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

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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

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bonexptip 3 years 2021-08-29T20:49:19+00:00 1 Answers 8 views 0

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  1. Gerda
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    2021-08-29T20:50:46+00:00
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    \lambda=9.348\cdot 10^{-6} m = 9.348 \mu m

    \lambda=9.348\cdot 10^{-6} m = 9348 nm

    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:

    \lambda T = b

    where:

    \lambda is the peak wavelength

    T is the absolute temperature

    b=2.898\cdot 10^{-3} m\cdot K is the Wien’s constant

    In this problem, the temperature of the blackbody is:

    T = 310 K

    Therefore, solving for \lambda, we find the wavelength of the peak of the radiation emitted:

    \lambda=\frac{b}{T}=\frac{2.898\cdot 10^{-3}}{310}=9.348\cdot 10^{-6} m

    which can be rewritten in nanometers and microns, keeping in mind that:

    1 m = 10^{-6} \mu m\\1 m = 10^{-9} nm

    we have:

    \lambda=9.348\cdot 10^{-6} m = 9.348 \mu m

    \lambda=9.348\cdot 10^{-6} m = 9348 nm

    Part b) of the problem is exactly the same as part a), since the temperature is the same, the wavelength is the same.

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