Talk:Phy5645

Given that Planck's energy distribution equation is:

$\rho _{\text{Planck}}={\frac {2c^{2}}{\lambda ^{5}}}{\frac {h}{e^{\frac {hc}{\lambda kT}}-1}}$

show that in the long wavelength limit this reduces to the Rayleigh-Jeans equation:

$\rho _{\text{Rayleigh}}={\frac {2ckT}{\lambda ^{4}}}$

Solution:

Evaluate the limit:

$\displaystyle \lim _{\lambda \to \infty }$

by expanding the exponential in the denominator for first order in $\lambda \!$:

$e^{\frac {hc}{\lambda kT}}-1\approx (hc)/(\lambda kT)$

$\implies {\frac {2c^{2}}{\lambda ^{5}}}{\frac {h}{e^{\frac {hc}{\lambda kT}}-1}}\approx {\frac {2c^{2}}{\lambda ^{5}}}\left({\frac {h}{(hc)/(\lambda kT)}}\right)$

then

${\frac {2c^{2}}{\lambda ^{5}}}\left({\frac {h}{(hc)/(\lambda kT)}}\right)={\frac {2ckT}{\lambda ^{4}}}$