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A Light Emitting Diode (LED) is a semiconductor light source that emits light when current flows through it. Electrons in the semiconductor recombine with holes, releasing energy in the form of photons. As far as I know, an LED is not a blackbody.

In order to be a blackbody, an object must satisfy 2 conditions:

  1. It should absorb all em radiations that fall on it and be in thermal equilibrium.
  2. It should emit the maximum em radiation possible by a body of its mass.

I have four questions.

  • Which of the conditions are not satisfied in the case of LEDs? Is high temperature a necessary condition for an object to be a blackbody?
  • Does a burning incandescent lamp filament absorb all radiations that fall on it? If so, how do we know that?
  • How is an incandescent lamp bulb filament in thermal equilibrium? (It seems to be way hot than its surroundings).
  • Are all blackbodies hot objects? If an object is not hot does it automatically violate one of the above two conditions of a blackbody? I have seen people claiming that snow is a blackbody. How does that work?
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2 Answers 2

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"Why LED is not the black-body?" is directly implied from "How does the black-body work?"

It's the radiation due jiggling of electrons in an atom due to thermal energy. The LED has totally different working and hence you would not expect it to emit black-body tradition.

The black-body radiation is not necessary to be emitted body of hot temperature. It's given by $$u_\nu=\frac{8\pi h}{c^3}\frac{\nu^3}{e^{\beta h\nu}-1}$$ and is defined for all temperature. The so-called Cosmic Microwave background is blackbody radiation with temperature $2.7$K.

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  • $\begingroup$ //It's the radiation due jiggling of electrons in an atom due to thermal energy.// - This seems to be an inference. Nowhere in the definition it is said that the emitted radiation should be thermal. That's what confusing me. $\endgroup$
    – Sasikuttan
    Commented Oct 4, 2021 at 3:39
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    $\begingroup$ Black-body radiation is a thermal radiation. See en.wikipedia.org/wiki/Thermal_radiation $\endgroup$
    – Himanshu
    Commented Oct 4, 2021 at 4:17
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The jiggling referred to by Young Kindaichi is exactly the thermal energy of the atoms in the hot object. The distribution of wavelengths in the blackbody spectrum comes from the distribution of energies possessed by those atoms.

All objects above 0 kelvin jiggle with thermal energy, so all of them will emit a blackbody spectrum. At low temperatures the average energy is very low and the wavelengths are long. At high temperatures the average energy is higher and the wavelengths are shorter.

LEDs work by pushing electrons across a semiconductor junction from which monochromatic light is emitted, so they do not produce a blackbody spectrum.

Inside the light bulb is an inert gas and the hot filament is in thermal equilibrium with the gas right next to the filament. By sharing that radiation back and forth, one of the conditions for blackbody radiation is met, and the filament yields a blackbody spectrum with a peak corresponding to a temperature of about 2800K.

Because snow is above 0 kelvin, it will radiate a blackbody spectrum with a peak corresponding to the temperature of the snow- which might be lots cooler than a bulb filament, but still lots warmer than absolute zero.

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  • $\begingroup$ I understand that thermal energy is the energy of vibrating atoms. But my doubt is that, is it defined that the radiation emitted by a blackbody should be thermal, and not from any other means, say, spontaneous emission? I didn't see that in the definition. //By sharing the radiation back and forth// - I didn't understand this. //one of the conditions for blackbody radiation is met// - Shouldn't both conditions be satisfied in order to classify a body as blackbody? $\endgroup$
    – Sasikuttan
    Commented Oct 4, 2021 at 4:06
  • $\begingroup$ Snow may approach the conditions to emit a Planck function at IR wavelengths (obviously not at visible wavelengths) because it absorbs IR radiation. $\endgroup$
    – ProfRob
    Commented Oct 4, 2021 at 13:13
  • $\begingroup$ @curiouserandcuriouser almost all visible and IR radiation is via spontaneous emission. "Thermal emission" isn't a microscopic process. $\endgroup$
    – ProfRob
    Commented Oct 4, 2021 at 13:15
  • $\begingroup$ @ProfRob Oops! I mixed up classical physics with quantum mechanics there. Thanks for pointing it out. $\endgroup$
    – Sasikuttan
    Commented Oct 5, 2021 at 2:29

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