Two stars, both of which behave like ideal blackbodies, radiate t...

Question

Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature T and a diameter 3.0 times that of the hotter star. (a) What is the temperature of the hotter star in terms of T ? (b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?

Accepted Answer

Hey, everyone. Let's go through this practice problem. Arcturus and Pro A are two of the brightest stars in the night sky. Arcturus has a radius of 25.4 multiplied by the radius of the sun. And an effective temperature of T proc A has a radius of 2.5, multiplied by the radius of the sun and an effective temperature of 6530. Kelvins assume that the two stars act similarly to an ideal black body and that proc A emits a total power times greater than arc. Part one of the problem asks us to call arris temperature. Part two of the problem asks us to express the peak intensity wavelength of arc in terms of the peak intensity wavelength of proc A, we're given four multiple choice options to choose from. Option A part 1, 1652 Kelvins and part two. Lambda Max Aus equals 3.9 and multiplied by Lambda Max Proc A option B part 1 25 191. Kelvins part two. Lambda max Aus equals 2.51 multiplied by Lambda Max Proc on a option C part 1 31 100 Kelvins and part two Lambda max Aus equals 2.1 multiplied by Lambda Max Proc on A and option D part 1 48 100 Kel 61 Kelvins and part two lambda max artery equals 1.34, multiplied by Lambda max proc on a, let's start with part one of the problem which asks for artery's temperature. First of all, since we're looking for a relationship between the temperature that we're trying to find and the variables that were given, which include things like radius and a relationship between the power of each of the stars. Let's first recall that the formula for power is equal to the intensity of something divided by the surface area over which it's distributed. Also recall that the formula for intensity can be given by the Stefan Boltzmann law which states that the intensity is equal to the Stefan Boltzmann constant sigma multiplied by the temperature raised to the power of four. The problem tells us to assume that the stars are acting like an ideal black body. So these two equations are going to be valid for this problem. Also, for the sake of simplicity, in the way I'm going to be writing this out, I'm just going to use the subscript A to refer to variables related to arc. And I'm going to use the subscript P to refer to variables relating to proc on a first off, the problem tells us that proc on A emits a total power times greater than Arce. So writing that out into a formula, the power P from pros A is equal to 500 multiplied by the power emitted by Artus or P sub A. Now I'm going to rewrite this using the power formula. So 500 multiplied by the power from Artus. So that's the same as the intensity from AUS I sub A divided by the surface area of Artus A sub A and this is equal to the power from pros A. So the intensity from proc A divided by the surface area of pros A. Now we're not given the surface areas of the stars, but we are given the radii in terms of the radius of the sun. So we can figure this out by recalling that the surface area of a circle or the surface area of a sphere is equal to four multiplied by pi multiplied by the square of the sphere's radius. So we can substitute that into our power formula. I'm also going to put the intensity formula in for intensity so that we can get temperature into the equation. So 500 multiplied by the intensity of Arcturus. In other words, Sigma multiplied by the temperature of Arri T sub A, raise the power of four divided by the surface area of Archer or four pi R sub A squared is equal to the intensity from proc where the Stefan Bosman constant sigma multiplied the temperature proc T sub B raise the power of four or T sub P raised the power of four divided by the surface area of proc or four pi R sub P squared. And right off the bat, we can do some simplification here because both sides of the equation have sigma. So they cancel out and both sides of the equation also have four pi which can also be canceled out. So a more simplified way of writing this would be that 500 multiplied by T sub A rates, the power of four divided by R sub A rates, the power of two is equal to T sub P raised to the power of four divided by R sub P squared. And remember that our goal here is to solve for the temperature of arch duris. So I'm going to solve this equation for T sub A. The first thing I'm going to do is multiply both sides of the equation by R sub A squared and then divide both sides of the equation by 500 so that we can get the sub A raise the power of four on its own. So what we find doing that is that T sub A raised to the power of four is equal to T sub P raised to the power of four multiplied by R sub A squared divided by 500 multiplied by R sub P squared. It should also be noted that both of the radii were given are in terms of the radius of the sun. Now, if you want, you could just look up at the radius of the sun. The problem doesn't give it. So you'd have to look it up or find a table that includes it. But one way we could do this problem without having to worry about that at all is to do some simplification around that. So instead of writing into our equation, R sub A squared divided by R sub P squared, we could bake the R sub sun into the radii terms. So instead of writing R sub A, we could write 25.4 multiplied by R sub sun divided by 2.5, multiplied by R subs sun and put all that into a squared term. And then the R sub sos cancel out because they're being divided by each other. And the last thing we want to do to solve for T sub A is then raise both sides of the equation to the power of 1/4 so that we can get rid of this four exponent on the T sub A. So what we find is that T sub A is equal, the T sub P raised the power of four divided by all multiplied by the square of 25.4 divided by 2.5. And then all of that is raise the power of 1/4. So all that's left for us to do is take the temperature that the problem gave us for pros 6530 Kelvins. So I want to plug that into the calculator when we do the final expression. So that's 65 30 Kelvins. And what we find when we do this, when we put that into a calculator, we find a temperature for, for a of about Kelvins. And so with that, then is our answer to the first part of the problem. Now, let's focus on part two of the problem which asks us to express the peak intensity wavelength of artery in terms of the peak intensity wavelength of proc on a. Now, this is another part of the problem where it's very helpful that we're being asked to assume that the stars are ideal black bodies. Because if we assume that if we assume that both of the stars are ideal black bodies, then that means we can use Viens Law, which states that the peak wavelength Lambda max multiplied by the temperature is always going to be equal to the constant 2.9 multiplied by 10 to the power of negative three m Kelvins. Since Lambda max multiplied by T will always have this value. We can use this to find a relationship between the two stars using Viens Law. So instead we can write Lambda Max or arri multiplied by the temperature of AUS is equal to Lambda Max for pros on a multiplied by the temperature of proc A. So all we gotta do is solve the equation for Lambda max A by dividing both sides of the equation by T A. So Lambda max A is an equal to Lambda Max. He multiplied by the temperature of proc divided by the temperature of AUS. And we're given in the problem, the temperature of pros and we just found the temperature of arce in the first part of the problem for arc. So we can just put it, put the T sub P divided by T sub A into a calculator to find what that ratio is going to be. So, Lambda max A is equal to 65 Kelvins divided by and then a part A, we found 48 61 calvins and that is multiplied by Lambda max P. And if we put that into a calculator, then we find that 65 30 Kelvins divided by 48 61. Kelvins is equal to about 1.34. And that's what's being multiplied by Lambda Max sub P. And that's the ratio. So our answer for the second part of the problem is that the peak wavelength for artery is equal to 1.34 multiplied by the peak wavelength for pros A. And if we look back to our multiple choice options, we can see that option D agrees with our answers to both parts of the problem. For part one, the temperature of artery is 4861 Kelvins. And for part two of the problem. Lambda Max for Aus is equal to 1.34 multiplied by Lambda Max for pros on a and that is it for this problem. I hope this video helped you out. If you'd like more practice. Please check out some of our other tutoring videos which will give you more experience with these types of problems. But that's all for now. I hope you all have a lovely day. Bye-bye.

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