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Thermal Physics

Rayleigh-Jeans Law of Radiation

The Rayleigh-Jeans law expresses the spectral radiance of electromagnetic radiation emitted by a black body in thermal equilibrium as a function of wavelength at a given temperature. It is derived using classical statistical mechanics, specifically the law of equipartition of energy applied to standing waves in a cavity.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Understanding the formula for energy density in terms of wavelength and frequency, qualitative behavior at long wavelengths, and classical foundations based on equipartition of energy.

PGT extension

Detailed derivation of the density of modes in a cavity, application of the equipartition theorem, derivation of the Rayleigh-Jeans distribution, and analysis of the ultraviolet catastrophe.

Prerequisites

Concept of Temperature and thermal equilibrium
Law of equipartition of energy
Wave mechanics and stationary waves in a 3D cavity

Classical Origin and Assumptions

Based entirely on classical thermodynamics and classical electrodynamics without quantum assumptions.
Treats radiation inside a black body cavity as a collection of standing electromagnetic waves (modes of vibration).
Assumes that each standing wave mode acts as a harmonic oscillator in thermal equilibrium with the cavity walls.

Application of Equipartition Theorem

According to the classical law of equipartition of energy, each independent degree of freedom possesses an average thermal energy of k_BT.
Since an electromagnetic standing wave in a cavity has two independent transverse polarization states, each mode possesses an average energy of 2(1/2)k_B$T = k_{B}T$.
The average energy of a mode is independent of its frequency or wavelength in classical physics.

Density of States (Modes)

The number of standing wave modes per unit volume in a frequency interval from \nu to \nu + d\nu is given by g(\nu)d$\nu = \frac{8\pi\nu^{2}}{c^{3}} d\nu$.
In terms of wavelength, the density of states per unit volume is g(\lambda)d$\lambda = \frac{8\pi}{\lambda^{4}} d\lambda$.
The density of modes increases rapidly with frequency (proportional to \nu^{2}).

The Ultraviolet Catastrophe

Multiplying the density of states by the average energy per mode (k_BT) yields the Rayleigh-Jeans spectral energy density: U(\nu)d$\nu = \frac{8\pi\nu^{2}}{c^{3}} k_{B}T d\nu$.
As frequency \nu approaches infinity (or wavelength \lambda approaches zero), the predicted energy density approaches infinity.
This divergence at short wavelengths is known as the ultraviolet catastrophe, demonstrating the failure of classical physics to explain black-body radiation.

Derivations

Derivation of Rayleigh-Jeans Energy Density (Frequency Form)

Conditions: Isotropic black body cavity of volume V at absolute temperature T, using classical wave equation and equipartition theorem.

Consider standing electromagnetic waves inside a rectangular cavity of volume $V = L_{x}L_{y}L_{z}$. The allowed wave numbers satisfy n_{x}^{2} + n_{y}^{2} + n_{z}^{2} = \frac{4\nu^{2}V^{2/3}}{c^{2}}.
Calculate the number of modes in the frequency range \nu to \nu + d\nu by taking the volume of a spherical shell octant in mode space, multiplying by 2 for the two polarization states, and dividing by V to get the density of states per unit volume: g(\nu)d$\nu = \frac{8\pi\nu^{2}}{c^{3}} d\nu$.
Apply the classical law of equipartition of energy, assigning an average energy \langle E $\rangle = k_{B}T$ to each mode.
Multiply the density of states by the average energy per mode to obtain the spectral energy density: U(\nu)d$\nu = g(\nu)\langle E \rangle d\nu = \frac{8\pi\nu^{2}k_{B}T}{c^{3}} d\nu$.

Result: U(\nu)d$\nu = \frac{8\pi\nu^{2}k_{B}T}{c^{3}} d\nu$

Transformation to Wavelength Form

Conditions: Relating frequency distribution to wavelength distribution using \lambda = \frac{c}{\nu}.

Start with the frequency distribution function U(\nu)d$\nu = \frac{8\pi\nu^{2}k_{B}T}{c^{3}} d\nu$.
Substitute \nu = \frac{c}{\lambda}, which gives d$\nu = -\frac{c}{\lambda^{2}} d\lambda (ignoring the negative sign$ as it represents interval direction).
Substitute these expressions into the energy density equation: U(\lambda)d\lambda = U(\nu)\left|\frac{d\nu}{d\lambda}\right| d\lambda.
Simplify the algebraic expression to obtain the wavelength-dependent spectral energy density.

