Planck Bλ
Blackbody spectral radiance per unit wavelength (SI).
About this calculator
Planck's law was the equation that launched quantum mechanics: Max Planck derived it in 1900 by proposing, initially as a mathematical trick to fix a badly broken classical prediction, that a hot object can only emit or absorb light energy in discrete packets rather than continuously. Classical physics predicted that a blackbody (an idealized object that absorbs and re-emits all radiation perfectly) should radiate infinite energy at short wavelengths -- the "ultraviolet catastrophe" -- which plainly contradicted every real measurement. Planck's law, B_λ = 2hc²/(λ⁵(e^(hc/λkT) − 1)), fixes that by making short wavelengths exponentially suppressed rather than unbounded, matching real blackbody spectra at every wavelength and temperature. This calculator computes the spectral radiance B_λ -- how much power a blackbody radiates per unit area, per unit solid angle, per unit wavelength -- at any wavelength λ and temperature T you enter.
Two behaviors define the shape of the resulting curve. First, spectral radiance at any fixed wavelength always increases with temperature: a hotter object radiates more strongly at every wavelength, without exception, which is why heating an object makes it glow brighter across its whole visible spectrum, not just at one color. Second, the wavelength at which radiance PEAKS shifts to shorter wavelengths as temperature rises (Wien's displacement law) -- which is why a heated metal glows first dull red, then orange, then white as it gets hotter: the peak of its emission is sliding from infrared through the visible spectrum toward blue. The Sun's surface temperature of 5,778 K puts its emission peak almost exactly at 500 nm, in the green-blue part of the visible spectrum -- not a coincidence, since the human eye evolved to be most sensitive near where sunlight is brightest.
Inputs
Results
B_λ (W·sr⁻¹·m⁻³)
26,375,669,866,614.81
How to Use This Calculator
- Enter the blackbody temperature T (K) -- Sun surface ~5778 K, room temperature ~293 K.
- Enter the wavelength lambda (m) -- visible light: 380-700 nm = 3.8e-7 to 7e-7 m.
- Review spectral radiance in W per sr per m3.
- Change T (K) and watch the chart's peak slide across the fixed 100 nm-50,000 nm window -- that shift is Wien's displacement law.
- Compare spectral radiance at different temperatures to understand why hotter objects appear bluer.
How the result changes with T (K)
| T (K) | B_λ (W·sr⁻¹·m⁻³) |
|---|---|
| 2,889 | 180,036,222,681.54 |
| 4,334 | 4,990,603,313,928.01 |
| 8,667 | 142,940,235,649,968.84 |
| 14,445 | 602,036,989,327,157.4 |
What each input means
- T (K)
- Sun ~5778 K.
- λ (m)
- 500 nm green. Range covers 100 nm (near UV) to 50,000 nm (far infrared), where real thermal/blackbody sources actually emit.
How this is calculated
Formula
B_λ = 2hc² / (λ⁵ × (e^(hc/λkT) − 1))Engine last updated . Checked against 2 independently-derived tests — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.
Frequently Asked Questions
Why does the ultraviolet catastrophe matter here?
Because it's the specific problem Planck's law was invented to fix, and understanding it explains why the formula has the shape it does. Classical physics (the Rayleigh-Jeans law) predicted that blackbody radiance should increase without bound as wavelength gets shorter, which would mean any warm object emits infinite ultraviolet energy -- clearly false. Planck's law instead has an exponential term in the denominator that suppresses short-wavelength emission, matching real measurements and correctly predicting that spectral radiance rises to a peak and then falls at shorter wavelengths.
Why does a hotter object glow a different color, not just brighter?
Because raising temperature doesn't just scale radiance up uniformly across all wavelengths -- it shifts WHERE the emission peaks, toward shorter wavelengths, per Wien's displacement law. A moderately hot object's peak sits in the infrared (invisible, felt as heat), a hotter object's peak moves into red visible light, and a very hot object's peak moves through orange, yellow, and eventually toward blue-white. This is why blacksmiths and glassblowers have historically judged temperature by color -- the color directly tracks where the emission peak has shifted to.
Why does the Sun's peak emission land almost exactly in visible green-blue light?
Because the Sun's surface temperature, about 5,778 K, happens to place its Wien's-law emission peak at roughly 500 nm -- squarely in the visible spectrum near green-blue. This isn't coincidence from the biological side: human eyes evolved under sunlight and are most sensitive to wavelengths close to where solar radiance is strongest, which is also close to where Earth's atmosphere is most transparent. Enter 5778 for T (K) and 5e-7 for λ (m) in this calculator to see that near-peak value directly.
Does spectral radiance ever decrease as temperature increases, at a fixed wavelength?
No -- at any single fixed wavelength, Planck's law is strictly increasing in temperature with no exceptions. What changes with temperature is where the OVERALL curve peaks across wavelengths, not whether a given wavelength's radiance can go down as the object gets hotter. If a specific wavelength's radiance seems to "fall behind" as temperature rises, it's because the peak has moved past it toward shorter wavelengths, not because that wavelength's own radiance value decreased.
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