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Thermal Physics
Black-Body Radiation
A black body is an idealized physical body that absorbs all incident electromagnetic radiation regardless of frequency or angle of incidence. The thermal radiation emitted by a black body depends solely on its absolute temperature and constitutes black-body radiation.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Stefan-Boltzmann law, Wien's displacement law, qualitative understanding of black body emission spectra, and simple numericals on radiation cooling.
PGT extension
Planck's radiation law, Rayleigh-Jeans law, ultraviolet catastrophe, derivation of Stefan-Boltzmann law from Planck's distribution, and Wien's distribution law.
Prerequisites
Concept of Temperature and Zeroth Law of Thermodynamics
Degrees of Freedom and Law of Equipartition of Energy
Basic Wave Optics and Electromagnetic Spectrum
Characteristics of Black-Body Radiation
A perfectly black body is theoretically realized by a hollow enclosure with a small pinhole, known as Ferry's black body.
Radiation emerging from the pinhole is cavity radiation and is completely independent of the material of the walls of the enclosure.
The emission spectrum is continuous, containing all wavelengths from zero to infinity, unlike atomic line spectra.
Spectral Emissive Power and Kirchhoff's Law
Spectral emissive power E(λ, T) is the amount of radiant energy emitted per unit surface area per unit time per unit wavelength interval at temperature T.
Monochromatic absorptive power a(λ) is the fraction of incident radiation of wavelength λ absorbed by the surface.
Kirchhoff's law states that the ratio of the spectral emissive power to the spectral absorptive power for any given wavelength at a fixed temperature is a constant for all bodies, equal to the emissive power of a black body at that same temperature: E(λ, T) / a(λ) = E_b(λ, T).
Planck's Quantum Hypothesis
Classical physics (Rayleigh-Jeans and Wien's laws) failed to explain the complete black-body spectrum, leading to the ultraviolet catastrophe at short wavelengths.
Max Planck postulated that electromagnetic energy is emitted or absorbed in discrete packets called quanta, having energy E = hν.
Planck's radiation law successfully accounts for the entire black-body spectrum across all wavelengths and temperatures without divergence.
Solar Constant and Surface Temperature of the Sun
The solar constant represents the total radiant energy received from the Sun per unit time on a unit surface area exposed perpendicular to the Sun's rays at the mean earth-sun distance.
Assuming the Sun acts as a black body, its effective surface temperature can be estimated using the Stefan-Boltzmann law and the inverse square law of radiation.
Typical calculated surface temperature of the Sun is approximately 5780 K to 6000 K.
Derivations
Derivation of Wien's Displacement Law from Planck's Law
Conditions: Finding the wavelength corresponding to maximum spectral emissive power.
Start with Planck's radiation law for spectral emissive power: E_b(λ, T) = \frac{2\pi h c^2}{\lambda^5 (e^{hc/\lambda k_B T} - 1)}.
Let $x = \frac{hc}{\lambda k_B T}$. Differentiate E_b(λ, T) with respect to λ and set \frac{dE_b}{d\lambda} = 0 to find the maximum.
This yields the transcendental equation: (5 - x)e^x - 5 = 0.
Solving this numerical equation gives x ≈ 4.965.
Substituting back x gives λ_m $T = \frac{hc}{4.965 k_B} = \text{constant} = b$.
Result: \lambda_m $T = b \approx 2.898 \times 10^{-3} \text{ m K}$
Derivation of Stefan-Boltzmann Law from Total Energy Density
Conditions: Integrating Planck's radiation law over all wavelengths from zero to infinity.
Total emissive power of a black body is given by $E = \int_{0}^{\infty} E_b(λ$, T) d\lambda.
Substitute Planck's expression for E_b(λ, T): $E = \int_{0}^{\infty} \frac{2\pi h c^2}{\lambda^5 (e^{hc/\lambda k_B T} - 1)} d\lambda$.
Substitute $x = \frac{hc}{\lambda k_B T}$, leading to $E = \frac{2\pi k_B^4 T^4}{h^3 c^2} \int_{0}^{\infty} \frac{x^3}{e^x - 1} dx$.
Evaluate the standard definite integral \int_{0}^{\infty} \frac{x^3}{e^x - 1} d$x = \frac{\pi^4}{15}$.
Collect constants into the Stefan-Boltzmann constant: $\sigma = \frac{2\pi^5 k_B^4}{15 c^2 h^3}$.
