The distribution of intensity of radiation with frequency for a blackbody, an ideal object that is both a perfect absorber and a perfect emitter of radiation and emits same intensity in all directions, is given by the Planck radiation law. Let \(B(f,T)\) denote the spectral radiance, i.e., the power in the radiation of frequency \(f\) per unit solid angle and per unit of area normal to the propagation in the radiation emitted by an object at thermal equilibrium at temperature \(T\) (Kelvin). The SI unit of \(B\) will be \(\text{W.m}^{-2}.\text{Sterad}^{-1}.\text{Hz}^{-1}\text{.}\) According to Plank’s law, the spectral radiance is given by
\begin{equation}
B(f,T) = \frac{2h}{c^2} \frac{f^3}{\exp(hf/k_BT) - 1}\ \ \ \text{(area normal to propagation)}\tag{23.13}
\end{equation}
Although spectral radiance \(B\) is independent of direction in space, the energy in the direction that is an angle \(\theta\) from the normal direction will have a factor of \(\cos\theta\) due to projection of the tilted area on the normal. Furthermore, we need to multiply it by the frequency range to convert it to power flowing per unit area. That is, the power flux in the direction \(\theta\) from normal will be
\begin{equation}
d\Phi(f,T, \theta) = B(f,T)\,\cos\theta\,d\Omega\,df.\tag{23.14}
\end{equation}
Writing the solid angle element \(d\Omega = \sin\theta\,d\theta\,d\phi\) and integrating over \(0\le \theta\le \theta/2\) and \(0\le \phi\le 2\pi\text{,}\) we get total power per unit area emitted in radiation of frequency \(0\le f \lt \infty\text{.}\)
\begin{equation}
d\Phi = \dfrac{2\pi h}{c^2}\dfrac{f^3}{\exp{\left(\dfrac{hf}{k_B T}\right)} - 1} df\ \ (0\le f \lt \infty) \tag{23.15}
\end{equation}
where \(c (3 \times 10^8\ \text{m/s})\) is the speed of light, \(h (6.63\times 10^{-34} \text{J.s})\) the Planck constant and \(k_B (1.38 \times 10^{-23} \text{J/K})\) the Boltzmann constant.
You can obtain the Stephan Boltzmann law, stated in Eq.
(23.7), from the Planck distribution, given in Eq.
(23.15), by integrating the later from
\(f = 0\) to
\(\infty\text{.}\)
\begin{equation*}
\Phi = \dfrac{2\pi h}{c^2}\left( \frac{k_B T}{h} \right)^4\,\int_0^\infty\, \frac{y^3\, dy}{e^y - 1},
\end{equation*}
where \(y = \dfrac{hf}{k_B T}\text{.}\) The integral evaluates to \(\pi^4/15\text{.}\) Combining all the constants in to one symbok, we get
\begin{equation*}
\Phi = \sigma\, T^4,
\end{equation*}
with
\begin{equation*}
\sigma = \frac{2\pi^5}{15}\,\frac{h}{c^2}\,\left( \frac{k_B}{h} \right)^4 = 5.7\times 10^{-8}\, \text{W/m}^2\text{K}^4.
\end{equation*}
By asking the frequency at which maximum of the spectral radiance in Eq.
(23.13), you can derive Wein’s law. Thus, let
\(y=hf/k_BT\) in that equation and simplify to
\begin{equation*}
B(f,T) = A \frac{y^3}{\exp(y) - 1},
\end{equation*}
where \(A\) stands for all the constant factors in the resulting equation that does not have either \(f\) or \(T\text{.}\) Then, you can find maximum of \(B\) with respect to \(y\text{.}\)
\begin{equation*}
\frac{dB}{df} = 0.
\end{equation*}
This gives the following condition
\begin{equation*}
e^{y} = \frac{1}{1 - \frac{1}{3}y}.
\end{equation*}
This can be solved by plotting left and right side on the same graph and finding the place(s) they intersect. They intersect at
\begin{equation*}
y \approx 2.82 \equiv y_0
\end{equation*}
That is when
\begin{equation*}
hf/k_BT \approx y_0 \longrightarrow f = a\, T,\ \ a = k_B\,y_0/h.
\end{equation*}
To write this as a condition on wavelength, we just need the fundamental equation.
\begin{equation*}
f \lambda = c, \ \ \text{speed of light, i.e., radiation here}.
\end{equation*}
This leads to the Wein’s displacement law
\begin{equation*}
\lambda = \frac{b}{T},
\end{equation*}
with
\begin{equation*}
b = \frac{c}{a} = \frac{hc}{k_By_0} \approx 2.9\times 10^{-3}\,\text{K.m}.
\end{equation*}
