Using the Plank’s theory, show that ρ(ν)dν = 8πν2 c 3 hν e hν kBT − 1 dν
ρ(ν)=8πhν3c3(1ehνkT−1),\rho(\nu)=\frac{8\pi h\nu^3}{c^3}(\frac 1{e^{\frac{h\nu}{kT}}-1}),ρ(ν)=c38πhν3(ekThν−11),
ρ(ν)dν=8πν2c3(hνehνkT−1)dν.\rho(\nu)d\nu=\frac{8\pi \nu^2}{c^3}(\frac {h\nu} {e^{\frac{h\nu}{kT}}-1})d\nu.ρ(ν)dν=c38πν2(ekThν−1hν)dν.
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