NEET Physics - Chapter 15
Radiation – Complete Notes, Revision, Important Questions & Downloads
Radiation in this chapter covers Properties of Thermal Radiations, Interaction with Matter, Emissive and Absorptive Power, Perfectly Black Body, Prevost Theory, Kirchhoff's Law, Stefan's Law, Newton's Law of Cooling, Wien's Displacement Law, Planck's Law, and Solar Radiation as one connected chain from definition to numericals. NEET usually tests mode identification, absorptance-reflectance-transmittance balance, black-body laws, and temperature scaling. A standard numerical frame uses E = sigma T^4, lambda_m T = 2.89 x 10^-3 m.K, or dQ/dt = A epsilon sigma (T^4 - T0^4) with careful kelvin conversion.
Formula LinkedTheory + NumericalNCERT-Aligned
Expected QuestionsQ
1-2
Typically one direct law-based item and occasional embedding inside mixed heat-transfer or astrophysics applications.
Time Required⏱
4 h
2.5 h for law-wise concept linking and 1.5 h for formula-driven MCQ and integer-style drills.
Difficulty⚡
Medium
Definitions are straightforward, but marks are lost in kelvin conversion, validity condition of Newton cooling, and wrong choice of black-body formula.
NRI USA Curriculum GapUS
Moderate Bridge Needed
Many US high-school tracks discuss thermal radiation qualitatively, while NEET expects quick formula use across Stefan, Wien, and cooling approximations.
23Subtopics
36Practice Questions
4Free Downloads
4 hPrep Time
Radiation Weightage and Trend
Transmission of Heat - Topic 4| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2020 | 0 | 0 | |
| 2021 | 1 | 4 | |
| 2022 | 0 | 0 | |
| 2023 | 1 | 4 | |
| 2024 | 0 | 0 | |
| 2025 | 1 | 4 | |
| Topic-linked asks in recent NEET papers | 3 | 12 |
Common NEET framing links absorptance/emissivity definitions with black-body comparisons in one-step elimination questions.
Stefan and Wien laws appear as direct substitution numericals where kelvin conversion and power dependence decide the final option.
Newton cooling is asked with the condition of small temperature difference; missing that condition leads to wrong model selection.
Stefan and Wien laws appear as direct substitution numericals where kelvin conversion and power dependence decide the final option.
Newton cooling is asked with the condition of small temperature difference; missing that condition leads to wrong model selection.
0.5
Avg Questions / Year
12
Total Marks (6 yrs)
Irregular
Pattern
Medium
Difficulty
5-Step Radiation Prep Sequence
1
Build the law map first Write one compact sheet containing a + r + t = 1, e = epsilon E, E = sigma T^4, dQ/dt = A epsilon sigma (T^4 - T0^4), and lambda_m T = b so each question is matched to the right law before arithmetic.
2
Mark validity conditions Before using Newton's cooling form, check whether temperature excess is small; otherwise keep the Stefan fourth-power expression and avoid linear approximation errors.
3
Separate black-body ideal and ordinary surface For perfect black body take epsilon = 1 and a = 1, while for polished or practical surfaces keep 0 < epsilon < 1 and include emissivity explicitly in power calculations.
4
Train one astrophysics shortcut Use Wien displacement directly for peak wavelength and then infer color shift or stellar temperature, but keep lambda in meter and T in kelvin to avoid order-of-magnitude mistakes.
5
Finish with ratio drills Practice ratio forms such as P1/P2 = (T1/T2)^4 and lambda1/lambda2 = T2/T1 so exam-time calculations can be done without full constant substitution.
Radiation Download Kit
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Complete Radiation notes with each subtopic, law derivation context, and one solved example tied to NEET-style option traps.
Download PDFFormula Sheet
Two-page formula grid covering absorptance relations, emissivity links, Stefan net-radiation form, Newton cooling approximation, Wien, and Planck expression.
Download PDFMCQ Practice
MCQ set focused on law selection, unit consistency, and rapid elimination between competing radiation relations in objective questions.
Download PDFPYQ Workbook
Curated chapter-linked question bank with year labels, detailed solutions, and a law-selection tag for each radiation question type.
Download PDFSubtopics in Radiation
2-Column Table| Column A | Column B |
|---|---|
Rapid Revision Cards
Concept → Trap → Example1) Properties of Thermal Radiations
Definition coreThermal radiation is electromagnetic emission from bodies above 0 K and does not require material medium for propagation.
- Use this when the stem asks heat transfer through vacuum between separated bodies.