Result: $E = \sigma T^4$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Stefan-Boltzmann Law
$$E = \sigma T^4$$
E = Total emissive power, $\sigma = Stefan-Boltzmann constant$, T = Absolute temperature W m^-2
For an ideal black body
Rate of Heat Radiation
$$Q/t = e \sigma A (T^4 - T_0^4)$$
e = Emissivity, A = Surface area, T = Body temperature, $T_0 = Surroundings temperature$ W
Body placed in an enclosure at uniform temperature T_0
Wien's Displacement Law
$$\lambda_m T = b$$
\lambd$a_m = Wavelength of maximum emission$, b = Wien's constant m K
$E_{\lambda} = Emissive power$, $a_{\lambda} = Absorptive power$, $E_{b\lambda} = Black body emissive power$ Dimensionless ratio for a, W m^-3 for E
Thermal equilibrium
Rate of Cooling
$$-\frac{dT}{dt} = \frac{4 \sigma e A T_0^3}{mc} (T - T_0)$$
m = Mass, c = Specific heat capacity, T - $T_0 = Small temperature difference$ K s^-1
Approximate form of Newton's law of cooling derived from Stefan's law
Graphs and experiments
Black-Body Spectral Emissive Power vs Wavelength
Draw: A family of bell-shaped curves for different absolute temperatures T_3 > T_2 > T_1. Each curve peaks at a specific wavelength \lambda_m and approaches zero asymptotically at very large and very small wavelengths.
Interpret: As temperature increases, the peak shifts to shorter wavelengths (Wien's law) and the total area under the curve increases drastically (Stefan's law, proportional to T^4).
Exam use: Frequently asked in conceptual questions regarding the shift of peak wavelength with temperature and verification of Wien's law.
Comparison of Classical Laws vs Planck's Law
Draw: A graph plotting E_b(λ) versus λ, showing Planck's law curve matching experimental data, Rayleigh-Jeans curve diverging to infinity at short wavelengths (ultraviolet catastrophe), and Wien's law matching only at short wavelengths.
Interpret: Illustrates the failure of classical equipartition of energy at high frequencies and the success of Planck's quantum hypothesis.
Exam use: Crucial for PGT conceptual questions distinguishing classical mechanics failures from quantum theory.
Ferry's Black Body Experimental Setup
Draw: A double-walled hollow spherical metal sphere coated inside with lampblack, having a small conical projection opposite to a tiny pinhole aperture.
Interpret: Any radiation entering through the pinhole undergoes multiple internal reflections with absorption at each bounce, yielding an effective absorptivity close to unity (0.98-0.99).
Exam use: Tested in theory questions on practical realization of an ideal black body.
Solved numericals
TGT core
The spectral emissive power of a black body has its maximum at a wavelength of 500 nm when its temperature is 5800 K. Calculate the temperature of the black body if the peak wavelength shifts to 600 nm.
State Wien's displacement law: \lambda_{m1} $T_1 = \lambda_{m2} T_2$.
Substitute the given values: 500 \text{ nm} \times 5800 \text{ K} = 600 \text{ nm} \times T_2.
Solve for T_2: $T_2 = \frac{500 \times 5800}{600}$.
Calculate final value: $T_2 = \frac{29000}{6} \approx 4833.33 \text{ K}$.
Answer: 4833.33 K
PGT extension
A spherical black body of radius 0.1 m is maintained at 400 K inside an evacuated enclosure maintained at 300 K. Find the net rate of heat loss by radiation. (Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4})$
Calculate the surface area of the sphere: $A = 4 \pi r^2 = 4 \pi (0.1)^2 = 0.04 \pi \text{ m}^2$.
Write the net rate of heat radiation formula: \frac{dQ}{dt} = \sigma A (T^4 - T_0^4).
Substitute the given values: \frac{dQ}{dt} = 5.67 \times 10^{-8} \times 0.04 \pi \times (400^4 - 300^4).
If the absolute temperature of a black body is doubled, by what factor does the total rate of energy emission per unit area increase?
State Stefan-Boltzmann law: $E = \sigma T^4$.
Let initial temperature be $T_1 = T$ and initial emission be $E_1 = \sigma T^4$.
Let final temperature be $T_2 = 2T$, so final emission is $E_2 = \sigma (2T)^4 = 16 \sigma T^4$.
Find the ratio: \frac{E_2}{$E_1 = 16$.
Answer: 16
Assumptions and limitations
The Stefan-Boltzmann law assumes the radiating body has an emissivity of exactly 1; real bodies require an emissivity factor e.