Result: U(\lambda)d$\lambda = \frac{8\pi k_{B}T}{\lambda^{4}} d\lambda$

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Spectral Energy Density (Frequency)	$$U(\nu)d\nu = \frac{8\pi\nu^{2}k_{B}T}{c^{3}} d\nu$$	U(\nu) = energy density per unit frequency, \nu = frequency, k_B = Boltzmann constant, T = absolute temperature, c = speed of light J m^{-3} Hz^{-1}	Valid only at very low frequencies (long wavelengths)
Spectral Energy Density (Wavelength)	$$U(\lambda)d\lambda = \frac{8\pi k_{B}T}{\lambda^{4}} d\lambda$$	U(\lambda) = energy density per unit wavelength, \lambda = wavelength, k_B = Boltzmann constant, T = absolute temperature J m^{-4}	Valid only at very long wavelengths
Density of States (Frequency)	$$g(\nu)d\nu = \frac{8\pi\nu^{2}}{c^{3}} d\nu$$	g(\nu) = number of standing wave modes per unit volume per unit frequency m^{-3} Hz^{-1}	Isotropic 3D cavity
Average Energy per Mode (Classical)	$$\langle E \rangle = k_{B}T$$	\langle E \rangle = mean energy per mode, k_B = Boltzmann constant, T = temperature J	Equipartition theorem for 2 degrees of freedom per mode
Spectral Radiance (Wavelength)	$$I(\lambda)d\lambda = \frac{c}{4\pi} U(\lambda)d\lambda = \frac{2ck_{B}T}{\lambda^{4}} d\lambda$$	I(\lambda) = emitted power per unit area per unit solid angle per unit wavelength W m^{-3} sr^{-1}	Hemispherical emission from black body surface
Relationship between U(\nu) and U(\lambda)	$$U(\nu)d\nu = U(\lambda)d\lambda \text{ where } \lambda\nu = c$$	Transformation of variables between frequency and wavelength domains Dimensionally consistent	General continuum relation
Boltzmann Constant	$$k_{B} = \frac{R}{N_{A}}$$	R = universal gas constant, $N_{A} = Avogadro number$ J K^{-1}	Universal constant
Speed of Electromagnetic Waves	$$c = \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}$$	\m$u_{0} = permeability of free space$, \varepsilo$n_{0} = permittivity of free space$ m s^{-1}	Vacuum

Graphs and experiments

Rayleigh-Jeans Law vs Experimental Black-Body Spectrum

Draw: A plot of spectral radiance (or energy density) on the y-axis against wavelength on the x-axis. Draw the experimental bell-shaped curve peaking at a finite wavelength, and overlay the Rayleigh-Jeans theoretical curve which matches at long wavelengths but shoots to infinity as \lambda \to 0.

Interpret: The Rayleigh-Jeans curve successfully accounts for black-body radiation only in the far-infrared and long-wavelength limit. It diverges catastrophically at ultraviolet and shorter wavelengths, highlighting the breakdown of classical equipartition.

Exam use: Frequently asked in conceptual questions identifying the failure of classical physics and the origin of the ultraviolet catastrophe.

Density of Modes vs Frequency

Draw: A parabolic curve showing g(\nu) proportional to \nu^{2} increasing monotonically without bound.

Interpret: Illustrates that an infinite number of high-frequency modes can exist within a finite cavity volume in classical mechanics.

Exam use: Tested in derivations of cavity radiation modes and partition functions.

Log-Log Comparison of Radiation Laws

Draw: A log U vs log \lambda plot showing Rayleigh-Jeans as a straight line with slope -4 at high wavelengths, compared with Wien's law and Planck's complete law.

Interpret: Demonstrates the asymptotic limits of Planck's law, reducing to Rayleigh-Jeans at high \lambda and Wien's law at low \lambda.

Exam use: Advanced PGT level question analysis of limiting forms of Planck's radiation law.

Solved numericals

TGT core

According to the Rayleigh-Jeans law, how does the spectral energy density U(\lambda) of black-body radiation at a constant temperature depend on wavelength \lambda?

