MixedJEE Physics · Original learning card15 original chapter questions

Emissivity, Absorptivity, and Kirchhoff's Law

Kirchhoff's law of thermal radiation states that for a body in thermodynamic equilibrium, spectral emissivity equals spectral absorptivity at each wavelength and direction; a blackbody has both equal to one.

Why this shows up in the exam

Surface coatings and thermal shields · Bodies inside black enclosures · Relating cooling curves to surface finish

Learn the idea

At thermal equilibrium a body's spectral emissivity equals its absorptivity, while surface finish controls both. A good absorber is also a good emitter at the same wavelength and temperature; a real body inside an enclosure both emits and absorbs radiation.

🧠 Memory hook: Good absorbers are good emitters under matching conditions.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • e_lambda = a_lambda — Kirchhoff equality at equilibrium for matching wavelength and direction
  • 0 <= e <= 1 — gray-body total emissivity bound
  • P_abs = a I_incident A_projected — absorbed incident power for the stated illumination geometry

How to approach it

  1. 1Identify source, surroundings, and surface properties
  2. 2Use matching spectral or gray quantities
  3. 3Balance absorbed and emitted powers without silently setting e to one

Common slip-ups that cost marks

  • •Equating emissivity and absorptivity across unrelated wavelengths
  • •Ignoring absorbed enclosure radiation
  • •Assuming every dark-looking surface is an ideal blackbody

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 15

A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.

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