Planck's Law of Blackbody Radiation
1. Introduction
Classical physics predicted the Rayleigh–Jeans law, which led to the ultraviolet catastrophe at high frequencies.
In 1900, Max Planck proposed a quantum hypothesis to explain blackbody radiation correctly.
Planck assumed that energy is emitted or absorbed in discrete packets
called quanta.
2. Planck's Quantum Hypothesis
Energy of an oscillator of frequency \( \nu \) is quantized:
\[
E = n h \nu \quad (n = 0,1,2,3,\dots)
\]
- \(h\) = Planck’s constant
- \(\nu\) = frequency of radiation
3. Average Energy of Oscillator
Using Bose–Einstein statistics, the average energy is:
\[
\langle E \rangle = \frac{h\nu}{e^{\beta h\nu} - 1}
\]
where
\[
\beta = \frac{1}{k_B T}
\]
4. Density of States
Number of electromagnetic modes between \( \nu \) and \( \nu + d\nu \):
\[
g(\nu) d\nu = \frac{8\pi \nu^2}{c^3} d\nu
\]
5. Planck's Radiation Formula
Energy density per unit frequency interval:
\[
u(\nu) d\nu =
\frac{8\pi h \nu^3}{c^3}
\frac{1}{e^{h\nu/k_B T} - 1}
\, d\nu
\]
Stefan's Law :
It states that the total energy radiated per unit area per unit time by a black body at any temperature T is directly proportional to the fourth power of it's absolute temperature .
Mathematically: $E \propto T^{4}$
$$
\begin{aligned}
& \text { or } \\
& E / A=\sigma T^{4} \quad\left(\sigma=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2}-K^{4}\right. \text { called as Stefan's Constant.) }
\end{aligned}
$$
Proof :
Energy density in the frequency range $f$ and $f+d f$
$$
\begin{aligned}
& \frac{d}{d \lambda}\left(E_{\lambda}\right)=\frac{d}{d \lambda}\left(\frac{8 \pi h c \lambda^{-5}}{e^{\frac{h c}{\lambda k T}}-1}\right)=0 \\
& \Rightarrow-5 \lambda^{-6}\left(e^{\frac{h c}{\lambda k T}}-1\right)^{-1}+\lambda^{-5}\left(e^{\frac{h c}{\lambda k T}}-1\right)^{-2} e^{\frac{h c}{\lambda k T}} \frac{h c}{k T} \lambda^{-2}=0 \\
& \Rightarrow-\frac{5}{\lambda^{6}} \frac{1}{e^{\frac{h c}{\lambda k T}}-1}+\frac{1}{\lambda^{7}} \frac{h c}{k T\left(e^{\frac{h c}{\lambda k T}}-1\right)^{2}}=0 \\
& \Rightarrow \frac{5}{\lambda^{6}}=\frac{1}{e^{\frac{h c}{\lambda k T}}-1}\left(\frac{h c}{k T}\right) \frac{1}{\lambda^{7}} \\
& \Rightarrow e^{\frac{h c}{\lambda k T}}-1=\frac{e^{\frac{h c}{\lambda k T}} h c}{5 \lambda k T} \\
& \Rightarrow e^{\frac{h c}{\lambda k T}}\left(1-\frac{h c}{5 \lambda k T}\right)=1 \\
& \Rightarrow e^{x}\left(1-\frac{x}{5}\right)=1 \\
& \Rightarrow x=4.965 \\
& \Rightarrow \frac{h c}{\lambda k T}=4.965 \\
& \Rightarrow \lambda_{m} T=\frac{4.965 k}{h c}=b=2 \cdot 898 \times 10^{-3} m-K \\
& \therefore \lambda_{m} \alpha \frac{1}{T}
\end{aligned}
$$
This is Planck’s Law of Blackbody Radiation.
