multiply rather than divide by wavelength. Longer wavelength means lower frequency and less energy per photon.
Find the energy of a $550$ nm photon in eV.
$E = \dfrac{1240}{550} = 2.25\ \text{eV}$
$hc = 1240$ eV·nm.
Convert it to joules.
$E = 2.25 \times 1.60 \times 10^{-19} = 3.61 \times 10^{-19}\ \text{J}$
Joules per eV.
Find its frequency.
$f = \dfrac{3.00 \times 10^8}{550 \times 10^{-9}} = 5.45 \times 10^{14}\ \text{Hz}$
$c/\lambda$.
Check with $E = hf$.
$6.63 \times 10^{-34} \times 5.45 \times 10^{14} = 3.61 \times 10^{-19}\ \text{J}$
Agrees.
Find its momentum.
$p = \dfrac{6.63 \times 10^{-34}}{550 \times 10^{-9}} = 1.21 \times 10^{-27}\ \text{kg·m/s}$
$h/\lambda$.
An LED bulb emits $2.0$ W of light, mostly near $550$ nm. Find the energy per photon.
$E = 3.61 \times 10^{-19}\ \text{J}$
From the first example.
Find the photons per second.
$N = \dfrac{2.0}{3.61 \times 10^{-19}} = 5.5 \times 10^{18}\ \text{per second}$
Power over energy per photon.
Find the flux $3.0$ m away, spread over a sphere.
$\dfrac{5.5 \times 10^{18}}{4\pi \times 3.0^2} = 4.9 \times 10^{16}\ \text{per m}^2\text{ per s}$
Area of a sphere of radius $3.0$ m.
Find how many enter a $5.0$ mm pupil.
$4.9 \times 10^{16} \times \pi(0.0025)^2 = 9.6 \times 10^{11}\ \text{per s}$
Pupil area.
Compare with the eye's dark-adapted threshold.
$\text{about } 10 \text{ photons}$
The eye has enormous range.
Explain why the light looks steady.
$\text{a trillion photons each second}$
Individual arrivals are invisible at this rate.
A carbon–carbon bond in DNA needs about $3.6$ eV to break. Find the longest wavelength that can break it.
$\lambda = \dfrac{1240}{3.6} = 344\ \text{nm}$
Ultraviolet.
Find the energy of a $300$ nm UV-B photon.
$E = \dfrac{1240}{300} = 4.1\ \text{eV}$
Enough to break the bond.
Find the energy of a $10\ \mu$m infrared photon.
$E = \dfrac{1240}{10{,}000} = 0.124\ \text{eV}$
Far too little.
Compare the counts for equal energy delivered.
$\dfrac{4.1}{0.124} = 33$
Thirty-three infrared photons carry what one UV photon does.
Explain why the infrared only warms.
$\text{each photon too weak to break a bond}$
It sets molecules vibrating instead.
Find the wavelength limit for a $2.3$ eV color change in film.
$\lambda = \dfrac{1240}{2.3} = 540\ \text{nm}$
Why darkrooms use red light.
Summarize the rule.
$\text{damage depends on energy per photon}$
Not on total energy.
Write the photon energy.
$E = \dfrac{hc}{\lambda}$
Energy from wavelength.
Substitute the wavelength.
$E = \dfrac{1240}{405}$
In eV·nm over nm.
Evaluate the energy.
Each photon from a lamp carries $2$ eV. A second lamp emits light of half the wavelength. How much energy does each of its photons carry?
Complete the worked solution: a photon has wavelength $400$ nm. With $hc = 1240$ eV·nm, $1$ eV $= 1.60 \times 10^{-19}$ J and $h = 6.63 \times 10^{-34}$ J·s, find its energy in eV, its energy in units of $10^{-19}$ J, and its momentum in units of $10^{-27}$ kg·m/s.
Divide $hc$ by the wavelength.
$E = \dfrac{1240}{\lambda} =$ e
In eV.
Convert to joules.
$E = E_{\text{eV}} \times 1.60 \times 10^{-19} =$ j
In units of $10^{-19}$ J.
Divide Planck's constant by the wavelength.
$p = \dfrac{h}{\lambda} =$ p
In units of $10^{-27}$ kg·m/s.
Check that energy equals momentum times $c$.
$E = pc$
True for any massless particle.
Match each photon quantity to its formula.
| $hf$ | $hc/\lambda$ | $h/\lambda$ | $P/hf$ | |
|---|---|---|---|---|
| energy from frequency | ||||
| energy from wavelength | ||||
| momentum | ||||
| photons per second |
Light has wavelength $1000$ nm. With $hc = 1240$ eV·nm, $c = 3.00 \times 10^8$ m/s and $h = 6.63 \times 10^{-34}$ J·s, fill in the photon energy in eV, the frequency in units of $10^{14}$ Hz, and the photon momentum in units of $10^{-27}$ kg·m/s.
| value | |
|---|---|
| photon energy (eV) | |
| frequency (10¹⁴ Hz) | |
| momentum (10⁻²⁷ kg·m/s) |
A faint pulse of light contains exactly $8$ photons, all of wavelength $L$ nm. With $hc = 1240$ eV·nm, write the pulse's total energy, in eV, as a formula in $L$.
Answer:
A $5$ mW laser emits light of wavelength $532$ nm. With $h = 6.63 \times 10^{-34}$ J·s and $c = 3.00 \times 10^8$ m/s, how many photons does it emit each second, in units of $10^{15}$?
Answer: × 10¹⁵ photons per second
A solar engineer in Tempe, Arizona, compares cell materials. A photon can free an electron in gallium arsenide only if its energy is at least the band gap, $1.42$ eV. With $hc = 1240$ eV·nm, what is the longest wavelength the material can use, in nm?
Answer: nm cutoff
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A faint pulse of light contains exactly $6$ photons, all of wavelength $L$ nm. With $hc = 1240$ eV·nm, write the pulse's total energy, in eV, as a formula in $L$.
Answer:
You can compute photon energies. Explain to someone why no amount of red light can sunburn you, but ultraviolet can.
21. Your turn: find the energy in eV of a photon from a $405$ nm violet laser., step 3
$E = 3.06\ \text{eV}$
Near the violet end of vision.