F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Quantum MechanicsPhotons: Light Comes in Packets
F2.1 F2.2 F3.1 F1.1About 65 minutes
Learning goals- I can describe Einstein's photon model of light.
- I can calculate the energy of a photon with E = hf = hc/λ, in joules and in electron-volts.
- I can calculate how many photons a light source emits each second.
- I can explain why ultraviolet light can damage skin when brighter visible light does not.
Words to know
photon
A quantum (packet) of electromagnetic energy. Its energy is E = hf.
photon energy
E = hf = hc/λ; higher frequency (shorter wavelength) means more energy per photon.
electron-volt (eV)
The energy an electron gains when it moves through a potential difference of 1 V: 1 eV = 1.60 × 10⁻¹⁹ J.
intensity
The power of light per unit area. In the photon model it depends on the number of photons arriving per second.
photon flux
The number of photons passing per second (sometimes per unit area).
ionizing radiation
Radiation with photons energetic enough (roughly 10 eV or more) to knock electrons out of atoms.
Light, one packet at a time
In 1905, the same year he published special relativity, Albert Einstein proposed something even more daring than Planck. Planck had said that the oscillators in hot matter exchange energy in quanta. Einstein said that light itself is made of quanta. A beam of light is a stream of separate energy packets, later called photons.
Each photon of frequency f carries energy
E = hf = hc/λ
A photon is absorbed all at once or not at all. You cannot have half a photon. Einstein was led to this idea by studying the statistics of blackbody radiation, and he used it to explain the photoelectric effect (the next two lessons). It was the photoelectric work, not relativity, that the 1921 Nobel Prize in Physics recognized.
Remember from earlier units
c = fλ links frequency and wavelength. In the unit Gravitational, Electric and Magnetic Fields you used the work done on a charge, W = qΔV. The electron-volt comes straight from that formula.
A unit sized for photons: the electron-volt
Photon energies in joules are awkward numbers like 3.6 × 10⁻¹⁹ J. Physicists use the electron-volt (eV): the energy an electron gains when it is accelerated through a potential difference of 1 V.
E = qΔV = (1.60 × 10⁻¹⁹ C)(1 V), so 1 eV = 1.60 × 10⁻¹⁹ J.
- To change J → eV, divide by 1.60 × 10⁻¹⁹.
- To change eV → J, multiply by 1.60 × 10⁻¹⁹.
A handy shortcut: hc = 1.24 × 10³ eV·nm. Then E (in eV) = 1240 ÷ λ (in nm). For example, 620 nm red light: E = 1240 ÷ 620 = 2.00 eV.
Worked exampleExample 1: Energy of a green photon
Find the energy of a photon of green light, λ = 550 nm, in joules and in electron-volts.
- E = hc/λ = (6.63 × 10⁻³⁴ J·s)(3.00 × 10⁸ m/s) ÷ (5.50 × 10⁻⁷ m).
- E = 3.62 × 10⁻¹⁹ J.
- In eV: 3.62 × 10⁻¹⁹ J ÷ 1.60 × 10⁻¹⁹ J/eV = 2.26 eV.
- Shortcut check: 1240 ÷ 550 = 2.25 eV. (The tiny difference comes from rounding the constants.)
3.62 × 10⁻¹⁹ J, or about 2.26 eV.
Worked exampleExample 2: Counting photons from a laser pointer
A red laser pointer (650 nm) has an output power of 5.0 mW. How many photons does it emit each second?
- Energy per photon: E = hc/λ = (6.63 × 10⁻³⁴)(3.00 × 10⁸) ÷ (6.50 × 10⁻⁷) = 3.06 × 10⁻¹⁹ J.
- Power is energy per second: 5.0 mW = 5.0 × 10⁻³ J/s.
- Photons per second: N = P ÷ E = (5.0 × 10⁻³ J/s) ÷ (3.06 × 10⁻¹⁹ J) = 1.6 × 10¹⁶ photons/s.
About 1.6 × 10¹⁶ photons every second. That huge number is why light from ordinary sources looks smooth and continuous.
Worked exampleExample 3: From energy to wavelength
A photon has energy 3.0 eV. Find its wavelength and name its colour.
- Use the shortcut: λ = hc/E = (1240 eV·nm) ÷ 3.0 eV = 413 nm.
- Check in SI: E = 3.0 × 1.60 × 10⁻¹⁹ = 4.8 × 10⁻¹⁹ J; λ = (6.63 × 10⁻³⁴)(3.00 × 10⁸) ÷ (4.8 × 10⁻¹⁹) = 4.1 × 10⁻⁷ m.