$$
\begin{aligned}
\Rightarrow u(f) d f&=\frac{E(f) d f}{V}\\
&=\frac{8 \pi h f^{3} d f}{c^{3}\left(e^{\frac{h f}{k T}}-1\right)} \\
& \left.=8 \pi h\left(\frac{k T}{h}\right)^{3} \frac{k T}{h} \int_{0}^{\infty} \frac{x^{3} d x}{c^{3}\left(e^{x}-1\right)} \quad \text { Putting } \frac{h f}{k T}=x \Rightarrow f=x k T / h, \Rightarrow d f=d x k T / h\right) \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} \frac{x^{3} d x}{\left(e^{x}-1\right)} \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} \frac{x^{3} e^{-x} d x}{e^{-x}\left(e^{x}-1\right)} \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} \frac{x^{3} e^{-x} d x}{\left(1-e^{-x}\right)} \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} x^{3} e^{-x}\left(1-e^{-x}\right)^{-1} d x \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} x^{3} e^{-x}\left(1+e^{-x}+e^{-2 x}+e^{-3 x}+\ldots\right) d x \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty}\left(x^{3} e^{-x}+x^{3} e^{-2 x}+x^{3} e^{-3 x}+x^{3} e^{-4 x}+\ldots\right) d x \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}}\left(6\left(1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+\ldots\right)\right) \text { using } \int_{0}^{\infty} x^{3} e^{-a x} d x=\frac{6}{a^{4}} \\
& =\left(\frac{8 \pi^{5} k^{4}}{15 c^{3} h^{3}} T^{4}\right. \\
& \therefore u_{f} \alpha T^{4}
\end{aligned}
$$
Determination of Stefan's Constant :
$$
\begin{aligned}
& \Rightarrow \frac{1}{4} c u_{f}=\sigma T^{4} \\
& \Rightarrow \frac{1}{4} c \frac{8 \pi^{5} k^{4}}{15 c^{3} h^{3}} T^{4}=\sigma T^{4} \\
& \Rightarrow \sigma=\frac{2 \pi^{5} k^{4}}{15 c^{2} h^{3}}=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2}-K
\end{aligned}
$$
Number of Photons in Volume $\mathbf{V}$ at any temperature T is :
$$
\begin{aligned}
N&=\int_{0}^{\infty} n(f) d f \\
& =\int_{0}^{\infty} \frac{8 \pi V f^{2} d f}{c^{3}\left(e^{\frac{b f}{k T}}-1\right)} \\
& =\frac{8 \pi V}{c^{3}} \int_{0}^{\infty} \frac{f^{2} d f}{\left(e^{\frac{H f}{h_{T}}}-1\right)} \\
& =\frac{8 \pi}{c^{3}}\left(\frac{k T}{h}\right)^{2 \infty} \frac{x^{2} d x}{\left(e^{x}-1\right)}\left(\frac{k T}{h}\right) \quad \text { Substituiting } \frac{h f}{k T}=x \\
& =8 \pi\left(\frac{k T}{h c}\right) \int_{0}^{3 \infty} \frac{x^{2} d x}{\left(e^{x}-1\right)} \\
& =8 \pi\left(\frac{k T}{h c}\right) \int_{0}^{3 \infty} \frac{x^{2} e^{-x} d x}{e^{-x}\left(e^{x}-1\right)} \\
& =8 \pi\left(\frac{k T}{h c}\right)^{3 \infty} \int_{0}^{2} e^{-x}\left(1-e^{-x}\right)^{-1} d x \\
& =8 \pi\left(\frac{k T}{h c}\right)^{3 \infty} \int_{0}^{2} x^{2} e^{-x}\left(1+e^{-x}+e^{-2 x}+e^{-3 x}+e^{-4 x}+\ldots\right) d x \\
& =8 \pi\left(\frac{k T}{h c}\right)^{3 \infty} \int_{0}^{2}\left(x^{2} e^{-x}+x^{2} e^{-2 x}+x^{2} e^{-3 x}+x^{2} e^{-4 x}+x^{2} e^{-5 x}+\ldots\right) d x \\