Rayleigh-Jeans law assumes classical equipartition of energy holds for all modes of standing electromagnetic waves in a cavity, which fails at high frequencies (ultraviolet catastrophe).
Wien's distribution law is valid only for the high-frequency (short wavelength) limit of the black-body spectrum.
Common misconceptions
Common error: A black body must appear pitch black under all conditions.
Correct: A black body absorbs all incident radiation, but when heated, it emits thermal radiation and can glow brightly (e.g., incandescent lamp filament, stars).
Color and appearance depend entirely on emission at high temperature, not on reflection.
Common error: Wien's displacement law means the wavelength of maximum emission is the only wavelength emitted.
Correct: A black body emits a continuous spectrum containing all wavelengths; \lambda_m is merely the wavelength corresponding to the maximum spectral emissive power.
The term 'displacement' refers to the shifting of the peak wavelength with temperature change, not a single monochromatic emission.
Quick revision
A black body is a perfect absorber and a perfect emitter of radiation.
Stefan-Boltzmann Law: Total emissive power E is proportional to T^4.
Wien's Displacement Law: \lambda_m T = constant.
Planck's quantum theory resolved the ultraviolet catastrophe by introducing discrete energy quanta E = hν.
Must remember
A perfectly black body is theoretically realized by a hollow enclosure with a small pinhole, known as Ferry's black body.
Kirchhoff's law states that $E_{\lambda, T} / a_{\lambda, T} = E_{b\lambda, T}$ for all bodies at a fixed temperature.
Stefan-Boltzmann law gives total emissive power as $E = \sigma T^4$ for an ideal black body.
Wien's displacement law states $\lambda_m T = b$, meaning the peak wavelength shifts inversely with absolute temperature.
Planck's quantum hypothesis resolved the ultraviolet catastrophe by assuming energy is emitted or absorbed in discrete quanta $E = h\nu$.
Glossary
Spectral Emissive PowerAmount of radiant energy emitted per unit surface area per unit time per unit wavelength interval at temperature T, denoted by $E(\lambda, T)$.
Monochromatic Absorptive PowerFraction of incident radiation of a specific wavelength $\lambda$ absorbed by a surface, denoted by $a(\lambda)$.
Solar ConstantTotal radiant energy received from the Sun per unit time on a unit surface area exposed perpendicular to rays at the mean earth-sun distance.
Ultraviolet CatastropheFailure of classical Rayleigh-Jeans law to predict energy emission at short wavelengths, leading to an infinite energy divergence.
Rayleigh-Jeans Law vs Wien's Distribution Law
Feature
Rayleigh-Jeans Law
Wien's Distribution Law
Equation
$E_b(\lambda, T) = \frac{8\pi c k_B T}{\lambda^4}$
$E_b(\lambda, T) = \frac{2\pi h c^2}{\lambda^5} e^{-\frac{hc}{\lambda k_B T}}$
Valid Region
Long wavelengths (low frequencies)
Short wavelengths (high frequencies)
Ultraviolet Failure
Fails at short wavelengths (diverges)
Fails at long wavelengths
Practice Quiz
Choose an option to reveal the explanation.
1. What is the theoretical realization of a perfectly black body?
Explanation: A hollow enclosure with a small pinhole acts as a perfect absorber because any entering radiation undergoes multiple internal reflections and gets trapped, realizing an absorptive power close to 1.
2. According to Kirchhoff's Law, the ratio of spectral emissive power to spectral absorptive power for any body at a constant temperature is:
Explanation: Kirchhoff's law states that $E_\lambda / a_\lambda = E_{b\lambda}$, meaning the ratio equals the emissive power of an ideal black body at the same temperature.
3. The total emissive power of a black body is proportional to which power of its absolute temperature $T$?
Explanation: According to the Stefan-Boltzmann law, the total emissive power $E = \sigma T^4$.
4. Wien's displacement law is mathematically expressed as:
Explanation: Wien's displacement law states that the product of the wavelength of maximum emission $\lambda_m$ and the absolute temperature $T$ is a constant ($b$).
5. Which classical law failed at short wavelengths, resulting in the ultraviolet catastrophe?
Explanation: The Rayleigh-Jeans law predicted energy density going to infinity as wavelength approaches zero at short wavelengths, known as the ultraviolet catastrophe.
6. Max Planck postulated that electromagnetic energy is emitted or absorbed in discrete packets called:
Explanation: Planck introduced the concept of energy quanta with energy $E = h\nu$ to explain black-body radiation.
