[tex]\lambda=9.348\cdot 10^{-6} m = 9.348 \mu m[/tex]
[tex]\lambda=9.348\cdot 10^{-6} m = 9348 nm[/tex]
Explanation:
This problem can be solved by using Wien's displacement law, which states that the peak wavelength and the temperature of a black body are inversely proportional to each other; mathematically:
[tex]\lambda T = b[/tex]
where:
[tex]\lambda[/tex] is the peak wavelength
T is the absolute temperature
[tex]b=2.898\cdot 10^{-3} m\cdot K[/tex] is the Wien's constant
In this problem, the temperature of the blackbody is:
T = 310 K
Therefore, solving for [tex]\lambda[/tex], we find the wavelength of the peak of the radiation emitted:
[tex]\lambda=\frac{b}{T}=\frac{2.898\cdot 10^{-3}}{310}=9.348\cdot 10^{-6} m[/tex]
which can be rewritten in nanometers and microns, keeping in mind that:
[tex]1 m = 10^{-6} \mu m\\1 m = 10^{-9} nm[/tex]
we have:
[tex]\lambda=9.348\cdot 10^{-6} m = 9.348 \mu m[/tex]
[tex]\lambda=9.348\cdot 10^{-6} m = 9348 nm[/tex]
Part b) of the problem is exactly the same as part a), since the temperature is the same, the wavelength is the same.
