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(b) A black body in vacuum emits maximum radiation at wavelength 0.60x10-6 m_ Estimate the (i) Surface temperature (iiJMonochromatic emissive power (iii)Total Emi...

Question

(b) A black body in vacuum emits maximum radiation at wavelength 0.60x10-6 m_ Estimate the (i) Surface temperature (iiJMonochromatic emissive power (iii)Total Emissive power:

(b) A black body in vacuum emits maximum radiation at wavelength 0.60x10-6 m_ Estimate the (i) Surface temperature (iiJMonochromatic emissive power (iii)Total Emissive power:



Answers

Consider a black body of surface area 20.0 $\mathrm{cm}^{2}$ and temperature 5000 $\mathrm{K}$ (a) How much power does it radiate? (b) At what wavelength does it radiate most intensely? Find the spectral power per wavelength interval at (c) this wavelength and at wavelengths of (d) 1.00 $\mathrm{nm}$ (an $\mathrm{x}$ - or gamma ray), (e) 5.00 $\mathrm{nm}$ (ultraviolet light or an x-ray), (f) 400 $\mathrm{nm}$ (at the boundary between UV and visible light), (g) 700 $\mathrm{nm}$ (at the boundary between visible and infrared light), (h) 1.00 $\mathrm{mm}$ (infrared light or a microwave), and (i) 10.0 $\mathrm{cm}$ (a microwave or radio wave). (j) Approximately how much power does the object radiate as visible light?

In this problem, We're gonna talk about black body radiation. So there are three main, uh, laws that we have to remember. First One is the Stephan Boatman Law. Stephan boats one wall, and it tells us that the power radiated by a black body is able to sigma. That's that multiple Carson's times the area of the body times t to the four where sigma is equal to 5.57 times 10 to the miners, eight watts per meter squared Calvin to the fourth. Then we have means law. Yeah, that tells us, uh, let me just write this year, Venus Law, It tells us that the wavelength Linda Max, for which the spectral energy density is maximum for given temperature T is given by this law. So all of the max I'm sees equal to B where B is equal to 2.9 times since the minus three, Um, meter Calvi. And finally you have blanks law. Uh, and it tells us that the power for unit area radiated by a black body for a given value off for for a given frequency is equal to I above the tee times d loved it. Where I have learned the tea is equal to mhm. Two pi h c squared overland the fifth times the exponential off age See over Linda K B t minus. What? Okay, so now I can go onto the exercise. So what we have to consider is a black body that has an area of 20 square centimeters. And is that a temperature of 5000? Calvin? And our goal in question A is to find the total power that it radiates. So this is Sigma ate. The fourth Sigma is 5.57 times 10, the miners eight. So what's per meter squared? Going to the fourth times the area that's 20 square centimeters. So 20 times has been minus 4 m squared. Ah, times 5000 to the fourth. Counting to the fourth e should be should before. So, Pete, is it? What, you 7.9 times 10 to the fourth. Once this is the answer to question it. Men in question be way have to calculate what is the wavelength? London, um at which the black body radiates most intensely so long that is equal. To be divided by T. B is 2.9 times 10 to the minus three, while t is 5000 county. So this is 5.8 times 10 to the minus seven years. Or, uh, 580. No, no. Just gonna write it like this. Okay, then can go on to question. See? And here goes to find what is the spectral power for wavelength? That is we want to calculate, um, D p uh, the team, actually, on the TDP he London. And according to a formula from Plank's law, this one here, uh, if we integrate over the area we obtained that this is equal to I have London, she times the area. Okay, Um, so this is two pi age C square over time. The fifth times the exponential of HC overland. The key bt minus one times the area A Okay. And this is two pi. And actually, I'm gonna right here the value, Um, and I'm gonna just leave Lunda as, uh, and unknown to have h five times 6.63 times. 10 to the minus 34. Jewel, second times. See, that's three times 10 to the eighth. Commuters for second square times. The area that's to any time set to the minus 4 m square and divided by love The Square I wanted to the fifth times the exponential off H C over K B k b t. So if Syria Okay, BT it's 6.63 times 10 to the minus 34 times three times 78 Divided by K. Uh, maybe that's three at 1.38 times 10 to the minus. 23 Jews were calving times 5000 Calvin and this is equal to 2.88 times 10 to the minus 6 m. So this is E to the 2.88 times 10 to the minus six, divided by Lunda minus one. So the PD London is equal to eight. In our case, all we have to do is to substitute Lunda for 5.8 times since the minus 7 m. And this is equal to 8.2 times 10 to the 10 watts remediate. Great. Okay, Now what I'm gonna do is that I'm gonna use this formula here, so I'm just gonna substitute Lunda over and over again for the next, uh, exercises. Okay, so in question, Do you have to do the same thing. But for Lambda equals one centimeter, so DP The Lunda is equal to 1.28 times 10 to the minus 1224 watts for meter thinning. Question E lambda is equal to five nanometers, so DPD London is 1.68 times 10 to the minus 2 to 7 watts per meter. In question. F Linda is 400 nanometers. So this is the first visible part of the spectrum on this is equal to 5.47 times 10 to the 10th. What's the meter? Noticed how it increases in question. G Monday $700 meters. So DP The London is 7.41 times since the 10th watts per meter in question age. Well, there is one millimeter. NDP starts decreasing significantly. So this is what is your point? 26. What spirit meter in question. I, Linda, is 10 centimeters, So dp the London is 2.6. I'm sent to the minus nine. What's for later? Okay. And finally in question J, we have to calculate the total power radiated by this but body in the visible spectrum. Eso uh, ideally what we would do is to integrate. So the idea is to integrate BP the Lambda over London in the visible range between 400 m and 700 m. Um, integrating it to say is a lot more difficult. Thanlyin doing an approximation, and that's what I'm gonna do. So we have is the p the London as a function off London. And, um, we're all interested in the visible spectrum, so I'm just gonna drive here. So between 400 and 700 millimeters, we have the values of DPD Lunda off. Um, for 400 it's 5.70 47 times 10 to the 10th. What's for meter? And then when? When the value of lambda is 580 we get to a value off the D. P. D Lamba that is equal to eight point old, too. Times 10 to the 10th. And when it goes to 700 nanometers, the value is seven point. And right here, seven point Ah, 41. Yeah. So let me just draw the points here. So at one point right here, another one up here, another one here, and I'm gonna make an approximation gonna treat this uh, these two sections as straight lines. Yeah, and we want to go with the area. This area here. In order to do so, I'm gonna separate into to trap designs such that the area of the first one, This one here call it a one. A one is equal to the base that, uh, 580 minus 400 nanometers, but times Ah, this first height here. That and that is four, uh, 5.47. Plus this height here, that's 8.2 times into the 10 watts per meter and divided by two. So everyone is equal to 12,141. What's? And a two iss 700 minus 580 nanometers times, uh, eight. Hm. 0.2 plus 7.41. Divided by two times into the 10th watts per meter. So a two is equal to 9258. What? So the total area is the approximation for the power, so that they won't was a to that's approximately 21.4 kilowatt. Now, this is an approximation, so I should leave very precise. The best estimation that I can give is that it's something around 20 kilowatts, which is the answer to exercise

