In our Higher Secondary textbook, it is clearly stated that Wein provided with the formula of Blackbody radiation based on the thermodynamics' law below $$ E_{\lambda} d\lambda = 8 \pi hc\lambda^{-5}e^{-\frac {hc}{k \lambda T}} d\lambda$$

Later on, Rayleigh-Jeans gave us the followed equation within the range of $\lambda + d\lambda$ $$ E_{\lambda} d\lambda = \frac {8\pi kT}{\lambda^{4}} d \lambda$$

For any shorter wavelength, Wein's formula matches with the experimental result but not consistent for longer wavelength whereas formula given by Rayleigh-Jeans makes a good performance for longer wavelength but no longer works for the experimental data used for analyzing energy density for comparably shorter wavelength.

I did somewhat a research for my curiosity. Now if I merge both equations by showing them equal for any optimized range of wavelength with its precise scale for any particular temperature '$T$' $$ 8 \pi hc\lambda^{-5}e^{-\frac {hc}{k \lambda T}} d\lambda = \frac {8\pi kT}{\lambda^{4}} d \lambda $$ $$ \implies kT \lambda = hc e^{-\frac {hc}{k \lambda T}} $$ $$ \implies \lambda = \frac {hc}{kT}e^{-\frac {hc}{k \lambda T}}$$ $$ \implies \ln \lambda = \ln (\frac {hc}{kT}) - \frac {hc}{k \lambda T} \ln e$$ $$ \implies \frac{d}{d \lambda} (\ln \lambda) = \frac {d}{d \lambda} {[ \ln (\frac {hc}{kT})]} - \frac {d}{d \lambda} (\frac {hc}{k \lambda T})$$ $$ \implies \frac {1}{\lambda} = -\frac {hc}{kT}(-\frac {1}{ \lambda^{2}})$$ $$ \therefore \lambda = \frac {hc}{kT}$$

And if we apply the value of $\lambda$ into Wein's formula, then we get Rayleigh-Jeans' formula and vice versa. So My Question is:

"Does it mean that in some certain points, both formula works the same for a certain range of wavelength equal to the numerical value of $\frac{hc}{kT}$ in such a way that experimental result can predict it with lots of clearance?"


1 Answer 1


This is a hard question, I cannot provide a full answer but maybe this mathematical insight will be helpful.

When you equate the two formulas and solve for $\lambda$. you are actually finding the points where two curves intersect. Now the formula $$ \lambda = \frac {hc}{kT}e^{-\frac {hc}{k \lambda T}}$$ defines the intersection implicitly and by the plots, there will be one (edit: at most one) numerical $\lambda$ where this holds.

Therefore, this is a single point defining formula and not a function of $\lambda$ that can be differentiated.

Perhaps if $\lambda$ was also a function of another parameter $s$, the formula would define a differentiable function $\lambda(s)$ via the implicit function theorem, and then differentiation would be possible.

Just my 2c.


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