Planck Blackbody Radiation Calculator

Every object with a temperature above absolute zero radiates electromagnetic energy. The Planck blackbody radiation law tells you how much energy is emitted at each wavelength, for any temperature T.

B(λ, T) = (2hc² / λ⁵) × 1 / (exp(hc / (λ k_B T)) − 1)

B is the spectral radiance — power per unit area, per steradian (solid angle), per unit wavelength. SI units are W·m⁻²·sr⁻¹·m⁻¹. That’s a mouthful, so most people quote it as W/(m²·sr·nm) when wavelength is in nanometres.

Why Planck mattered historically. In 1900, classical physics predicted that a blackbody’s radiation should grow without bound at short wavelengths (the “ultraviolet catastrophe”). Real measurements disagreed wildly with theory above the visible. Max Planck got the right curve by assuming, against every physical instinct of the time, that energy is exchanged in discrete packets E = hf. He thought the trick was a mathematical convenience. It turned out to be the founding observation of quantum mechanics.

Three things the curve tells you:

The peak. Hotter objects peak at shorter wavelengths. Wien’s displacement law (derivable from this formula) says λ_peak × T = 2.898 × 10⁻³ m·K. The Sun (~5778 K) peaks at ~501 nm, in the green of the visible spectrum. A red-hot stove element at ~1000 K peaks at ~2900 nm, deep in the infrared. Earth at 288 K peaks at ~10 μm. This is the entire reason infrared cameras work for night vision.

The total power. Integrate B(λ, T) over all wavelengths and you get the Stefan-Boltzmann law: P/A = σT⁴, where σ = 5.67 × 10⁻⁸ W/(m²·K⁴). Doubling temperature multiplies total emission by 16. This is why a tungsten lamp filament at 3000 K is so much brighter per square millimetre than your skin at 310 K, even though both are emitting blackbody radiation.

The colour. The visible part of the curve determines apparent colour. Cooler stars (3000 K) look reddish; the Sun looks white-ish (with a slight yellow tint when filtered through atmosphere); Sirius at ~9940 K looks bluish-white. Photographers and lighting designers call this colour temperature, and they buy bulbs labelled 2700 K (warm white) or 5000 K (daylight) to match.

Worked example. A tungsten filament at T = 3000 K, λ = 500 nm (green light). hc/(λk_BT) = (6.626e-34 × 3e8) / (500e-9 × 1.381e-23 × 3000) ≈ 9.6. So the denominator (exp(9.6) − 1) ≈ 14,754. Numerator 2hc²/λ⁵ ≈ 3.81e13. B ≈ 2.58e9 W/(m²·sr·m). That’s per metre of wavelength, so per nanometre it’s 2.58 W/(m²·sr·nm) — the green-light spectral radiance from the filament.

Caveats. Real objects are never perfect blackbodies. Their effective emission is reduced by an emissivity factor ε ≤ 1. For most matte surfaces ε ≈ 0.85 to 0.95. Polished metals drop to ε ≈ 0.05 — which is exactly why a polished copper bottom on a pan loses heat slowly.