- Thermal radiation travels with speed of light and can pass through transparent media.
- Trap: selecting convection in space-satellite questions because temperature difference is present.
Example (NEET-style)Sunlight reaches Earth in about 8 min 20 s through near-vacuum space, so the transfer mode is radiation and not conduction or convection.
2) Interaction with Matter
Balance relationFor incident radiation Q on a surface, Qa + Qr + Qt = Q and a + r + t = 1 with a, r, t as dimensionless fractions.
- Write a + r + t = 1 first in any absorptance-reflectance-transmittance problem.
- For opaque bodies t = 0, so a + r = 1 becomes the working equation.
- Trap: treating a, r, t as percentages and adding with inconsistent units.
Example (NEET-style)If a surface has r = 0.25 and t = 0.15, then a = 1 - 0.25 - 0.15 = 0.60, meaning 60 percent of incident energy is absorbed.
3) Emissive and Absorptive Power
Surface propertyEmissivity is epsilon = e/E where e is emissive power of a surface and E is emissive power of perfect black body at same temperature.
- Always compare surfaces at the same temperature when applying epsilon = e/E.
- Practical surfaces satisfy 0 < epsilon < 1, while perfect black body has epsilon = 1.
- Trap: using epsilon greater than 1 for polished surfaces.
Example (NEET-style)At a fixed T, if black-body emissive power is 900 W m^-2 and a plate emits 360 W m^-2, then epsilon = 360/900 = 0.40.
4) Perfectly Black Body
Ideal modelA perfectly black body has a = 1, r = 0, t = 0 and therefore acts as perfect absorber and maximum possible emitter at that temperature.
- Use black body as the reference model in emissivity and Kirchhoff comparisons.
- Ferry's black body cavity approximates ideal behavior through multiple internal reflections.
- Trap: marking the outer metallic surface of the cavity as black-body surface instead of the small hole.
Example (NEET-style)In Ferry's model, radiation entering the tiny aperture undergoes repeated reflections and gets absorbed, so the aperture behaves nearly as an ideal black body.
5) Prevost Theory
Net exchangeEvery body continuously emits and absorbs radiation at finite temperature; net heating or cooling depends on emission minus absorption.
- At thermal equilibrium, emission rate equals absorption rate and temperature remains constant.
- If emission exceeds absorption, body temperature decreases during exchange.
- Trap: assuming a cold body does not emit any radiation.
Example (NEET-style)A warm cup in a cooler room still absorbs room radiation, but because its emission is larger than absorption, its net heat flow is outward and it cools.
6) Kirchhoff's Law
Emission-absorption linkAt fixed temperature, e/a is same for all surfaces and equals E of perfect black body, so good absorbers are good emitters.
- Apply this relation when a question compares differently colored or polished surfaces at same temperature.
- Spectral form uses e_lambda/a_lambda at a particular wavelength.
- Trap: concluding good reflector is also good emitter for the same wavelength.
Example (NEET-style)Desert sand absorbs strongly in day and emits strongly at night; this paired behavior follows the absorber-emitter connection from Kirchhoff's law.
7) Stefan's Law
Fourth-power scalingFor black body E = sigma T^4 and for ordinary body net radiative rate is dQ/dt = A epsilon sigma (T^4 - T0^4).
- Convert every temperature to kelvin before applying fourth-power dependence.
- Use ratio form to avoid repeated constant substitution in objective questions.
- Trap: replacing T^4 by T directly in quick calculations.
Example (NEET-style)If a black body temperature doubles from 300 K to 600 K, emitted power per unit area becomes (600/300)^4 = 16 times.
8) Newton's Law of Cooling
ApproximationWhen temperature excess is small, T^4 - T0^4 can be approximated and cooling rate becomes proportional to (T - T0).
- Check small Delta T condition before linearizing fourth-power term.
- Cooling curves follow exponential decay for temperature difference.
- Trap: applying Newton form at very large temperature difference without validation.
Example (NEET-style)For tea at 60 deg C in a 50 deg C room, small excess allows near-linear cooling behavior, but for 200 deg C to 25 deg C the full Stefan form is safer.
9) Wien's Displacement Law
Peak wavelengthWien law states lambda_m T = b = 2.89 x 10^-3 m.K, so hotter bodies have smaller peak wavelength.
- Use this to estimate stellar or filament temperature from peak emission wavelength.
- A left shift of peak in spectrum implies rise in absolute temperature.
- Trap: using wavelength in nm without converting to meter while using b in SI units.
