Answer
See explanation.
Work Step by Step
a. Use Wien’s displacement law, equation 39.21.
$$T =\frac{2.90\times10^{-3}m \cdot K}{\lambda_m}$$
$$T =\frac{2.90\times10^{-3}m \cdot K}{10\times10^{-6}m }=2.90\times10^{2}K$$
b. Use Wien’s displacement law, equation 39.21.
$$T =\frac{2.90\times10^{-3}m \cdot K}{\lambda_m}$$
$$T =\frac{2.90\times10^{-3}m \cdot K}{600\times10^{-9}m }=4.83\times10^{3}K$$
c. Use Wien’s displacement law, equation 39.21.
$$T =\frac{2.90\times10^{-3}m \cdot K}{\lambda_m}$$
$$T =\frac{2.90\times10^{-3}m \cdot K}{100\times10^{-9}m }=2.90\times10^{4}K$$
Most materials would melt or vaporize long before reaching 29000 kelvin.
