22 Chapter 1 Blackbody Radiation^ Image Plane Intensity, and Units
= hc/qX =
= kT/q - 0.026
at room temperature, 300 K, the thermal energy kT is
eV, while the "photon" energy at a wavelength of 1 jim or 10,000
angstrom units, is 48 times as large or 1.24 eV. This is why there is
effectively zero visible light radiated by room temperature objects.
A final help in performing radiation calculations is an expression for
the fractional amount of radiated power from a blackbody occurring at
wavelengths shorter than a given value X. This may be obtained by
evaluating the expression
7t^ J^ (ey - V kT
obtained from Eq. (1.17). Here
has a maximum value of unity and
decreases as the minimum photon energy increases from zero. Since, as
we will learn, most detectors only respond above a given photon energy
(or below a given
wavelength), the appropriate spectrum-
limited radiance may be obtained by taking the product,
emissivity e should of course be an appropriate weighted average.
Since F(x) falls off almost exponentially at high photon energies, a
choice of the emissivity at the cut-off wavelength usually gives an accu-
rate answer. The function
is plotted in Fig. 1.13.
Example: Calculate the power in W/m^ radiated by a unit emis-
sivity surface at temperature 2000 K at wavelengths less than I jUm.
and we then find x = hv/kT from hv
^ 1.24 eV and kT =
0.026(2000/300) = 0.17 eV The argument, x, becomes 7.3 and
The final radiance is then 6 xW or