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Why does a thermal source emit across many wavelengths?
A hot source can emit a broad continuum rather than one pure color.
Sources for this view
An ideal blackbody’s spectral shape depends on its temperature. Stellar spectra often have a broadly thermal envelope, while real atmospheres add absorption and other departures from the ideal curve.
Sources for this view
Explore the model (optional)
EXPLORE THE MODEL
The thermal shape of light
Bλ = (2hc²/λ⁵) / [exp(hc/λkT) − 1]
Each curve is divided by its own sampled peak. This view compares shape; brightness cannot be compared.
Read sampled numerical values
| Vacuum wavelength (nm) | 6000 K | 3000 K reference |
|---|---|---|
| 100 | 1.441e-5 | 1.777e-14 |
| 390 | 0.888 | 0.061 |
| 680 | 0.78 | 0.713 |
| 970 | 0.402 | 1 |
| 1,260 | 0.206 | 0.857 |
| 1,550 | 0.113 | 0.635 |
| 1,840 | 0.066 | 0.452 |
| 2,130 | 0.041 | 0.321 |
| 2,420 | 0.027 | 0.231 |
| 2,710 | 0.018 | 0.168 |
| 3,000 | 0.013 | 0.125 |
Use the normalized view to compare shape and the fixed radiance scale to compare brightness. Hotter ideal emitters peak at shorter wavelengths and emit more at every wavelength. A thermal shape alone cannot establish elemental composition.
What this model can establish
Blackbody spectral radiance per unit vacuum wavelength. An ideal continuum is not a measured stellar spectrum; atmospheres, dust, spectral lines and instrument response are omitted. The peak depends on the spectral convention.
Optional: check your understanding
MAKE THE CONNECTION
A model spectrum peaks at 483 nm. Does it emit only at 483 nm?