& =8 \pi\left(\frac{k T}{h c}\right)^{3} \times 2 \times \sum \frac{1}{n^{3}} \quad \text { using } \int_{0}^{\infty} x^{m} e^{-a x} d x=\frac{m!}{n^{m+1}}
\end{aligned}
$$
6. Limiting Cases
6.1 Low Frequency Limit (hν ≪ kBT)
$$
\begin{aligned}
& \text { or } \\
& E / A=\sigma T^{4} \quad\left(\sigma=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2}-K^{4}\right. \text { called as Stefan's Constant.) }
\end{aligned}
$$
Proof :
Energy density in the frequency range $f$ and $f+d f$
$$
\begin{aligned}
& \frac{d}{d \lambda}\left(E_{\lambda}\right)=\frac{d}{d \lambda}\left(\frac{8 \pi h c \lambda^{-5}}{e^{\frac{h c}{\lambda k T}}-1}\right)=0 \\
& \Rightarrow-5 \lambda^{-6}\left(e^{\frac{h c}{\lambda k T}}-1\right)^{-1}+\lambda^{-5}\left(e^{\frac{h c}{\lambda k T}}-1\right)^{-2} e^{\frac{h c}{\lambda k T}} \frac{h c}{k T} \lambda^{-2}=0 \\
& \Rightarrow-\frac{5}{\lambda^{6}} \frac{1}{e^{\frac{h c}{\lambda k T}}-1}+\frac{1}{\lambda^{7}} \frac{h c}{k T\left(e^{\frac{h c}{\lambda k T}}-1\right)^{2}}=0 \\
& \Rightarrow \frac{5}{\lambda^{6}}=\frac{1}{e^{\frac{h c}{\lambda k T}}-1}\left(\frac{h c}{k T}\right) \frac{1}{\lambda^{7}} \\
& \Rightarrow e^{\frac{h c}{\lambda k T}}-1=\frac{e^{\frac{h c}{\lambda k T}} h c}{5 \lambda k T} \\
& \Rightarrow e^{\frac{h c}{\lambda k T}}\left(1-\frac{h c}{5 \lambda k T}\right)=1 \\
& \Rightarrow e^{x}\left(1-\frac{x}{5}\right)=1 \\
& \Rightarrow x=4.965 \\
& \Rightarrow \frac{h c}{\lambda k T}=4.965 \\
& \Rightarrow \lambda_{m} T=\frac{4.965 k}{h c}=b=2 \cdot 898 \times 10^{-3} m-K \\
& \therefore \lambda_{m} \alpha \frac{1}{T}
\end{aligned}
$$
This is Planck’s Law of Blackbody Radiation.
$$
\begin{aligned}
\Rightarrow u(f) d f&=\frac{E(f) d f}{V}\\
&=\frac{8 \pi h f^{3} d f}{c^{3}\left(e^{\frac{h f}{k T}}-1\right)} \\
& \left.=8 \pi h\left(\frac{k T}{h}\right)^{3} \frac{k T}{h} \int_{0}^{\infty} \frac{x^{3} d x}{c^{3}\left(e^{x}-1\right)} \quad \text { Putting } \frac{h f}{k T}=x \Rightarrow f=x k T / h, \Rightarrow d f=d x k T / h\right) \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} \frac{x^{3} d x}{\left(e^{x}-1\right)} \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} \frac{x^{3} e^{-x} d x}{e^{-x}\left(e^{x}-1\right)} \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} \frac{x^{3} e^{-x} d x}{\left(1-e^{-x}\right)} \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} x^{3} e^{-x}\left(1-e^{-x}\right)^{-1} d x \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} x^{3} e^{-x}\left(1+e^{-x}+e^{-2 x}+e^{-3 x}+\ldots\right) d x \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty}\left(x^{3} e^{-x}+x^{3} e^{-2 x}+x^{3} e^{-3 x}+x^{3} e^{-4 x}+\ldots\right) d x \\
