[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.