The exitance (power per unit area per unit wavelength) from a blackbody (Box 19-1) is given by the Planck distribution:

where _ is wavelength, T is temperature (K), h is Planck's constant, c is the speed of light, and k is Boltzmann's constant. The area under each curve between two wavelengths in the blackbody graph in Box 19-1 is equal to the power per unit area (W/m2) emitted between those two wavelengths. We find the area by integrating the Planck function between wavelengths λ1 and λ2:

For a narrow wavelength range, ˆ†Î», the value of Mλ is nearly constant and the power emitted is simply the product Mλˆ†Î».
(a) Evaluate M_ at__2.00 _m and at __10.00 _m at T _ 1 000 K.
(b) Calculate the power emitted per square meter at 1 000 K in the interval _ _ 1.99 _m to _ _ 2.01 _m by evaluating the product M___, where __ _ 0.02 _m.
(c) Repeat part (b) for the interval 9.99 to 10.01 _m.
(d) The quantity [M_ (_ _ 2 _m)]/[M_ (_ _ 10 _m)] is the relative exitance at the two wavelengths. Compare the relative exitance at these two wavelengths at 1 000 K with the relative exitance at 100 K. What does your answer mean?

Transcribed Image Text:

LTİC" Az Power emitted- 1 Mn dλ A2