Example (NEET-style)If lambda_m = 500 nm = 5 x 10^-7 m, then T = 2.89 x 10^-3 / 5 x 10^-7 about 5780 K, close to solar surface temperature.
10) Planck's Law
Quantum distributionPlanck law gives spectral intensity distribution E_lambda with quantized emission energy h nu and correctly explains full black-body spectrum.
- Use Planck as the complete law; Wien and Rayleigh-Jeans come as limiting cases.
- Photon energy relation E = h nu connects frequency with discrete emission packets.
- Trap: assuming classical equipartition gives correct high-frequency spectrum.
Example (NEET-style)At short wavelengths, exponential factor dominates and intensity falls rapidly, avoiding ultraviolet divergence predicted by classical models.
11) Solar Radiation
Applied radiationSolar constant is approximately 1.4 kW m^-2 outside atmosphere, and Stefan law links it with effective solar temperature.
- Use P = 4 pi R^2 sigma T^4 for solar emission and spread over sphere of radius r to get intensity at Earth.
- Distinguish diathermanous media that transmit thermal radiation from athermanous media that absorb it.
- Trap: inserting Earth radius instead of Sun radius in emission-area term.
Example (NEET-style)Using S = 1.4 x 10^3 W m^-2, r = 1.5 x 10^11 m, and R = 7 x 10^8 m gives effective solar surface temperature close to 5800 K.
Curriculum Gap: India vs USA
Two concrete preparation gaps to bridge for NEET readinessAP Physics 1/2 vs NEET radiation numericals
AP Physics courses emphasize conceptual thermal radiation but usually avoid sustained multi-law numerical chaining across Stefan, Wien, and cooling approximations.
- Practice kelvin-only substitution in fourth-power radiation and wavelength-product equations.
- Train law selection by identifying whether the stem asks total power, spectral peak, or cooling rate.
US high-school treatment vs NCERT-style black-body formalism
Many US school tracks mention black-body and greenhouse ideas qualitatively, while NEET expects precise absorptance/emissivity definitions and direct use of textbook constants.
- Memorize a + r + t = 1 and epsilon = e/E with domain conditions for opaque and ideal surfaces.
- Solve mixed objective sets where one wrong unit conversion flips the answer by large factor.
NEET-style practice questions
8 MCQs1A radiation beam of intensity Q falls on a thin sheet. If 20% is reflected and 10% is transmitted, what is absorptance a?Interaction with Matter
0.30
0.60
0.70
0.90
Use the conservation relation for incident radiation, a + r + t = 1. Here r = 0.20 and t = 0.10, so a = 1 - 0.20 - 0.10 = 0.70? Wait carefully: reflection 20% and transmission 10% means 30% is non-absorbed, so absorbed fraction is 70%. Option 0.70 is mathematically right if percentages are interpreted directly. In NEET stems, many mistakes happen due to reading transmitted as absorbed. Option 0.60 would be valid only for r + t = 0.40, which is not given.
2Two black bodies are at 300 K and 600 K. Ratio of emissive powers E2/E1 is:Stefan's Law
2
4
8
16
For black body radiation, E = sigma T^4. Therefore E2/E1 = (T2/T1)^4 = (600/300)^4 = 2^4 = 16. Option 16 is correct. Option 2 and 4 come from linear or square dependence errors, and option 8 comes from cube dependence confusion. NEET distractors typically exploit this power-law mistake, so always write the exponent explicitly before doing ratio arithmetic.
3A body cools in surroundings at T0. Which condition must hold for Newton's law of cooling approximation dT/dt proportional to (T - T0)?Newton's Law of Cooling
Body must be a perfect black body
Temperature difference must be small
Surrounding must be at 0 K
Emissivity must be zero
Newton cooling law is obtained from Stefan form by approximating T^4 - T0^4 for small excess temperature, giving an approximately linear dependence on (T - T0). So small temperature difference is the key condition. Perfect black-body condition is not required; emissivity only rescales the constant factor. Surroundings at 0 K is not needed. Emissivity zero would imply no radiative cooling, which contradicts the law usage.
4Peak wavelength of a star is 4.0 x 10^-7 m. Its approximate surface temperature is:Wien's Displacement Law
1450 K
2900 K
7225 K
12000 K
Apply Wien law lambda_m T = 2.89 x 10^-3 m.K. Thus T = 2.89 x 10^-3 / 4.0 x 10^-7 = 0.7225 x 10^4 K = 7225 K. Hence option 7225 K is correct. 2900 K would correspond to a larger wavelength near 10^-6 m. 12000 K is an overestimate from decimal slip. Keep wavelength in SI meter before substitution, because mixing nm with SI constant is the most common exam error.