Recall the Rayleigh-Jeans formula for spectral energy density in terms of wavelength: U(\lambda) = \frac{8\pi k_{B}T}{\lambda^{4}}.
Identify the variables held constant: temperature T, Boltzmann constant k_{B}, and constants 8, \pi.
Establish the proportionality: U(\lambda) \propto \lambda^{-4}.

Answer: U(\lambda) is inversely proportional to the fourth power of wavelength (U(\lambda) \propto \lambda^{-4}).

PGT extension

Using the Rayleigh-Jeans law, calculate the spectral energy density U(\nu) inside a cavity at temperature 1000 K for a frequency of 1.0 \times 10^{8} Hz (Take $k_{B} = 1.38 \times 10^{-23} J K^{-1}$, $c = 3 \times 10^{8} m s^{-1})$.

Write the Rayleigh-Jeans formula: U(\nu) = \frac{8\pi\nu^{2}k_{B}T}{c^{3}}.
Substitute the given values: $\nu = 10^{8} Hz$, T = 1000 K, $k_{B} = 1.38 \times 10^{-23} J K^{-1}$, $c = 3 \times 10^{8} m s^{-1}$, \pi \approx 3.1416.
Compute the numerator: 8 \times 3.1416 \times (10^{8})^{2} \times (1.38 \times 10^{-23}) \times $1000 = 8 \times 3.1416 \times 10^{16} \times 1.38 \times 10^{-20} \approx 3.472 \times 10^{-2}$.
Compute the denominator: (3 \times 10^{8})^{3} = 2.7 \times 10^{25}.
Divide numerator by denominator: U(\nu) = \frac{3.472 \times 10^{-2}}{2.7 \times 10^{25}} \approx 1.286 \times 10^{-27} J m^{-3} Hz^{-1}.

Answer: 1.29 \times 10^{-27} J m^{-3} Hz^{-1}

PGT extension

Show how the Rayleigh-Jeans law can be obtained as a limiting case of Planck's radiation law U(\nu)d$\nu = \frac{8\pi h \nu^{3}}{c^{3}} \frac{1}{e^{h\nu / k_{B}T} - 1} d\nu when h\nu \ll k_{B}T$.

Write Planck's law: U(\nu) = \frac{8\pi h \nu^{3}}{c^{3}} \frac{1}{e^{h\nu / k_{B}T} - 1}.
Apply the condition h\nu \ll k_{B}T (or \nu \to 0), which means the exponent \frac{h\nu}{k_{B}T} is much less than 1.
Expand the exponential term using Taylor series: e^{h\nu / k_{B}T} \approx 1 + \frac{h\nu}{k_{B}T}.
Substitute the expansion into the denominator: e^{h\nu / k_{B}T} - 1 \approx \left(1 + \frac{h\nu}{k_{B}T}\right) - $1 = \frac{h\nu}{k_{B}T}$.
Substitute back into Planck's formula: U(\nu) \approx \frac{8\pi h \nu^{3}}{c^{3}} \left(\frac{k_{B}T}{h\nu}\right).
Cancel the Planck constant h and frequency term \nu to arrive at the Rayleigh-Jeans expression: \frac{8\pi\nu^{2}k_{B}T}{c^{3}}.

Answer:

Assumptions and limitations

Assumption: Every standing wave mode in the cavity is an independent harmonic oscillator with mean thermal energy equal to k_{B}T according to classical equipartition theorem. Limitation: Fails at high frequencies because high-energy quantum states (h\nu \ge k_{B}T) cannot be thermally excited.
Assumption: Radiation interacts continuously with matter in arbitrary amounts of energy. Limitation: Energy is actually quantized in discrete packets (photons) of magnitude E = h\nu, as introduced by Planck.

Common misconceptions

Common error: The Rayleigh-Jeans law is valid across all frequencies for black-body radiation.

Correct: It is valid only in the low-frequency (long-wavelength) limit.

At high frequencies, it predicts infinite energy density (ultraviolet catastrophe), which contradicts experimental observation.

Common error: The ultraviolet catastrophe is a mathematical error in calculus.

Correct: It is a fundamental failure of classical statistical mechanics.

The divergence arises from assigning k_{B}T energy to an infinite number of high-frequency modes without quantum constraints.

Common error: Rayleigh and Jeans used quantum mechanics in their derivation.

Correct: The derivation is entirely classical, utilizing Maxwell's electromagnetic wave theory and classical thermodynamics.

Planck introduced quantization later to resolve the failure of the Rayleigh-Jeans law.