& =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}}\left(6\left(1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+\ldots\right)\right) \text { using } \int_{0}^{\infty} x^{3} e^{-a x} d x=\frac{6}{a^{4}} \\
& =\left(\frac{8 \pi^{5} k^{4}}{15 c^{3} h^{3}} T^{4}\right. \\
& \therefore u_{f} \alpha T^{4}
\end{aligned}
$$
Determination of Stefan's Constant :
$$
\begin{aligned}
& \Rightarrow \frac{1}{4} c u_{f}=\sigma T^{4} \\
& \Rightarrow \frac{1}{4} c \frac{8 \pi^{5} k^{4}}{15 c^{3} h^{3}} T^{4}=\sigma T^{4} \\
& \Rightarrow \sigma=\frac{2 \pi^{5} k^{4}}{15 c^{2} h^{3}}=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2}-K
\end{aligned}
$$
Number of Photons in Volume $\mathbf{V}$ at any temperature T is :
$$
\begin{aligned}
N&=\int_{0}^{\infty} n(f) d f \\
& =\int_{0}^{\infty} \frac{8 \pi V f^{2} d f}{c^{3}\left(e^{\frac{b f}{k T}}-1\right)} \\
& =\frac{8 \pi V}{c^{3}} \int_{0}^{\infty} \frac{f^{2} d f}{\left(e^{\frac{H f}{h_{T}}}-1\right)} \\
& =\frac{8 \pi}{c^{3}}\left(\frac{k T}{h}\right)^{2 \infty} \frac{x^{2} d x}{\left(e^{x}-1\right)}\left(\frac{k T}{h}\right) \quad \text { Substituiting } \frac{h f}{k T}=x \\
& =8 \pi\left(\frac{k T}{h c}\right) \int_{0}^{3 \infty} \frac{x^{2} d x}{\left(e^{x}-1\right)} \\
& =8 \pi\left(\frac{k T}{h c}\right) \int_{0}^{3 \infty} \frac{x^{2} e^{-x} d x}{e^{-x}\left(e^{x}-1\right)} \\
& =8 \pi\left(\frac{k T}{h c}\right)^{3 \infty} \int_{0}^{2} e^{-x}\left(1-e^{-x}\right)^{-1} d x \\
& =8 \pi\left(\frac{k T}{h c}\right)^{3 \infty} \int_{0}^{2} x^{2} e^{-x}\left(1+e^{-x}+e^{-2 x}+e^{-3 x}+e^{-4 x}+\ldots\right) d x \\
& =8 \pi\left(\frac{k T}{h c}\right)^{3 \infty} \int_{0}^{2}\left(x^{2} e^{-x}+x^{2} e^{-2 x}+x^{2} e^{-3 x}+x^{2} e^{-4 x}+x^{2} e^{-5 x}+\ldots\right) d x \\
& =8 \pi\left(\frac{k T}{h c}\right)^{3} \times 2 \times \sum \frac{1}{n^{3}} \quad \text { using } \int_{0}^{\infty} x^{m} e^{-a x} d x=\frac{m!}{n^{m+1}}
\end{aligned}
$$
6. Limiting Cases
6.1 Low Frequency Limit (hν ≪ kBT)
Using approximation \( e^x \approx 1 + x \):
\[
u(\nu) \approx \frac{8\pi \nu^2}{c^3} k_B T
\]
This reduces to the Rayleigh–Jeans law.
6.2 High Frequency Limit (hν ≫ kBT)
Then:
\[
u(\nu) \approx \frac{8\pi h \nu^3}{c^3}
e^{-h\nu/k_B T}
\]
This reduces to Wien’s law.
7. Wien’s Displacement Law
Maximum intensity occurs at:
\[
\lambda_{max} T = \text{constant}
\]
Thus, peak wavelength decreases as temperature increases.
8. Stefan–Boltzmann Law
Total energy density:
\[
u = a T^4
\]
Total emitted power:
\[
E = \sigma T^4
\]
where \( \sigma \) is the Stefan–Boltzmann constant.
9. Importance of Planck's Law
- Resolved ultraviolet catastrophe.
- Introduced concept of energy quantization.
- Foundation of quantum mechanics.
10. Important Result
Planck’s law successfully explains the entire blackbody spectrum
at all frequencies and temperatures.
It marks the birth of quantum theory.