About 410 nm, violet light near the edge of the visible spectrum.
Two beams of the same power. The violet beam needs fewer photons because each violet photon carries more energy (here 12 × 2 units = 8 × 3 units). Picture counts are for illustration only.Brightness versus colour
The photon model splits two ideas that the wave model mixed together:
- The energy of each photon depends only on frequency (colour): E = hf.
- The brightness (intensity) depends on how many photons arrive per second.
This explains a puzzle about sunburn. A bright red lamp can pour lots of energy onto your skin, but each red photon has only about 1.8 eV, too little to break the chemical bonds in skin cells' DNA. Ultraviolet photons from the Sun (for UV-B, about 4 eV each) can break those bonds one photon at a time. That is why a cloudy summer day can still burn you, and why sunscreen is designed to block UV rather than visible light.
At the high end, X-ray and gamma photons (thousands to millions of eV) can knock electrons right out of atoms. That is ionizing radiation, which you met in Grade 11 when studying nuclear decay. At the low end, a radio or Wi-Fi photon carries about 10⁻⁷ eV to 10⁻⁵ eV, millions of times too little to break a bond.
Worked exampleExample 4: Why radio waves are not ionizing
An FM radio station broadcasts at 100 MHz. Find the energy of one of its photons in eV and compare it with the roughly 10 eV needed to ionize an atom.
- E = hf = (6.63 × 10⁻³⁴ J·s)(1.00 × 10⁸ Hz) = 6.63 × 10⁻²⁶ J.
- In eV: 6.63 × 10⁻²⁶ ÷ 1.60 × 10⁻¹⁹ = 4.1 × 10⁻⁷ eV.
- Ratio: 10 eV ÷ 4.1 × 10⁻⁷ eV ≈ 2 × 10⁷.
Each FM photon has about 4.1 × 10⁻⁷ eV, roughly 20 million times too little to ionize an atom. Adding more photons does not help: in ordinary light, each photon acts alone.
Just for funCERN likes to say that 1 TeV (10¹² eV) is about the kinetic energy of a flying mosquito. The Large Hadron Collider squeezes several TeV into a single proton, which is a lot of mosquito for one particle.
Hands-on activityGlow-in-the-dark colour test
You need: a glow-in-the-dark sticker, toy or star (from a craft or dollar store), a dark room, and two light sources: a red light (a phone screen showing a full red image, or a red bike light) and a blue or violet light (a phone screen showing full blue, or a UV/black-light flashlight if you have one). Never shine any UV light into eyes.
- Leave the sticker in a dark drawer for an hour so it stops glowing.
- In the dark, hold the red light 5 cm from the sticker for 30 seconds. Turn it off. Does the sticker glow?
- Wait until it is dark again. Repeat with the blue/violet light for 30 seconds at the same distance.
- Try making the red light brighter or holding it longer.
What to look for: blue or violet light makes the sticker glow; red light, even bright red light for a long time, hardly does. The glow material needs photons above a certain energy (blue/violet photons, about 2.6 eV to 3.1 eV) to excite its electrons. Red photons (about 1.8 eV to 2.0 eV) are each too weak, and adding more weak photons does not add up. That is the photon idea in your hand. (Phone screens are not perfectly pure colours, so a very faint glow after red is possible.)
Your turn
1Find the energy, in joules and eV, of a photon of wavelength 450 nm.
E = (6.63 × 10⁻³⁴)(3.00 × 10⁸) ÷ (4.50 × 10⁻⁷) = 4.42 × 10⁻¹⁹ J = 4.42 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = 2.76 eV.
2Find the energy of a photon of frequency 5.0 × 10¹⁴ Hz in eV.
E = hf = (6.63 × 10⁻³⁴)(5.0 × 10¹⁴) = 3.3 × 10⁻¹⁹ J = 2.1 eV.
3A photon has energy 2.48 eV. What is its wavelength?
λ = 1240 ÷ 2.48 = 500 nm (blue-green).
4Convert 12.5 eV to joules.
12.5 × 1.60 × 10⁻¹⁹ = 2.00 × 10⁻¹⁸ J.
5A 1.0 mW green laser (532 nm) is on for 1.0 s. How many photons leave it?
E = 1240 ÷ 532 = 2.33 eV = 3.74 × 10⁻¹⁹ J. N = 1.0 × 10⁻³ J ÷ 3.74 × 10⁻¹⁹ J = 2.7 × 10¹⁵ photons.
6Which photon has more energy, one of red light or one of blue light? By what factor, if the wavelengths are 650 nm and 450 nm?