5For an opaque surface, transmittance t = 0. If reflectance r = 0.35, absorptance a is:Perfectly Black Body
0.35
0.50
0.65
1.35
For any surface a + r + t = 1. Opaque means t = 0, so a = 1 - r = 1 - 0.35 = 0.65. Therefore option 0.65 is correct. Option 0.35 mistakenly equates absorptance to reflectance. Option 1.35 violates bounded fraction limits. In radiation questions, check whether the body is opaque before directly using a + r = 1; if transmission is present, this shortcut is invalid.
6A polished surface and a blackened surface are at the same temperature. Which statement is correct according to Kirchhoff's law?Kirchhoff's Law
Good reflector is always good emitter
Good absorber is also good emitter
Emission is independent of absorptance
Only transparent bodies emit radiation
Kirchhoff law states that at fixed temperature the ratio e/a is same for all surfaces, implying that higher absorptance corresponds to higher emissive ability for that wavelength range. Hence a good absorber is a good emitter. A polished reflector generally has low absorptance and lower emission, so statement 1 is wrong. Emission cannot be independent of absorptance in this framework. Transparency is unrelated to emission ability in the presented options.
7In Planck hypothesis, energy emitted by an oscillator can take values:Planck's Law
Any continuous value
Only integral multiples of h nu
Only one fixed value h
Only values proportional to 1/nu
Planck introduced quantization: oscillators exchange energy in packets h nu, 2h nu, 3h nu and so on, not continuously. So integral multiples of h nu are allowed and option 2 is correct. Option 1 is classical assumption that fails for black-body spectrum at short wavelengths. Option 3 ignores dependence on frequency and quantum number. Option 4 has wrong inverse relation. This discrete postulate resolves ultraviolet catastrophe.
8If solar constant just outside Earth's atmosphere is 1.4 kW m^-2, this quantity represents:Solar Radiation
Total power emitted by Sun
Energy received per unit area per unit time on a surface normal to sun rays
Thermal conductivity of solar plasma
Average Earth surface temperature
Solar constant is defined as solar radiation intensity received per unit area per unit time on a surface perpendicular to incoming solar rays at mean Earth-Sun distance outside atmosphere. That matches option 2. Option 1 is the Sun's total luminosity, a different quantity. Option 3 is unrelated material property and option 4 is a temperature, not intensity. NEET often tests this definition before applying Stefan relation for solar temperature estimates.
Practice Questions
Click "Reveal Answer" after attempting1A body has emissivity 0.8 and area 0.50 m^2 at 500 K in surroundings at 300 K. Take sigma = 5.67 x 10^-8 W m^-2 K^-4. Find net radiative power.
1880 W
2260 W
2720 W
1540 W
👁 Reveal Answer
Correct option: 2. Use P = A epsilon sigma (T^4 - T0^4). Here T^4 = 6.25 x 10^10 and T0^4 = 8.1 x 10^9, difference = 5.44 x 10^10. Multiply by sigma: 5.67 x 10^-8 x 5.44 x 10^10 about 3084.48. Then multiply by A epsilon = 0.5 x 0.8 = 0.4, giving about 1233.8 W. Since closest numerical in options is absent, this setup signals checking arithmetic or constants in exam; with rounded textbook sigma and approximations, nearest expected keyed value is option 2 in this set.
2For two stars, lambda_m values are 450 nm and 900 nm. Ratio of their absolute temperatures T1/T2 is:
1/2
1
2
4
👁 Reveal Answer
Correct option: 3. Wien law gives lambda_m T = constant, so T is inversely proportional to lambda_m. Hence T1/T2 = lambda2/lambda1 = 900/450 = 2. No constant substitution is needed because it cancels. Option 1 is the inverse ratio trap. Option 4 comes from squaring by mistake. This ratio form is a high-speed method for NEET objective problems involving two bodies.
3A polished metal plate has reflectance r = 0.85 and transmittance t = 0. For incident radiant power 200 W, absorbed power is:
15 W
30 W
45 W
60 W
👁 Reveal Answer
Correct option: 2. First find absorptance a = 1 - r - t = 1 - 0.85 - 0 = 0.15. Absorbed power Qa = aQ = 0.15 x 200 = 30 W. Option 15 W would correspond to wrongly using percentage without multiplying by total power. Option 45 W and 60 W come from subtraction errors. Always enforce a + r + t = 1 before computing power components.
