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Photon Energy

Calculates the energy of a single photon from its wavelength, using Planck's constant and the speed of light.

OpticsPhotonicsFundamental

Photon Energy CalculatorE = h·c / λ

E = h · c / λ
E = photon energy (J)  ·  h = Planck's constant (J·s)  ·  c = speed of light (m/s)  ·  λ = wavelength (m)
⟹ SolveE, h, c, λ
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J·s
m/s
m
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Photon Energy
E: h: c: λ:
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Photon Energy (log scale)
Low (< 1e-19 J) Medium (1e-19–1e-17 J) High (> 1e-17 J)
E = h·c / λ  ·  h = 6.62607015 × 10⁻³⁴ J·s, c = 2.99792458 × 10⁸ m/s.

Interpretation

E = h·c/λ. Energy of a photon from its wavelength. Used in quantum optics, photovoltaics, and spectroscopy.

E = h*c/λ
Photon Energy

Variables

SymbolQuantityUnit
EPhoton energyJ or eV
hPlanck's constantJ*s
cSpeed of lightm/s
λWavelengthm

What it means

The energy of a photon is inversely proportional to its wavelength. This is fundamental in quantum mechanics and photonics. It is used to determine whether a photon has enough energy to excite an electron (as in solar cells or photodetectors), to calculate the energy levels in atoms, and to interpret spectra. Understanding this is essential for anyone working with light‑matter interactions and quantum optics.

Worked example

Photon Energy – Two Detailed Examples

Real‑World
Scenario: A blue LED emits light at λ = 400 nm. The photon energy E = hc/λ = (6.626×10⁻³⁴ J·s × 3×10⁸ m/s) / (400×10⁻⁹ m) = 4.97×10⁻¹⁹ J, which is approximately 3.1 eV. This energy is sufficient to excite electrons in some semiconductors, which is why blue LEDs were a breakthrough. The physicist uses this to understand the bandgap of the material.
ParameterValue
λ (nm)400
1E = (6.626e-34 × 3e8) / (400e-9) = 4.97e-19 J
2Convert to eV: 4.97e-19 / 1.602e-19 ≈ 3.10 eV
Result 3.1 eV ✓ Photon energy
Scenario: An infrared laser operating at λ = 1550 nm (1.55 µm) has photon energy E = hc/λ = (6.626e-34 × 3e8) / (1550e-9) = 1.28e-19 J ≈ 0.80 eV. This low energy is in the near‑infrared and is commonly used in fiber‑optic communications. The engineer considers that the energy is below the bandgap of silicon, so silicon detectors are not efficient at this wavelength, influencing the choice of photodetector material.
ParameterValue
λ1550
1E = (6.626e-34 × 3e8) / (1550e-9) = 1.28e-19 J
2≈ 0.80 eV
Result 0.80 eV ✓ Infrared photon
Insight: Photon energy is inversely proportional to wavelength. Shorter wavelengths (blue/UV) carry more energy, while longer wavelengths (IR) carry less. This is fundamental in photovoltaics, photodetectors, and spectroscopy.

Common mistakes

  • Photon energy: E = h·c / λ – where h is Planck’s constant, c is speed of light.
  • Units: h in J·s, c in m/s, λ in m → E in J. Or use h = 4.136×10⁻¹⁵ eV·s and c = 3.00×10⁸ m/s to get energy in eV.
  • Alternative: E = h·f – where f is frequency (Hz).
  • Quantum nature: Light energy is quantised in photons.
  • Photoelectric effect: The photon energy must exceed the work function to eject electrons.

Applications

Photon energy, E = h·c/λ, relates the energy of a photon to its wavelength. This is fundamental in photonics, quantum optics, and spectroscopy. Engineers use it to design photodetectors, solar cells, and optical communication systems. By calculating photon energy, they can determine whether a photon has enough energy to be absorbed by a semiconductor, to excite an electron, or to cause a photochemical reaction. This formula is also used in medical imaging (PET, X‑ray) and in astronomy to interpret spectra. Understanding photon energy is essential for any work involving light‑matter interactions.

  • Design of photodetectors and solar cells (bandgap matching)
  • Spectroscopic interpretation of atomic and molecular lines
  • Optical communication – energy per bit calculations
  • Medical imaging (X‑ray, gamma) – energy and dose
  • Education on quantum nature of light

Frequently Asked Questions

Q01What is the Photon Energy formula used for?
A01

It calculates the energy of a single photon from its wavelength: E = h c / λ.

Q02What do the variables E, h, c, and λ represent?
A02

E = photon energy (Joules).
h = Planck's constant (6.626×10⁻³⁴ J·s).
c = speed of light (3.0×10⁸ m/s).
λ = wavelength (m).

Q03What is the energy in electron‑volts (eV) for a given wavelength?
A03

E (eV) = 1240 / λ (nm), because hc ≈ 1240 eV·nm.

Q04How is the photon energy derived?
A04

From quantum mechanics: E = hν, and ν = c/λ, so E = hc/λ.

Q05What is the difference between photon energy and total light power?
A05

Photon energy is the energy per photon; total power is the product of photon energy and the photon flux (number of photons per second).

Q06Give a worked example using the photon energy formula.
A06

λ = 500 nm (green light). E = (6.626×10⁻³⁴ × 3.0×10⁸) / (500×10⁻⁹) = 3.975×10⁻¹⁹ J = 2.48 eV.

Q07What are the common pitfalls when applying the photon energy formula?
A07

  • Confusing photon energy with total energy of a light beam.
  • Forgetting to convert wavelength to metres when using SI units.
  • Using the wrong value of h or c.

Q08How does photon energy relate to the photoelectric effect?
A08

If photon energy exceeds the work function of a material, electrons are emitted; the excess energy becomes kinetic energy.

Q09What is the significance of photon energy in spectroscopy?
A09

It determines the transitions between energy levels in atoms and molecules, giving characteristic spectra.

Q10How does photon energy change with wavelength?
A10

Photon energy is inversely proportional to wavelength; shorter wavelengths (e.g., UV) have higher energy.