Question 40.3: The Ultraviolet Catastrophe Find the total intensity of a bl......
The Ultraviolet Catastrophe Find the total intensity of a black body as predicted by Rayleigh–Jeans’s law (Eq. 40.4).
I_\lambda=\frac{2 \pi c}{\lambda^4} k_{ B } T \quad \quad (40.4)INTERPRET and ANTICIPATE
We know that Rayleigh–Jeans’s law predicts at short wavelengths a higher spectral intensity than is observed (Fig. 40.5). So we expect that the total intensity will be greater than what is observed.
SOLVE
The total intensity comes from integrating the spectral intensity over all wavelengths (Eq. 40.2).
Substitute Rayleigh–Jeans’s formula (Eq. 40.4).
I=\int_0^{\infty}\left(\frac{2 \pi c}{\lambda^4} k_{ B } T\right) d \lambdaPull out the constants, integrate, and substitute limits.
\begin{aligned}I & =2 \pi c k_{ B } T \int_0^{\infty} \frac{d \lambda}{\lambda^4}=-\left.\frac{2 \pi c k_{ B } T}{3} \frac{1}{\lambda^3}\right|_0 ^{\infty} \\I & =\frac{2 \pi c k_{ B } T}{3}\left(\frac{-1}{\infty}+\frac{1}{0}\right) \rightarrow \infty\end{aligned}CHECK AND THINK
We just found that the total intensity predicted by Rayleigh–Jeans’s formula is infinite, which, as expected, is greater than the total intensity observed in a black body. Had we integrated Planck’s law instead, we would have found that the total intensity is given by I(T)=\sigma T^4 where \sigma=\frac{2 \pi^5 k_{ B }^4}{15 h^3 c^2} , a finite value consistent with Equation 40.3, when ε = 1.