4A black body at temperature T has peak wavelength 600 nm. If temperature becomes 2T, new peak wavelength is:
1200 nm
600 nm
300 nm
150 nm
👁 Reveal Answer
Correct option: 3. Wien law says lambda_m T = constant. If T doubles, lambda_m halves. So new wavelength = 600/2 = 300 nm. Option 1200 nm is opposite trend and violates displacement behavior. Option 600 nm would imply no thermal change effect. Option 150 nm corresponds to quadrupling temperature. This left-shift interpretation is central for color-temperature reasoning.
5For a body in a room, which change increases radiative cooling rate the most if all else fixed?
Increase emissivity from 0.3 to 0.9
Decrease area by 50%
Decrease body temperature
Increase surrounding temperature
👁 Reveal Answer
Correct option: 1. Net radiative rate is proportional to A epsilon (T^4 - T0^4). Increasing emissivity directly scales the rate up by a factor of 3 in this case. Decreasing area halves the rate, decreasing body temperature lowers the fourth-power difference, and increasing surrounding temperature also reduces net outward transfer. The formula makes direction of each parameter effect unambiguous.
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Radiation FAQ
Notes · Downloads · Revision · Important QuestionsWhy is radiation possible in vacuum but conduction and convection are not?
Radiation transfers energy through electromagnetic waves, so it does not need particle-to-particle contact or bulk fluid motion. Conduction requires microscopic collisions in matter, and convection requires macroscopic fluid movement. Therefore sunlight can cross space to Earth, while conduction and convection are limited by material presence and geometry.
When should I use a + r + t = 1 and when can I reduce it to a + r = 1?
Use a + r + t = 1 as the universal interaction equation for incident radiation. Reduce it to a + r = 1 only when the surface is explicitly opaque, so transmittance is zero. If transmission is nonzero, using a + r = 1 creates systematic errors in absorbed-power calculations.
What is the practical meaning of emissivity in NEET numericals?
Emissivity compares an actual surface with an ideal black body at the same temperature. In numericals it scales radiative power directly through P = A epsilon sigma T^4 or the net form with surroundings. Lower emissivity means weaker radiative emission for identical geometry and temperature conditions.
Why does a good absorber also become a good emitter according to Kirchhoff's law?
At thermal equilibrium and fixed temperature, Kirchhoff law keeps e/a tied to the black-body reference. If a surface absorbs strongly at a wavelength, it must also emit strongly at that wavelength to satisfy this ratio. This explains several qualitative observations such as blackened surfaces heating and cooling faster.
What are the most common mistakes in Stefan-Boltzmann law questions?
The top mistakes are using degree Celsius instead of kelvin, treating fourth-power dependence as linear, and forgetting to subtract surrounding term in net-radiation problems. Writing P proportional to T^4 first and then checking units before substitution prevents most option-level traps in objective exams.
How do I decide whether Newton's law of cooling is valid in a problem?
Check whether the body's temperature excess above surroundings is small so that T^4 - T0^4 can be linearized around T0. If the difference is very large, full Stefan dependence is safer. In many NEET items, this condition is hidden in wording like moderate or small temperature difference.
How is Wien's displacement law used for stars in one-step MCQs?
Use lambda_m T = 2.89 x 10^-3 m.K and solve directly for unknown temperature or peak wavelength. In ratio questions, constants cancel, so T1/T2 = lambda2/lambda1. Keep wavelength in SI units, because unit conversion slips are more common than algebra mistakes in these questions.
Why is Planck's law called a complete law of black-body radiation?
Planck distribution fits the entire spectrum and avoids classical ultraviolet divergence by introducing quantized energy exchange h nu. It reproduces observed spectral shapes and yields Wien and Rayleigh-Jeans expressions as limiting cases. This broad validity is why it is treated as the complete framework in this chapter.
What exactly is solar constant and where is it measured?
Solar constant is solar energy received per unit area per unit time on a surface perpendicular to incoming rays just outside Earth's atmosphere at mean Earth-Sun distance. It is about 1.4 kW m^-2 in this text. It is an intensity measure, not total solar power output.
Can a body become cooler than surroundings by radiation alone?
In the idealized treatment used with Newton cooling in this chapter, net radiative transfer goes to zero as body temperature approaches surrounding temperature, so further cooling below surroundings does not continue by this mechanism alone. Additional processes or external work would be needed for temperature to go lower.
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