Unit 5 · Lesson 1a

1aBlackbody Radiation and Planck's Hypothesis

Discover how the failure of classical physics to explain glowing objects led Planck to quantize energy and launch the quantum revolution.

Planck's quantization hypothesis was the spark that ignited quantum physics — without it, we could not explain lasers, semiconductors, LEDs, or any modern electronic device.

Lesson Overview

A blackbody is an idealized object that absorbs all incident radiation and re-emits energy based solely on its temperature. Classical physics predicted infinite energy at high frequencies — the ultraviolet catastrophe. In 1900, Max Planck resolved this crisis by proposing that energy is emitted in discrete packets called quanta, each with energy E = nhf. This revolutionary idea launched modern quantum physics.

Key Concepts

Blackbody

An ideal object that absorbs all radiation and emits based on temperature alone

Ultraviolet Catastrophe

Classical prediction of infinite energy at high frequencies — experimentally wrong

Planck's Quantization

Energy emitted in quanta: E = nhf (n = integer, h = 6.626×10⁻³⁴ J·s, f = frequency)

Wien's Displacement Law

λ_max · T = 2.898×10⁻³ m·K — peak wavelength shifts with temperature

Stefan-Boltzmann Law

Total power radiated: P = σAT⁴ (σ = 5.67×10⁻⁸ W/m²·K⁴)

Planck's Constant

h = 6.626×10⁻³⁴ J·s — fundamental constant of quantum physics

Example 1

A blackbody has a surface temperature of 5800 K (similar to the Sun). Find the peak wavelength of its emission using Wien's displacement law.

Answer:λ_max = (2.898×10⁻³ m·K) / T = (2.898×10⁻³) / 5800 ≈ 5.0×10⁻⁷ m = 500 nm. This is green-yellow visible light, consistent with the Sun's spectrum.
Example 2

Calculate the energy of a single quantum of light with frequency 6.0×10¹⁴ Hz using Planck's formula.

Answer:E = hf = (6.626×10⁻³⁴ J·s)(6.0×10¹⁴ Hz) = 3.98×10⁻¹⁹ J ≈ 4.0×10⁻¹⁹ J.
Example 3

A star emits peak radiation at λ_max = 290 nm (ultraviolet). What is its surface temperature?

Answer:T = (2.898×10⁻³ m·K) / λ_max = (2.898×10⁻³) / (290×10⁻⁹) ≈ 10,000 K. This is a very hot blue-white star.
Example 4

Explain why classical physics predicted the ultraviolet catastrophe and how Planck's hypothesis resolved it.

Answer:Classical theory allowed oscillators to have any energy, predicting power ∝ f², which diverges at high f. Planck required energy in units of hf — at high frequencies, hf ≫ kT so oscillators are rarely excited, cutting off the high-frequency divergence and matching experiment.
Example 5

A blackbody sphere of radius 0.10 m is at 1000 K. Calculate the total power it radiates. (σ = 5.67×10⁻⁸ W/m²·K⁴)

Answer:A = 4πr² = 4π(0.10)² = 0.1257 m². P = σAT⁴ = (5.67×10⁻⁸)(0.1257)(1000)⁴ = (5.67×10⁻⁸)(0.1257)(10¹²) ≈ 7.1 W.
Guided Problem 1

A glowing coal has a peak emission wavelength of 1.45 μm. What is its surface temperature?

Hint: Use Wien's law: T = (2.898×10⁻³ m·K) / λ_max. Convert μm to m first.

Guided Problem 2

What is the energy of a quantum of infrared radiation with frequency 3.0×10¹³ Hz?

Hint: Apply E = hf with h = 6.626×10⁻³⁴ J·s.

Guided Problem 3

If a blackbody's temperature doubles, by what factor does its total radiated power increase?

Hint: Use the Stefan-Boltzmann law P ∝ T⁴. If T → 2T, what happens to T⁴?

Guided Problem 4

Why does a red-hot object appear red rather than white or blue?

Hint: Think about Wien's law — at lower temperatures, the peak wavelength is longer (red end of spectrum).

Guided Problem 5

A photon has energy 2.48 eV. Convert to joules and find its frequency. (1 eV = 1.6×10⁻¹⁹ J)

Hint: Convert eV to J, then use f = E/h.

Key Vocabulary

Blackbody

An idealized object that absorbs all incident electromagnetic radiation and emits radiation purely based on its temperature.

Example: The Sun approximates a blackbody at ~5800 K, emitting peak radiation in the visible range.

Ultraviolet Catastrophe

The failure of classical physics to predict blackbody radiation — it incorrectly predicted infinite energy at high frequencies.

Example: The Rayleigh-Jeans law matched experiment at low frequencies but diverged catastrophically in the ultraviolet.

Quantum

The smallest discrete unit of energy for a given frequency: E = hf, where h is Planck's constant.

Example: A photon of green light (f ≈ 6×10¹⁴ Hz) carries one quantum of energy ≈ 4×10⁻¹⁹ J.

Wien's Displacement Law

The peak wavelength of blackbody radiation is inversely proportional to temperature: λ_max · T = 2.898×10⁻³ m·K.

Example: Hotter stars emit peak radiation at shorter (bluer) wavelengths than cooler stars.

Interactive Practice — 5 Questions

1

What does Wien's displacement law relate?

2

Planck's hypothesis states that energy is emitted:

3

A blackbody at 3000 K emits peak radiation at approximately:

4

The Stefan-Boltzmann law states that total radiated power is proportional to:

5

The ultraviolet catastrophe refers to:

Independent Practice

1

A blackbody has a peak emission wavelength of 725 nm. Calculate its surface temperature using Wien's displacement law.

2

Calculate the energy (in joules and eV) of a quantum of violet light with frequency 7.5×10¹⁴ Hz.

3

Explain in your own words why classical physics failed to describe blackbody radiation and how Planck's hypothesis fixed the problem.

4

A blackbody at 500 K has its temperature raised to 1000 K. By what factor does its total radiated power increase?

5

★ The cosmic microwave background (CMB) radiation has a peak wavelength of about 1.06 mm. Calculate the temperature of the universe corresponding to this radiation and explain its cosmological significance.

Challenge
⚠️

Common Mistakes

Confusing Wien's law (peak wavelength) with Stefan-Boltzmann law (total power).

Wien's law: λ_max · T = constant (tells you the color). Stefan-Boltzmann: P = σAT⁴ (tells you the total brightness).

Thinking Planck's constant h has units of joules.

h = 6.626×10⁻³⁴ J·s (joule-seconds) — it has units of energy × time, also called action.

💡

Math Tips

📌

For Wien's law, always convert wavelength to meters before calculating. For Stefan-Boltzmann, remember T must be in Kelvin and the exponent is 4 — a small temperature change causes a large power change.