\[B_{\lambda}=\frac{2hc^2}{\lambda^5\left(e^{\frac{hc}{\lambda kT}}-1\right)}\]
Variables
B[Bλ]spectral radiance (W·m^-3·sr^-1)
hPlanck constant (J·s)
cspeed of light (m/s)
lambda[λ]wavelength (m)
kBoltzmann constant (J/K)
Tabsolute temperature (K)
Description
What is this formula?
Planck's blackbody spectral radiance formula calculates the electromagnetic radiation emitted by an ideal blackbody as a function of wavelength and temperature. It was a foundational result of quantum mechanics and solved the ultraviolet catastrophe problem.
When to use it
Use this formula when analyzing thermal radiation, blackbody emission spectra, stars, infrared systems, and quantum physics phenomena.
Example
Data:
Temperature:
T=5800 K
Wavelength:
λ=500×10^-9 m
Formula:
Bλ=(2hc²)/(λ^5(e^(hc/(λkT))-1))
Substitution:
Bλ=(2(6.626×10^-34)(3×10^8)^2)/((500×10^-9)^5(℮^((6.626×10^-34×3×10^8)/(500×10^-9×1.38×10^-23×5800))-1))
Result:
Bλ≈2.64×10^13 W·m^-3·sr^-1
Applications
- Astrophysics
- Infrared thermography
- Thermal radiation analysis
- Stellar physics
- Quantum mechanics
- Remote sensing