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