Blue. E ∝ 1/λ, so the ratio is 650 ÷ 450 = 1.44.
7Why can a dim UV lamp cause a chemical change that a bright red lamp cannot?
The change needs a certain energy delivered to one molecule. Each UV photon has enough energy; each red photon does not, and red photons cannot combine their energies. Brightness only changes the number of photons.
More practice: warm-up
1Write the two forms of the photon energy equation.
3If a photon's frequency doubles, what happens to its energy?
4If a photon's wavelength doubles, what happens to its energy?
5What does the brightness of a beam of a single colour depend on in the photon model?
The number of photons arriving per second.
More practice: on your own
1A microwave oven uses 2.45 GHz radiation. Find the photon energy in eV.
E = (6.63 × 10⁻³⁴)(2.45 × 10⁹) = 1.62 × 10⁻²⁴ J = 1.0 × 10⁻⁵ eV. Far too little to ionize: microwaves heat food by making water molecules rotate, not by breaking bonds.
2A dental X-ray photon has energy 25 keV. Find its wavelength.
λ = 1240 eV·nm ÷ 25 000 eV = 0.050 nm (5.0 × 10⁻¹¹ m).
3A light bulb gives off 5.0 W of visible light. Treat it all as 550 nm. How many visible photons per second?
E = 3.62 × 10⁻¹⁹ J. N = 5.0 ÷ 3.62 × 10⁻¹⁹ = 1.4 × 10¹⁹ photons/s.
4How much energy in total is carried by 2.0 × 10¹⁸ photons of 400 nm light?
E per photon = 1240 ÷ 400 = 3.10 eV = 4.96 × 10⁻¹⁹ J. Total = 2.0 × 10¹⁸ × 4.96 × 10⁻¹⁹ = 0.99 J (about 1.0 J).
5Explain why the photon model does not contradict the smooth appearance of everyday light.
Ordinary sources emit enormous numbers of photons per second (10¹⁶ or more), so individual packets are far too many and too small to notice, just as sand looks smooth from far away.
Challenge
1Sunlight delivers about 1.0 × 10³ W/m² at the ground on a clear day. Assume an average photon energy equal to that of 550 nm light. About how many photons strike a 1.0 cm² patch of skin each second?
Power on 1.0 cm² (1.0 × 10⁻⁴ m²) = 0.10 W. N = 0.10 ÷ 3.62 × 10⁻¹⁹ = 2.8 × 10¹⁷ photons/s.
2A detector registers a pulse of 1.5 × 10⁻¹⁷ J from a flash of 600 nm light. How many photons were in the pulse? Why must the answer be a whole number?
E per photon = (6.63 × 10⁻³⁴)(3.00 × 10⁸) ÷ (6.00 × 10⁻⁷) = 3.32 × 10⁻¹⁹ J. N = 1.5 × 10⁻¹⁷ ÷ 3.32 × 10⁻¹⁹ ≈ 45 photons. It must be whole because light is absorbed in whole photons; the measured energy is an exact multiple of hf (to within measurement error).
3Show that hc ≈ 1240 eV·nm starting from h = 6.63 × 10⁻³⁴ J·s and c = 3.00 × 10⁸ m/s.
hc = 1.989 × 10⁻²⁵ J·m. Divide by 1.60 × 10⁻¹⁹ J/eV: 1.243 × 10⁻⁶ eV·m. Multiply by 10⁹ nm/m: 1243 eV·nm ≈ 1240 eV·nm. (With more precise constants it is 1239.8 eV·nm.)
Quiz
1. What is the energy of a photon with frequency 1.0 × 10¹⁵ Hz?
2. A beam of blue light is made twice as bright without changing its colour. What happens?
3. Which photon carries the most energy?
4. What is 4.0 eV in joules?
5. A photon has energy 2.0 eV. Its wavelength is about
6. Why is a very bright red lamp unable to cause sunburn while weaker UV can?
Big idea (F2.1, F2.2, F3.1): light is made of photons with energy E = hf = hc/λ. Colour (frequency) sets the energy of each photon; brightness sets how many photons arrive.
How to help: have your teen explain the glow-in-the-dark activity in terms of photons. Ask: "If I made the red light 10 times brighter, would it work then?" (No: each photon is still too weak.)
Common difficulties:
- Unit slips: nanometres to metres (× 10⁻⁹), and J ↔ eV (÷ or × 1.60 × 10⁻¹⁹).
- Using the 1240 eV·nm shortcut with λ in metres, or with E in joules.
- Thinking brighter light means more energetic photons.
- Small last-digit differences (2.25 eV vs 2.26 eV) come from rounding constants and are fine.