38-4 The Birth of Quantum Physics

38-4 The Birth of Quantum Physics#

Prompts

  • What is an ideal blackbody radiator? Define spectral radiancy \(S(\lambda)\) and its physical meaning.

  • Classical puzzle: The classical radiation law predicts \(S(\lambda) \propto 1/\lambda^4\) at short wavelengths. What goes wrong? Why was this called the “ultraviolet catastrophe”?

  • Write Planck’s radiation law for \(S(\lambda)\). What role does the factor \(hc/(\lambda kT)\) play? How did Planck’s law suggest that oscillator energies are quantized?

  • Wien’s law states \(\lambda_{\max} T = 2898\ \mu\text{m}\cdot\text{K}\). For a cavity at 2000 K, where is the peak wavelength? Is it in the visible or infrared?

  • For a narrow wavelength range \(\Delta\lambda\) at wavelength \(\lambda\), how do you approximate the power emitted in that range from a blackbody of area \(A\)?

Lecture Notes#

Overview#

  • A blackbody radiator emits thermal radiation that depends only on its temperature. The spectral radiancy \(S(\lambda)\) describes intensity per unit wavelength.

  • Classical physics predicted \(S(\lambda) \propto 1/\lambda^4\) at short wavelengths—the radiation “blows up” to infinity (the ultraviolet catastrophe), contradicting experiment.

  • Planck’s radiation law (1900) fits all wavelengths and temperatures. Its form implied that oscillator energies are quantized—the first hint of quantum physics.

  • Wien’s law \(\lambda_{\max} T = 2898\ \mu\text{m}\cdot\text{K}\) relates peak wavelength to temperature.

Blackbody radiation and spectral radiancy#

An ideal blackbody radiator emits thermal radiation that depends only on temperature—not on the material, surface, or geometry. The setup: a cavity with walls at uniform temperature \(T\). Atoms on the walls oscillate (thermal energy) and emit electromagnetic waves. A small hole samples the radiation without altering it.

Spectral radiancy \(S(\lambda)\) is the intensity per unit wavelength at wavelength \(\lambda\):

(358)#\[ S(\lambda) = \frac{\text{intensity}}{\text{unit wavelength}} = \frac{\text{power per unit area}}{\text{unit wavelength}} \]

For a narrow wavelength range \(d\lambda\), the power per unit area emitted in that range is \(S(\lambda)\,d\lambda\).

Classical failure: the ultraviolet catastrophe#

Classical physics (equipartition of energy among oscillators) predicted the Rayleigh–Jeans law:

(359)#\[ S(\lambda) = \frac{2\pi c k T}{\lambda^4} \]

At long wavelengths, this agrees with experiment. But at short wavelengths, \(S(\lambda) \to \infty\) as \(\lambda \to 0\)—the total radiated power would be infinite. This is the ultraviolet catastrophe: classical theory predicted that any hot object would radiate infinite energy at short wavelengths, which is absurd.

Physical intuition

Classical equipartition gives each oscillator mode an average energy \(kT\), independent of frequency. The number of modes per unit wavelength grows as \(1/\lambda^4\) at short \(\lambda\), so the predicted intensity grows without bound. Planck’s fix: restrict oscillator energies to discrete multiples of \(hf\), which suppresses high-frequency modes at low \(T\).

Planck’s radiation law#

In 1900, Planck devised a formula that fit all experimental data:

(360)#\[ S(\lambda) = \frac{2\pi c^2 h}{\lambda^5} \cdot \frac{1}{e^{hc/(\lambda k T)} - 1} \]

where \(h\) is the Planck constant, \(k\) is the Boltzmann constant, and \(T\) is the temperature in kelvins. The key factor is \(hc/(\lambda k T) = hf/(kT)\)—the ratio of photon energy \(hf\) to thermal energy \(kT\).

Planck’s derivation implied that the atomic oscillators in the cavity wall can have only discrete energies—integer multiples of \(hf\). This was the first suggestion that energies are quantized. Planck himself was reluctant to accept this; Einstein later explained it with the photon model (quantized radiation itself).

Radiated power: Integrating Eq. (360) over all wavelengths gives the total power per unit area. For an ideal blackbody of area \(A\) and emissivity \(\varepsilon = 1\):

(361)#\[ P = \sigma A T^4 \]

where \(\sigma \approx 5.67 \times 10^{-8}\ \text{W/m}^2\cdot\text{K}^4\) is the Stefan–Boltzmann constant (see Eq. 18-38).

For a narrow range \(\Delta\lambda\) at wavelength \(\lambda\), the power in that range is approximately \(S(\lambda)\,A\,\Delta\lambda\).

Wien’s law#

The wavelength \(\lambda_{\max}\) at which \(S(\lambda)\) is maximum (for a given \(T\)) satisfies Wien’s law:

(362)#\[ \lambda_{\max} T = 2898\ \mu\text{m}\cdot\text{K} \]

For example, at \(T = 2000\ \text{K}\), \(\lambda_{\max} \approx 1.5\ \mu\text{m}\)—in the infrared. As the temperature increases, the peak shifts to shorter wavelengths.

Example

Sun’s surface: If the Sun’s surface temperature is about 5800 K, then \(\lambda_{\max} \approx 2898/5800 \approx 0.5\ \mu\text{m}\), which is in the visible (green) range. That is why the Sun’s spectrum peaks in the visible.

Summary#

  • Blackbody radiator: emission depends only on \(T\); cavity with small hole samples thermal radiation.

  • Spectral radiancy \(S(\lambda)\): intensity per unit wavelength.

  • Classical failure: Rayleigh–Jeans predicts \(S \propto 1/\lambda^4\) at short \(\lambda\) → ultraviolet catastrophe.

  • Planck’s law Eq. (360): fits all data; first use of quantized oscillator energies.

  • Wien’s law \(\lambda_{\max} T = 2898\ \mu\text{m}\cdot\text{K}\): peak wavelength shifts with temperature.