Question

Using the Plank’s theory, show that ρ(ν)dν = 8πν2 c 3 hν e hν kBT − 1 dν

Answer 1

$\rho(\nu)=\frac{8\pi h\nu^3}{c^3}(\frac 1{e^{\frac{h\nu}{kT}}-1}),$

$\rho(\nu)d\nu=\frac{8\pi \nu^2}{c^3}(\frac {h\nu} {e^{\frac{h\nu}{kT}}-1})d\nu.$