7. If the absolute temperature of a black body is doubled, its total emissive power increases by a factor of:
Explanation: Since $E \propto T^4$, doubling $T$ increases emissive power by $2^4 = 16$ times.
8. The dimensions of Stefan-Boltzmann constant ($\sigma$) are:
Explanation: From $E = \sigma T^4$, $\sigma = E / T^4$ with units $\text{W m}^{-2}\text{K}^{-4} = \text{kg s}^{-3}\text{K}^{-4}$.
9. What is the typical estimated surface temperature of the Sun based on black-body approximation?
Explanation: Using the solar constant, Stefan-Boltzmann law, and inverse square law, the effective surface temperature of the Sun is calculated to be roughly 5780 K to 6000 K.
10. According to Wien's displacement law, if the temperature of a black body is increased, the wavelength corresponding to maximum emission ($\lambda_m$):
Explanation: Since $\lambda_m T = b$, $\lambda_m$ is inversely proportional to $T$; thus, increasing temperature shifts the peak to shorter wavelengths.
11. Planck's radiation law reduces to Rayleigh-Jeans law under the condition of:
Explanation: For very large $\lambda$ (long wavelengths), the exponential term $e^{hc/\lambda k_B T} \approx 1 + \frac{hc}{\lambda k_B T}$, reducing Planck's law to the Rayleigh-Jeans law.
12. A good absorber of radiation is:
Explanation: By Kirchhoff's law, a good absorber ($a_\lambda$ is high) is also a good emitter ($E_\lambda$ is high) at the same temperature.
13. The unit of spectral emissive power $E(\lambda, T)$ in SI is:
Explanation: Spectral emissive power is energy per unit area per unit time per unit wavelength, giving SI units of $\text{W m}^{-3}$.
14. Which law of radiation is valid only for short wavelengths (high frequencies)?
Explanation: Wien's distribution law was empirically derived for short wavelengths and holds well in that region.
15. The emissive power of a perfectly black body depends on:
Explanation: Cavity radiation from a black body depends solely on its absolute temperature $T$ and is independent of its material or size.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. If the temperature of a black body is increased from $27^\circ\text{C}$ to $327^\circ\text{C}$, the ratio of total radiated energy per second per unit area will be:
2. The wavelength corresponding to maximum energy emission in the spectrum of a black body at temperature $2000\text{ K}$ is $1.5^\circ\mu\text{m}$. What is the wavelength of maximum emission at $4000\text{ K}$?
Explanation: By Wien's displacement law, $\lambda_m T = \text{constant}$, so $\lambda_{m1} T_1 = \lambda_{m2} T_2 \implies 1.5 \times 2000 = \lambda_{m2} \times 4000 \implies \lambda_{m2} = 0.75^\circ\mu\text{m}$.
3. A spherical black body of radius $r$ radiates power $P$ at temperature $T$. If its radius is halved and temperature is doubled, the radiated power becomes:
Explanation: Power $P = (4\pi r^2) \sigma T^4$. When $r' = r/2$ and $T' = 2T$, $P' = 4\pi (r/2)^2 \sigma (2T)^4 = 4\pi (r^2 / 4) \sigma (16 T^4) = 4 \times (4\pi r^2 \sigma T^4) = 4P$... wait, calculation: $4 \times 16 / 4 = 16P$. Let's re-verify: $r^2$ gives $1/4$, $T^4$ gives $16$, product is $16/4 = 4P$. Let's check options: 4P is option C. Wait, let's recalculate carefully: $(1/4) \times 16 = 4$. So $P' = 4P$. Let's set correct_index to 2 (which is $4P$).
4. According to Stefan's law, the rate of loss of heat by a body of surface area $A$ at temperature $T$ placed in surroundings at temperature $T_0$ is given by:
Explanation: When a body is placed in an enclosure at temperature $T_0$, the net rate of heat radiation is $Q/t = e\sigma A (T^4 - T_0^4)$.
5. Planck's radiation formula is given by $E_b(\lambda, T) = \frac{2\pi h c^2}{\lambda^5 (e^{\frac{hc}{\lambda k_B T}} - 1)}$. At extremely high temperatures ($T \to \infty$), what does the exponential term $e^{\frac{hc}{\lambda k_B T}}$ approximate to?
Explanation: For small argument $x = \frac{hc}{\lambda k_B T}$ (when $T$ is very large), $e^x \approx 1 + x$, which leads to Rayleigh-Jeans law.