This exercise. We have the radius of the sun, which is 6.96 times 10 to the eighth meters. And I have the power output of the sun, which is 3.85 times center the 26 meters. I'm sorry. Uh, what? And we have to calculate in question A What is the temperature of the sun? So according to a staff of boatman law, the power of a black body is given by the Stefan Boatman. Constant signal times the area of the black body times the temperature to the fourth from where we can obtain the T equals p over single A to do 1/4. So let's subsitute while we already have. Here he is 3.85 times 10 to the 26th watts, divided by Sigma Sigma is just a Stephan Botes When Carson So that's 5.67 times since the minus eight. What per meter squared calving to the fourth times the area off the of the sun, which is the heir of us fear. So it's for pie times. The radius and the radius is, uh, six point 96 times 10 to the eighth, so it's four pi radius square and the temperature is 5780 calving. Well, this is the temperature of the sun. According to Stephan Botes, One law and in question be We have to calculate what is the wavelength off the of the light that's emitted with the greatest intensity by the sun and that we can calculate from Vince low. So we have that the temperature ty of a black body times the wavelength off the other life that's emitted with the maximum intensity is equal to 2.9 times 10 to the minus three meters calving. So the temperature start. So, Linda Max, we already know the temperature. So the London max is 2.9 times since of the minus three meters calving divided by 5.78 times sent to the third Calvi. That's yes, what we found in question. They and this is thank you. Zero point at 5.2 times 10 to the minus seven meters. Okay, so this is the way bling off the all the lights that's emitted with the greatest intensity by the sun According to means law

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.

This question you want to find estimate the surface temperature of the sun and uh power emitted by the sun. Okay. So to solve this problem in my area, given them uh we're playing of the radiation that has a maximum intensity. So we'll be using um my owns this basement law. Whoa. Oh okay. Which states that London max? He equals to 2.90. has 10 to the negative tree me to Calvin. So we can calculate the T. Really. Uh huh. Put in India Uh wait line which is firing 50 cm. Yeah. So by doing so we helped create that. We found that uh huh The face temperature of the sun is 5270 covering key. So this is the temperature of the sun. Next you'll be calculating the power immediate by the sun. Okay. So we've been using the formula P equals two sigma. E A E T. Do that for. Okay. So the power will be cinema is the Stefan Grossman constant, which is 5.67 times 10 to the negative. Eight. Okay. The E. Is emissivity one because you assume son is a perfect blackbody is a surface area of the sun, which is four pi R square. The radios of the sun is 6.96 times 10 to 8 leaders. Okay, then, uh, the face temperature to the power for. And then you calculate this, you get 2.67 times 10- 26. What? Okay. So this is the immediate power by the sun. Me. And that's all for this question.


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