Playground
Linear plot of E vs frequency with a visible-light color bar. Slider picks frequency; energy shown in eV and J.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Photon energyoutput Energy of a single photon | J | M*L^2*T^-2 | 1e-25 – 1e-13 | |
| Planck's constant Fundamental quantum of action | J*s | M*L^2*T^-1 | 6.62607015e-34 – 6.62607015e-34 | |
| Frequency Frequency of the electromagnetic wave | Hz | T^-1 | 100000000 – 100000000000000000000 |
Deep dive
Derivation
Planck modeled blackbody oscillators as only emitting/absorbing energy in integer multiples of hf. Fitting the blackbody spectrum required this quantization. Einstein (1905) extended it: each electromagnetic mode of frequency f carries energy in packets of hf, which are localized photons.
Experimental verification
Photoelectric effect (Millikan 1916), Compton scattering (1923), single-photon interference experiments, and photon-counting detectors in astronomy confirm E = hf to parts-per-billion accuracy.
Common misconceptions
- Brightness does not change single-photon energy — only photon count
- E = hf applies to the photon, not to a 'wave packet' of finite duration
- Frequency here is the linear frequency f, not angular frequency omega = 2*pi*f
Real-world applications
- Solar cells and photovoltaics
- X-ray imaging and radiation therapy
- LED and laser design
- Photon energy thresholds for chemical reactions
Worked examples
UV photon energy
Given:
- f:
- 1000000000000000
Find: E
Solution
E = hf = 6.62607015e-34 * 1e15 = 6.626e-19 J ≈ 4.14 eV
Red laser (650 nm)
Given:
- lambda:
- 6.5e-7
Find: E
Solution
E = hc/lambda ≈ 3.06e-19 J ≈ 1.91 eV
Scenarios
What if…
- scenario:
- What if h were larger?
- answer:
- Single photons would carry more energy — visible light would be ionizing and life as we know it impossible.
- scenario:
- What if you halved the frequency?
- answer:
- Photon energy halves — red light instead of UV, and bond-breaking chemistry turns off.
- scenario:
- What if you increase intensity but keep f fixed?
- answer:
- More photons per second, same E per photon — no change to photoelectric threshold.
Limiting cases
- condition:
- f = 5e14 Hz (visible green)
- result:
- E ≈ 3.31e-19 J ≈ 2.07 eV
- explanation:
- Visible photon — enough to drive photosynthesis.
- condition:
- f = 1e15 Hz (UV)
- result:
- E ≈ 6.63e-19 J ≈ 4.14 eV
- explanation:
- Above many chemical bond energies — hence ionization and sunburn.
- condition:
- f → 0 (radio)
- result:
- E → 0
- explanation:
- Radio photons carry vanishingly small energy per photon, but coherent waves still transmit information.
Context
Max Planck · 1900
Planck introduced energy quanta to fit blackbody radiation data; Einstein extended it in 1905 to explain the photoelectric effect.
Hook
Why does UV light burn your skin but red light doesn't?
Energy of light comes in discrete packets proportional to frequency. What is the energy of a single UV photon at 1e15 Hz?
Dimensions:
- lhs:
- E → [M*L^2*T^-2]
- rhs:
- [M*L^2*T^-1] * [T^-1] → [M*L^2*T^-2]
- check:
- Both sides are energy. ✓
Validity: Valid for all electromagnetic radiation, any frequency. E and f are single-photon quantities; beam intensity corresponds to photon number.