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Problem 4 Medium Difficulty

The radius of our Sun is $6.96 \times 10^{8} \mathrm{m},$ and its total power output is $3.85 \times 10^{26} \mathrm{W} .$ (a) Assuming the Sun's surface emits as a black-body, calculate its surface temperature. (b) Using the result of part (a), find $\lambda_{\max }$ for the Sun.

Answer

a. 5.78 \times 10^{3} K
b. 5.01 \times 10^{-7} m

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Top Physics 103 Educators
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Liev B.

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Zachary M.

Hope College

Jared E.

University of Winnipeg

Meghan M.

McMaster University

Video Transcript

Okay, So, um, we know that the radius of the sun is 6.96 times 10 to the eighth Is that Munir's air kilometers meters and its output is 3.85 times 10 to the 26th wants and way wanted get the surface temperature. So what we want to do is use wines. Displacement law. Um, so see, possible I think about it, cause minus displacement. Law says that land A T is equal 2.2989 times 10 to the minus two Mom meter Calvin, not Milica Wind meter Calvin. And we're not really given anything those given. So can we get any of these givens based on? That's okay. We're going to calculate the temperature, or, um, yeah, we want. So we need Ben to get the wavelength. So I bet we can get the wavelength. Ah, from this information, I just want I can think of a way to do it. And I just want to do it in the most direct way. Someone a pause and re read the text. OK, so I looked into a little more, and I realized we have to use formulas from outside this section. So we want to go to a chapter 11.5 point four and used to fund slot to get the temperature s. Oh, that's so that state. So we use this one later, but for now, what we want to use this to find solid says that power is equal to Sigma III to before. And then I'll talk about what that ISS means. Um or what? Those mean e to the core Isn't there like a sigma and their Ah, yeah, Sigma 80 to you before how many? Rearrange these. But it's okay. Multiplication is associative. So, um so basically, um So what we want to do is, uh, actually let me rewind and say what each of these things are. So Sigma is a constant. And then that's equal to 5.6696 times 10 to the minus eight. Was it 66 96 times 10 to the minus eight. And, um, A is gonna be the surface area of the emitting object. And so, in our case is gonna be equal to pi R. Excuse me. Four pi r squared. That's the surface area of the sun and, um, great and that's a missive. Ity is between their own one. I'm just going approximately this one, Thio, um, to to do this question. Um, uh, yeah, I guess I could look into it. Um, let's see him come for reading through the text. It says it's a perfect black body as a missive. Ity amiss it activity. I miss activity of one. Um, And so let's just say that it's a perfect black body and, um yeah, so, using the power, we can get tea so we can get that t is equal to pee over pie. Where pie are the sun squared? So that's the area. Um, times e sigma today Forget anything you sigma. And then this is my warped parenthesis e This is all raised to the 1/4 power. So I'm gonna possible I plug all of this into a calculator. Okay, I got five point 78 times tend to the third power Coben lips dropping, texting. Um, and now we want to get lambda. So just trying to squeeze in Landover here, Lambda is gonna be taking We're gonna take both sides of the equation and divided by the tea that we got over here. So, um, doing that? Let's see what I get with that. I got 517 centimeters. Ah, right in the optical or the visible middle of the visible spectrum. Probably not a coincidence, given that we humans use the sun. I mean, just double check my numbers. 208 I'm Rhea. Small typing error. So that's actually 501 not 5 17 Let me go fix that.

University of Washington
Top Physics 103 Educators
LB
Liev B.

Numerade Educator

Zachary M.

Hope College

Jared E.

University of Winnipeg

Meghan M.

McMaster University