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3.6 Thermal Radiation

Pre-Lecture Reading 3.6

Video Lecture

Supplementary Notes

Thermal Radiation

Figure 1
  • Higher temperatures correspond to faster particle speeds, and consequently faster bounce speeds, and consequently higher-frequency ripples, and consequently higher-frequency light: The distribution of light shifts to higher frequencies. This is called Wien's Law.
  • Higher temperatures correspond to faster particle motions, and consequently more particle bounces, and consequently more ripples in the electromagnetic field, and consequently more light: The distribution of light becomes more intense at all frequencies. This is called Stefan's Law.

Wien's Law

( 10 )
λpeak =
2.9 mm
(T/1 K)
( 11 )
λpeak =
2,900 nm
(T/1,000 K)
Example:
The sun's surface temperature is 5,800 K. Hence, it emits most of its light at
λpeak =
2,900 nm
(5,800 K/1,000 K)
= 500 nm,
which is in the visible part of the electromagnetic spectrum.
  • If λpeak is in the ultraviolet, X-ray, or gamma-ray parts of the electromagnetic spectrum, its thermal distribution will still cross the visible part of the spectrum, and will be blue in color.
  • If λpeak is in the near-infrared (infrared, but almost visible) part of the electromagnetic spectrum, its thermal distribution will still cross the red side of the visible part of the spectrum, and will consequently be red in color (e.g., the
    T = 1,000 distribution in Figure 1).
  • If λpeak is in the far-infrared or radio parts of the electromagnetic spectrum, its distribution will not cross the visible part of the spectrum, and will consequently be invisible (although detectable in the far-infrared and radio; e.g., the T = 300 distribution in Figure 1).

Stefan's Law

( 12 )
F = σT4
In this course, you will never need to use the Stefan–Boltzmann constant to solve a problem.
Example:
Person A has a fever and is 1.01 times hotter than Person B. The energy flux coming off of Person A is how many times greater than the energy flux coming off of Person B? Solution: Let TA and FA be the temperature and energy flux of Person A. Let TB and FB be the temperature and energy flux of Person B. Then, FA = σTA4 and FB = σTB4. Dividing the latter equation into the former equation yields:
FA
FB
=
σTA4
σTB4
=
TA
TB
4
= 1.014 ≈ 1.04.
Most of the math problems in this course are ratio problems.
FT4
Example:
Person A has a fever and is 1.01 times hotter than Person B. The energy flux coming off of Person A is 1.014 = 1.04 times greater than the energy flux coming off of Person B.

Assignment 3