Talk:Phy5645

Given that Planck's energy distribution equation is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho_{\text{Planck}} = \frac{2c^2}{\lambda^5}\frac{h}{e^\frac{hc}{\lambda k T}-1}}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho_{\text{Rayleigh}} = \frac{2ckT}{\lambda^4}}

Solution:

Evaluate the limit:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle\lim_{\lambda\to\infty}}

by expanding the exponential in the denominator for first order in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda \!} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^\frac{hc}{\lambda k T}-1 \approx (hc)/(\lambda kT) }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \implies \frac{2c^2}{\lambda^5}\frac{h}{e^\frac{hc}{\lambda k T}-1} \approx \frac{2c^2}{\lambda^5}\left(\frac{h}{(hc)/(\lambda kT)}\right) }

then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2c^2}{\lambda^5}\left(\frac{h}{(hc)/(\lambda kT)}\right) = \frac{2 c k T }{\lambda^4}}