Blackbody Radiation Calculator
Calculates blackbody spectral radiance, peak wavelength, and total radiated power using Planck's law and the Stefan–Boltzmann law. Use it when analysing thermal radiation from stars, furnaces, or infrared sensors.
About this calculator
A perfect blackbody absorbs all incident radiation and re-emits thermal radiation whose spectrum depends only on temperature T. Planck's law gives the spectral radiance: B(λ,T) = (2hc²/λ⁵) / (exp(hc/(λkT)) − 1), where h = 6.626×10⁻³⁴ J·s, c = 3×10⁸ m/s, and k = 1.381×10⁻²³ J/K. The peak wavelength follows Wien's displacement law: λ_max = 2.898×10⁻³ / T (m), so hotter objects peak at shorter, bluer wavelengths. Total power radiated per unit area is given by the Stefan–Boltzmann law: P = σT⁴, where σ = 5.67×10⁻⁸ W·m⁻²·K⁻⁴. Planck's quantisation hypothesis — that energy is emitted in discrete quanta E = hν — was the historical foundation of quantum mechanics, resolving the ultraviolet catastrophe predicted by classical physics.
How to use
Find the peak wavelength and total power for a star with surface temperature T = 5778 K (the Sun). Peak wavelength: λ_max = 2.898×10⁻³ / 5778 = 5.015×10⁻⁷ m = 501.5 nm — green-yellow visible light, consistent with solar observations. Total power per unit area: P = 5.67×10⁻⁸ × (5778)⁴ = 5.67×10⁻⁸ × 1.114×10¹⁵ ≈ 6.32×10⁷ W/m². Enter temperature = 5778 K, select peak wavelength to get ~502 nm, then select total power to get ~6.32×10⁷ W/m².
Frequently asked questions
What is the difference between Planck's law and the Stefan–Boltzmann law for blackbody radiation?
Planck's law describes how radiated power is distributed across wavelengths at a given temperature — it gives the full spectral shape. The Stefan–Boltzmann law is obtained by integrating Planck's law over all wavelengths, yielding the total power P = σT⁴ — a single number for the total emission. Wien's displacement law, a further simplification, identifies only the peak wavelength. For applications like designing thermal cameras or estimating stellar luminosity you often need only σT⁴ or λ_max, but for detailed spectral matching — e.g. calibrating spectrometers or designing selective emitters — the full Planck spectrum is required.
Why does the peak wavelength of blackbody radiation shift to shorter wavelengths at higher temperatures?
Wien's displacement law λ_max = 2.898×10⁻³/T shows an inverse relationship: doubling the temperature halves the peak wavelength. Physically, higher temperature means more thermal energy per oscillator, exciting higher-frequency (shorter-wavelength) modes more strongly. This is why a metal heated to ~800 K glows red, at ~3000 K (incandescent bulb filament) glows yellow-white, and a star at 30 000 K emits primarily ultraviolet radiation. The shift is also used in astronomy to measure stellar surface temperatures from colour alone.
How does the blackbody radiation formula resolve the ultraviolet catastrophe predicted by classical physics?
Classical Rayleigh–Jeans theory predicts spectral radiance proportional to kT/λ⁴, which diverges to infinity as wavelength decreases — the so-called ultraviolet catastrophe. Planck resolved this in 1900 by postulating that oscillators can only emit energy in discrete quanta E = hν = hc/λ. At short wavelengths (high frequencies), the quantum energy hν far exceeds kT, so the exponential term exp(hν/kT) becomes very large, driving the spectrum to zero. This quantum suppression at high frequencies perfectly matches experimental observations and was the founding insight of quantum mechanics.