Question 11: each part = [5 pts]Question 13: each part = [5 pts]Question 12: [5 pts]Question 14: [5 pts]Using the blackbody radiation calcul

Question 11: each part = [5 pts]Question 13: each part = [5 pts]Question 12: [5 pts]Question 14: [5 pts]Using the blackbody radiation calculator, at what wavelength does a temperature of 310 K radiate? Use the calculator to express in11.(a) nanometers &microns: ________________ nm[round to whole number](b) ________________ μm[round to 2decimals]12.Using the Planck Curve for Blackbody Radiation, the peak of the curve for 310 K is at what wavelength? ____________ μm

0 thoughts on “Question 11: each part = [5 pts]Question 13: each part = [5 pts]Question 12: [5 pts]Question 14: [5 pts]Using the blackbody radiation calcul”

  1. [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.

    Reply

Leave a Comment