Quick revision

Rayleigh-Jeans law gives U(\lambda)d$\lambda = \frac{8\pi k_{B}T}{\lambda^{4}} d\lambda$.
Derived using classical equipartition of energy (\langle E $\rangle = k_{B}T)$ and density of cavity modes.
Works well at long wavelengths (infrared and radio waves).
Fails completely at short wavelengths, leading to the ultraviolet catastrophe.
Planck's law reduces to Rayleigh-Jeans law in the limit h\nu \ll k_{B}T.

Must remember

The Rayleigh-Jeans law is derived entirely using classical thermodynamics and classical electrodynamics without any quantum assumptions.
According to the classical equipartition theorem, each electromagnetic standing wave mode in a cavity possesses an average thermal energy of $k_B T$.
The density of states per unit volume in terms of frequency is given by $g(\nu)d\nu = \frac{8\pi\nu^{2}}{c^{3}} d\nu$.
The Rayleigh-Jeans spectral energy density in terms of frequency is $U(\nu)d\nu = \frac{8\pi\nu^{2}k_{B}T}{c^{3}} d\nu$.
The Rayleigh-Jeans spectral energy density in terms of wavelength is $U(\lambda)d\lambda = \frac{8\pi k_{B}T}{\lambda^{4}} d\lambda$.
The Rayleigh-Jeans law is valid only at very low frequencies or long wavelengths.
At short wavelengths or high frequencies, the Rayleigh-Jeans law predicts infinite energy density, a failure known as the 'ultraviolet catastrophe'.

Glossary

$U(\nu)$Spectral energy density per unit frequency, representing energy per unit volume per unit frequency.

$U(\lambda)$Spectral energy density per unit wavelength, representing energy per unit volume per unit wavelength.

$g(\nu)$Density of states (number of standing wave modes per unit volume per unit frequency) in an isotropic 3D cavity.

Equipartition TheoremA classical statistical mechanics principle stating that each independent degree of freedom has an average thermal energy of $\frac{1}{2}k_B T$.

Ultraviolet CatastropheThe classical prediction that total radiated energy from a black body approaches infinity as wavelength approaches zero.

$\langle E \rangle$Mean energy per standing wave mode, equal to $k_B T$ in classical physics due to two transverse polarization states.

Frequency vs. Wavelength Rayleigh-Jeans Formulas

Feature	Frequency Domain ($U(\nu)$)	Wavelength Domain ($U(\lambda)$)
Spectral Energy Density	$U(\nu)d\nu = \frac{8\pi\nu^{2}k_{B}T}{c^{3}} d\nu$	$U(\lambda)d\lambda = \frac{8\pi k_{B}T}{\lambda^{4}} d\lambda$
Dependence	Proportional to $\nu^2$	Proportional to $\lambda^{-4}$
Validity Limit	Low frequencies ($\nu \to 0$)	Long wavelengths ($\lambda \to \infty$)
Failure Region	High frequencies ($\nu \to \infty$)	Short wavelengths ($\lambda \to 0$)

Practice Quiz

Choose an option to reveal the explanation.

1. Which theoretical framework forms the basis of the Rayleigh-Jeans law of radiation?

Explanation: The Rayleigh-Jeans law is derived entirely from classical physics, utilizing Maxwell's electromagnetic wave theory and the classical equipartition theorem without any quantum assumptions.

2. What is the average energy assigned to each standing wave mode in a black body cavity according to the classical Rayleigh-Jeans approach?

Explanation: Each standing wave mode has two independent transverse polarization states. Since each degree of freedom has $\frac{1}{2}k_B T$ of energy, a mode has $2 \times \frac{1}{2}k_B T = k_B T$.

3. The density of states per unit volume for electromagnetic standing waves in a frequency interval from $\nu$ to $\nu + d\nu$ in a 3D cavity is given by:

Explanation: In an isotropic 3D cavity considering two polarization states, the density of standing wave modes per unit volume per unit frequency is $g(\nu)d\nu = \frac{8\pi\nu^{2}}{c^{3}} d\nu$.

4. What is the Rayleigh-Jeans spectral energy density $U(\nu)d\nu$ expressed in terms of frequency $\nu$?

Explanation: Multiplying the density of states $g(\nu)d\nu = \frac{8\pi\nu^2}{c^3}d\nu$ by the average energy per mode $\langle E \rangle = k_B T$ gives $U(\nu)d\nu = \frac{8\pi\nu^2 k_B T}{c^3} d\nu$.

