Blackbody spectrum
Calculate the Planck spectrum, its peak and the visible energy fraction of an ideal blackbody.
About this tool
Ideal blackbody
This vacuum model uses emissivity 1 and temperatures from 500 to 12000 K. Examples are model temperatures, not exact measurements of particular stars. Calculate explicitly; editing clears the old result and Reset retains the temperature.
Spectrum and axes
Bλ = 2hc² / [λ⁵(exp(hc/(λkT))−1)]. Internally this is radiance per metre of wavelength. For W/(m²·sr·µm), multiply by 10⁻⁶. The vertical axis is linear and shows an explicit power-of-ten factor. The wavelength axis is logarithmic from 0.1 to 30 µm: equal horizontal distances mean equal wavelength ratios.
The curve remains Bλ, not λBλ. Its drawn area on this log axis is not an energy fraction. The display window is not the integration interval for total emission.
Results
The wavelength-density peak is λ_max = b/T, where b = hc/(kx₀) and x₀ = 4.965114231744276… is the positive nonzero root of 5(1−exp(−x₀)) = x₀. A frequency-density spectrum has a different peak. A single peak wavelength does not describe the perceived colour of the whole spectrum.
The chosen visible band is 380–780 nm. Its energy fraction is divided by the energy over all wavelengths, from zero to infinity; it is not weighted by human vision and is not a photon-count fraction. A convergent 64-term exponential series computes the fraction independently of plot samples. Very small values use scientific notation.
Total exitance into the outward hemisphere is M = σT⁴; integrated radiance is M/π. Visible exitance is the fraction times M. We derive σ = 2π⁵k⁴/(15h³c²) using current exact SI values: h = 6.62607015 × 10⁻³⁴ J·s, c = 299792458 m/s and k = 1.380649 × 10⁻²³ J/K. Material emissivity, detector response and colour matching are outside this model.
Sources
Historical numerical constants in these documents are replaced by the current SI definitions.