# 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

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

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

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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