5. How does the Rayleigh-Jeans spectral energy density $U(\lambda)$ vary with wavelength $\lambda$?

Explanation: The Rayleigh-Jeans wavelength distribution is given by $U(\lambda)d\lambda = \frac{8\pi k_B T}{\lambda^4} d\lambda$, showing that energy density is inversely proportional to the fourth power of wavelength.

6. Under what condition is the Rayleigh-Jeans law found to be in good agreement with experimental observations?

Explanation: The Rayleigh-Jeans law matches experimental black-body radiation curves successfully only in the limit of very low frequencies (or long wavelengths).

7. What is the 'ultraviolet catastrophe' associated with the Rayleigh-Jeans law?

Explanation: As $\nu \to \infty$, $U(\nu) \to \infty$, meaning classical theory predicts infinite total radiated energy, which is known as the ultraviolet catastrophe.

8. Which physical constant does NOT appear in the Rayleigh-Jeans spectral energy density formula $U(\nu)d\nu = \frac{8\pi\nu^{2}k_{B}T}{c^{3}} d\nu$?

Explanation: Planck's constant $h$ does not appear in the Rayleigh-Jeans formula because the derivation is completely classical and does not involve energy quantization.

9. If the absolute temperature $T$ of a black body cavity is doubled, how does the Rayleigh-Jeans energy density at a given frequency change?

Explanation: According to $U(\nu) = \frac{8\pi\nu^2 k_B T}{c^3}$, the energy density is directly proportional to absolute temperature $T$. Doubling $T$ doubles the energy density.

10. What is the dimension/unit of spectral energy density $U(\nu)$ per unit frequency?

Explanation: Energy density is energy per unit volume ($\text{J m}^{-3}$), and spectral energy density per unit frequency divides this by frequency, giving $\text{J m}^{-3} \text{Hz}^{-1}$.

11. Why do classical physics principles fail to explain black-body radiation across all frequencies?

Explanation: Classical statistics assigns average energy $k_B T$ to every vibrational mode regardless of frequency, leading to an infinite number of high-frequency modes absorbing all energy.

12. The relationship between spectral radiance $I(\lambda)$ and spectral energy density $U(\lambda)$ for isotropic black-body radiation is:

Explanation: Hemispherical spectral radiance emitted from a black body surface is related to energy density by $I(\lambda) = \frac{c}{4\pi} U(\lambda)$.

13. In terms of wavelength, what is the density of states $g(\lambda)d\lambda$ per unit volume in a 3D cavity?

Explanation: Transforming the density of states from frequency to wavelength using $\nu = \frac{c}{\lambda}$ yields $g(\lambda)d\lambda = \frac{8\pi}{\lambda^{4}} d\lambda$.

14. Which scientist successfully resolved the ultraviolet catastrophe by introducing the concept of energy quanta?

Explanation: Max Planck introduced the quantum hypothesis ($E = h\nu$) in 1900, deriving Planck's radiation law which correctly reduces to Rayleigh-Jeans at low frequencies and avoids divergence.

15. What is the average energy per mode in the high-frequency limit according to Planck's law, which contrasts with the classical Rayleigh-Jeans value of $k_B T$?

Explanation: Planck's quantum derivation gives the average energy per mode as $\langle E \rangle = \frac{h\nu}{e^{h\nu / k_B T} - 1}$, which approaches zero at high frequencies, resolving the ultraviolet catastrophe.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. The Rayleigh-Jeans law for spectral energy density of black-body radiation is valid only for:

Explanation: The classical Rayleigh-Jeans formula works well only in the long-wavelength (low-frequency) regime and fails catastrophically at short wavelengths.

2. According to the classical equipartition theorem, the average energy associated with each standing wave mode of an electromagnetic cavity is:

Explanation: Each standing wave has two transverse polarization states (degrees of freedom), each contributing $\frac{1}{2}k_B T$, resulting in a total average energy of $k_B T$ per mode.

3. The dimension of the density of states $g(\nu)$ per unit volume per unit frequency is:

Explanation: Density of states $g(\nu)d\nu$ gives the number of modes per unit volume ($m^{-3}$) per unit frequency interval ($Hz^{-1}$), giving units of $\text{m}^{-3} \text{Hz}^{-1}$.

