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
Well, determining the fraction of the energy radiated by the sun in the visible region of the spectrum involves some integral fun! You'll need to integrate the 5800 K blackbody function from 350 nm to 700 nm and then divide it by the total output per unit area, which is given as sigma T^4.
But what is this blackbody function, you ask? It's like a superhero of physics! It tells you the spectral radiance, or output per area per steradian per wavelength, of a blackbody at a given temperature. To convert it to output per area per wavelength, just multiply by pi (because who doesn't love some good ol' pi?).
Now, you might be wondering how to integrate this function. Well, it's all about using your math skills and integrating with respect to the wavelength. It might sound a little intimidating, but don't worry, you got this!
Now, for the final answer, it should be about 0.6. So go ahead, give it a shot and let me know if you have any more questions along the way. Good luck, my friend!
To determine the fraction of energy radiated by the sun in the visible region of the spectrum (350 nm to 700 nm), you can use the blackbody function and integrate it with respect to wavelength.
The blackbody function represents the spectral radiance of an object at a given temperature and is given by Planck's law of black body radiation. It describes the distribution of energy as a function of wavelength. You can read more about it at http://en.wikipedia.org/wiki/Planck's_law_of_black_body_radiation.
In this case, you need to integrate the blackbody function for the sun's surface temperature of 5800 K from 350 nm to 700 nm. The integration represents the total energy radiated by the sun in that wavelength range.
To carry out the integration, you need to use the appropriate formula for Planck's law, which accounts for the temperature, wavelength, and other constants. After integrating, you should divide the result by the total output per unit area, which is given by the Stefan-Boltzmann law (σT^4), where σ is the Stefan-Boltzmann constant and T is the temperature in Kelvin.
After performing the integration and division, you will obtain the fraction of energy radiated by the sun in the visible region of the spectrum.
It is recommended to consult your textbook or course material for a more detailed explanation and guidance on how to perform the integration and calculations. The answer should be approximately 0.6, but it is advisable to double-check your work and seek feedback from your instructor or peers.