Question 40.1: Thermal Radiation from Different Objects (A) Find the peak w......

(A) Conceptualize Thermal radiation is emitted from the surface of any object. The peak wavelength is related to the surface temperature through Wien’s displacement law (Eq. 40.2).

\lambda_{\operatorname*{max}}\ T=2.898\times10^{-3}\,\mathrm{m}\cdot\mathrm{K}           (40.2)

Categorize We evaluate results using an equation developed in this section, so we categorize this example as a substitution problem.

Solve Equation 40.2 for \lambda_{\max } :

\text { (1) } \lambda_{\max }=\frac{2.898 \times 10^{-3} \mathrm{~m} \cdot \mathrm{K}}{T}

Substitute the surface temperature:

\lambda_{\max }=\frac{2.898 \times 10^{-3} \mathrm{~m} \cdot \mathrm{K}}{308  \mathrm{~K}}=9.41  \mu \mathrm{m}

This radiation is in the infrared region of the spectrum and is invisible to the human eye. Some animals (pit vipers, for instance) are able to detect radiation of this wavelength and therefore can locate warm-blooded prey even in the dark.

(B) Substitute the filament temperature into Equation (1):

\lambda_{\max }=\frac{2.898 \times 10^{-3} \mathrm{~m} \cdot \mathrm{K}}{2000 \mathrm{~K}}=1.45  \mu \mathrm{m}

This radiation is also in the infrared, meaning that most of the energy emitted by a lightbulb is not visible to us.

(C) Substitute the surface temperature into Equation (1):

\lambda_{\max }=\frac{2.898 \times 10^{-3} \mathrm{~m} \cdot \mathrm{K}}{5800 \mathrm{~K}}=0.500  \mu \mathrm{m}

This radiation is near the center of the visible spectrum, near the color of a yellow-green tennis ball. Because it is the most prevalent color in sunlight, our eyes have evolved to be most sensitive to light of approximately this wavelength.