6. Two black bodies $A$ and $B$ have temperatures $T_A$ and $T_B$ respectively. If the maximum spectral emissive power for body $A$ occurs at wavelength $0.5^\circ\mu\text{m}$ and for body $B$ at $1.0^\circ\mu\text{m}$, the ratio $T_A / T_B$ is:
Attempt these questions under a ten-minute limit using a separate timer.
1. What happens to the total radiant energy emitted per second by a black body if its absolute temperature is tripled?
Explanation: Emissive power $E \propto T^4$. When $T$ is tripled, $E$ increases by $3^4 = 81$ times.
2. Which of the following represents Wien's constant ($b$) dimensionally?
Explanation: Since $\lambda_m T = b$, the dimensional formula is $[\lambda_m] [T] = [L][K] = [L K]$.
3. Ferry's black body consists of a double-walled hollow sphere with a small pinhole and a conical projection opposite to the pinhole. What is the purpose of the conical projection?
Explanation: The conical projection ensures that any incident ray entering through the pinhole is scattered and undergoes multiple internal reflections inside the cavity, preventing immediate reflection back out.
4. According to Kirchhoff's law, if a body absorbs radiation of a certain wavelength strongly at a given temperature, it will:
Explanation: High absorptive power ($a_\lambda$) implies high emissive power ($E_\lambda$) at that same wavelength and temperature.
5. The energy distribution curve of black-body radiation peaks at a specific wavelength $\lambda_m$. As temperature increases, the peak shifts towards:
Explanation: Wien's displacement law states $\lambda_m \propto 1/T$, so higher temperature results in a shift toward shorter wavelengths.
6. Why did Rayleigh-Jeans law fail at high frequencies (short wavelengths)?
Explanation: Rayleigh-Jeans law applied classical equipartition theorem to electromagnetic modes in a cavity, predicting that energy density grows as $\nu^2$ or $1/\lambda^4$, diverging to infinity at high frequencies.
7. Calculate the rate of heat emission per unit area for a black body at temperature $527^\circ\text{C}$. (Take Stefan's constant $\sigma = 5.67 \times 10^{-8}\text{ W m}^{-2}\text{K}^{-4}$)
Explanation: $T = 527 + 273 = 800\text{ K}$. $E = \sigma T^4 = 5.67 \times 10^{-8} \times (800)^4 = 5.67 \times 10^{-8} \times 4.096 \times 10^{11} \approx 2.32 \times 10^4\text{ W m}^{-2}$. Wait, let's recalculate: $8^4 = 4096$, $4096 \times 5.67 = 23224.32$, times $10^3$, so $2.32 \times 10^4$. Let's match options: option C is $1.47$, option B is $3.67$. Wait, let's check $T=800$: $8^4 = 4096$. $4096 \times 5.67 = 23224$. Let's re-evaluate options or adjust calculation. If $T = 500\text{ K}$ ($227^\circ\text{C}$): $5^4 = 625$, $625 \times 5.67 = 3543$. If option A is $7.35 \times 10^3$, let's check $T = 700\text{ K}$: $7^4 = 2401$, $2401 \times 5.67 \approx 13613$. Let's make sure the correct option matches a clean number or select based on standard textbook problem where $T=527^\circ\text{C} \implies 800\text{ K}$, $E = 23224\text{ W m}^{-2}$. Let's adjust options so that correct_index is correct for $2.32 \times 10^4$ or similar. Let's fix the options in this question: Options: A) $2.32 \times 10^4\text{ W m}^{-2}$, B) $1.16 \times 10^4\text{ W m}^{-2}$, C) $4.64 \times 10^4\text{ W m}^{-2}$, D) $5.67 \times 10^4\text{ W m}^{-2}$. Correct index 0.
8. The emissive power of a body depends upon:
Explanation: While black body emission depends only on temperature, the emissive power of any general body depends on both its temperature and the nature of its surface (emissivity/material).
9. Newton's law of cooling is a special approximate case of which broader law?
Explanation: Newton's law of cooling can be derived directly from Stefan-Boltzmann law by applying binomial expansion when the temperature difference between the body and surroundings is very small ($T - T_0 \ll T_0$).
10. Which of the following statements is true regarding the black-body emission spectrum?
Explanation: Unlike atomic emission spectra which are line spectra, a black body emits a continuous spectrum covering all wavelengths from zero to infinity.