Posted by **--pat--** on Sunday, August 5, 2007 at 7:49pm.

Determine the fraction of the energy radiated by the sun in the visible region of the spectrum (350 nm to 700 nm). (Assume the sun's surface temperature is 5800 K.)

any help is always appreciated

thanks

Solve by integrating the 5800 K blackbody function from 350 to 700 nm, and dividing by the total output per unit area, sigma T^4.

We will be glad to critique your work.

Note: The blackbody function is usually quoted as spectral radiance (output per area per steradian per wavelngth). You must multiply that by pi to get output per area per wavelength.

what is the blackbody function?

It is what you need to use to do this problem. Is there a textbook assiciated with your course?

You can read aboutit at

http://en.wikipedia.org/wiki/Planck's_law_of_black_body_radiation
am i supposed to integrate the function with respect to the wavelength?

yes. The answer should be about 0.6

## Answer this Question

## Related Questions

- physics - The sun is 1.5 1011 m from the earth. (a) Over what surface area is ...
- physics - the energy received by the earth from the sun is 1400w/m2. Assuming ...
- physics - The difference between energy radiated from the sun and energy ...
- physics - The difference between energy radiated from the sun and energy ...
- Physics - the temperature at the surface of the sun is approximatly at 5700K and...
- physics-sun - a 1 square meter solar panel receives a total of 1000Watts from ...
- physics - The furthest distance from the Sun to Earth is df=1.521E8 The shortest...
- Physics - At the upper surface of the earth's atmosphere, the time-averaged ...
- Physics - At the upper surface of the earth's atmosphere, the time-averaged ...
- Physics to Elena - At the upper surface of the earth's atmosphere, the time-...

More Related Questions