Given that Planck's energy distribution equation is:
ρ P l a n c k = 2 c 2 λ 5 h e h c λ k T − 1 {\displaystyle \rho _{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:
ρ R a y l e i g h = 2 c k T λ 4 {\displaystyle \rho _{Rayleigh}={\frac {2ckT}{\lambda ^{4}}}}
Solution:
Evaluate the limit:
lim λ → ∞ {\displaystyle \displaystyle \lim _{\lambda \to \infty }}
by expanding the exponential in the denominator for first order in lambda:
e h c λ k T − 1 ≈ ( c h ) / ( k λ t ) ⟹ {\displaystyle e^{\frac {hc}{\lambda kT}}-1\approx (ch)/(k\lambda t)\implies } 2 c 2 λ 5 h e h c λ k T − 1 ≈ 2 c 2 λ 5 ( h ( c h ) / ( k λ t ) ) {\displaystyle {\frac {2c^{2}}{\lambda ^{5}}}{\frac {h}{e^{\frac {hc}{\lambda kT}}-1}}\approx {\frac {2c^{2}}{\lambda ^{5}}}\left({\frac {h}{(ch)/(k\lambda t)}}\right)}
then
2 c 2 λ 5 ( h ( c h ) / ( k λ t ) ) = 2 c k t λ 4 {\displaystyle {\frac {2c^{2}}{\lambda ^{5}}}\left({\frac {h}{(ch)/(k\lambda t)}}\right)={\frac {2ckt}{\lambda ^{4}}}}