4. In the Rayleigh-Jeans law $U(\lambda)d\lambda = \frac{8\pi k_B T}{\lambda^4} d\lambda$, the energy density $U(\lambda)$ is proportional to:

Explanation: The formula explicitly shows $U(\lambda)$ is inversely proportional to $\lambda^4$, meaning $U(\lambda) \propto \lambda^{-4}$.

5. The ultraviolet catastrophe arises in classical black-body radiation theory because:

Explanation: Because the density of modes increases as $\nu^2$ and each mode is assigned energy $k_B T$, summing over infinite high-frequency modes yields infinite total energy density.

6. Which of the following expressions correctly represents the relationship between spectral energy density in frequency $U(\nu)$ and wavelength $U(\lambda)$?

Explanation: The energy contained in interval $d\nu$ must equal the energy in corresponding interval $d\lambda$, hence $U(\nu)d\nu = U(\lambda)d\lambda$ where $\lambda\nu = c$.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. What physical assumption distinguishes the derivation of Rayleigh-Jeans law from Planck's radiation law?

Explanation: The core difference is classical continuous equipartition energy ($\langle E \rangle = k_B T$) vs. quantum discrete energy levels (Planck's hypothesis).

2. Calculate the ratio of the density of states at frequency $2\nu$ to that at frequency $\nu$ in a 3D isotropic cavity.

Explanation: Since the density of states $g(\nu)$ is proportional to $\nu^2$, doubling the frequency increases $g(\nu)$ by a factor of $2^2 = 4$.

3. If the wavelength of observation for a black body is increased by a factor of 3, how does the Rayleigh-Jeans energy density $U(\lambda)$ change?

Explanation: Since $U(\lambda) \propto \lambda^{-4}$, increasing $\lambda$ by a factor of 3 results in $U(\lambda)$ changing by $(1/3)^4 = 1/81$ (decreases by 81).

4. Which of the following statements about standing wave modes in a cubic cavity of volume $V$ is correct?

Explanation: Density of states per unit volume is intensive, meaning the total number of modes in a cavity is directly proportional to its volume $V$.

5. In the context of the Rayleigh-Jeans law, what does the equipartition theorem fail to account for?

Explanation: Quantum mechanics shows high-frequency modes require more energy than $k_B T$ to be excited, effectively 'freezing out' high-frequency modes, which classical equipartition ignores.

6. The spectral radiance $I(\lambda)$ of a black body according to the Rayleigh-Jeans law is given by:

Explanation: Using $I(\lambda) = \frac{c}{4\pi} U(\lambda)$ and $U(\lambda) = \frac{8\pi k_B T}{\lambda^4}$, we get $I(\lambda) = \frac{c}{4\pi} \left(\frac{8\pi k_B T}{\lambda^4}\right) = \frac{2c k_B T}{\lambda^4}$.

7. Why is the Rayleigh-Jeans law considered an asymptotic limit of Planck's law at high temperatures or long wavelengths?

Explanation: Expanding the exponential term $e^{x} \approx 1 + x$ when $x = \frac{h\nu}{k_B T} \ll 1$ in Planck's law yields the Rayleigh-Jeans energy density formula.

8. What is the physical significance of the polarization factor 2 in the density of states calculation for electromagnetic waves?

Explanation: An electromagnetic wave in free space has two orthogonal transverse polarization states for every wave vector $\mathbf{k}$, doubling the number of modes.

9. If Wien's displacement law states $\lambda_{max} T = b$, how does the maximum energy density predicted by Rayleigh-Jeans at $\lambda_{max}$ scale with temperature $T$?

Explanation: Substituting $\lambda_{max} \propto T^{-1}$ into $U(\lambda_{max}) \propto T / \lambda_{max}^4$ gives $U(\lambda_{max}) \propto T \cdot (T^{-1})^{-4} = T^5$ (Wait: $U(\lambda) \propto T / \lambda^4$, so $U(\lambda_{max}) \propto T / (b/T)^4 \propto T^5$. Let's check options: $T^5$ is correct). Let's adjust correct index to 2.

10. Which of the following curves correctly depicts the failure of the Rayleigh-Jeans law compared to experimental black-body spectra?

Explanation: Experimental curves peak and drop to zero at short wavelengths, whereas Rayleigh-Jeans continually rises toward infinity (ultraviolet catastrophe).