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Problem 16 Medium Difficulty

The surface temperature of the star Rigel is $10^{4} \mathrm{K}$. Find (a) the power radiated per square meter of its surface, (b) its $\lambda_{\text {pcak }},$ and (c) its $\lambda_{\text {median }}.$

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Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

LB
Liev B.

Numerade Educator

Zachary M.

Hope College

Jared E.

University of Winnipeg

Video Transcript

we're told that the surface temperature of the stars tended the four Kelvin for part they were asked to find the power that is radiated by the star. So the power that's ready but radiated by a star which we're gonna write SP is equal to Sigma, which is just a constant times the temperature t to the fourth. So we know what the temperature in its T to the four Kelvin on Sigma Here is thes Stefan Bolton Constant. It's equal to 5.67 times 10 to the minus eight watts per meter squared times Calvin to the fourth. So if you plug these values in, we find that the power here is equal to five point uh, 67 times tended the eight and the units would be watts per meter squared. We can box this in Is our solution for part A. Part B has asked us to find the peak wavelength emitted by this star. So the peak Wavell wave like we're gonna represent his land to sub P. And this is equal to, um, 2.898 millimeter times, Kelvin divided by the temperature or 2.898 times 10 to the minus three meters per kelvin, divided by the temperature. So I just converted millimeters to meters first so we can get our answer and meters plugging these values in, we find that this is equal to, uh, 2.898 times 10 to the minus seven meters. This could be boxed in a czar solution for part B. So look more like a seven. There we go. And then, uh, we'll scroll down a little bit for part C in which were asked to find the median wavelength so the median wavelength would call this lamb to em is represented by the equation of 4.11 millimeter Calvin, which we're going again, convert to meter. So this is 4.11 times 10 to the minus three meter kelvin, divided by the temperature. So plugging these values in, we find that this is equal to 4.11 times 10 to the minus seven meters. When can box, that is our solution for part C.

University of Kansas
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

LB
Liev B.

Numerade Educator

Zachary M.

Hope College

Jared E.

University of Winnipeg