Portal:Radiation astronomy/Problems/6

Planck's equation
Planck's equation describes the amount of spectral radiance at a certain wavelength radiated by a black body in thermal equilibrium.

In terms of wavelength (λ), Planck's equation is written as


 * $$B_\lambda(T) =\frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda k_\mathrm{B}T}} - 1}$$

where B is the spectral radiance, T is the absolute temperature of the black body, kB is the Boltzmann constant, h is the Planck constant, and c is the speed of light.

This form of the equation contains several constants that are usually not subject to variation with wavelength. These are h, c, and kB. They may be represented by simple coefficients: c1 = 2h c2 and c2 = h c/kB.

By setting the first partial derivative of Planck's equation in wavelength form equal to zero, iterative calculations may be used to find pairs of (λ,T) that to some significant digits represent the peak wavelength for a given temperature and vice versa.


 * $$\frac{\partial B}{\partial \lambda} = \frac{c1}{\lambda^6}\frac{1}{ e^{\frac{c2}{\lambda T}} - 1}[\frac{c2}{\lambda T}\frac{1}{ e^{\frac{c2}{\lambda T}} - 1}e^{\frac{c2}{\lambda T}} - 5] = 0.$$

Or,


 * $$\frac{c2}{\lambda T}\frac{1}{ e^{\frac{c2}{\lambda T}} - 1}e^{\frac{c2}{\lambda T}} - 5 = 0.$$


 * $$\frac{c2}{\lambda T}\frac{1}{ e^{\frac{c2}{\lambda T}} - 1}e^{\frac{c2}{\lambda T}} = 5.$$

Use c2 = 1.438833 cm K.