Blackbody fraction of Radiation

In summary, the sun emits energy in the visible spectrum in a fraction that depends on its surface temperature. The function I(\lambda,T) that gives the intensity at all wavelength can be used to calculate this fraction.
  • #1
patapat
20
0

Homework Statement


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.)


Homework Equations


R=[tex]\sigma[/tex]T[tex]^{4}[/tex]

for some reason i can't make the sigma come down, but it's a constant=5.67x10[tex]^{-8}[/tex]W/m[tex]^{2}[/tex]K[tex]^{4}[/tex]

The Attempt at a Solution


I'm not sure if the blackbody radiation equation is relevant, but I'm not sure where to begin with this.
 
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  • #2
patapat said:

Homework Statement


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.)


Homework Equations


R=[tex]\sigma[/tex]T[tex]^{4}[/tex]

for some reason i can't make the sigma come down, but it's a constant=5.67x10[tex]^{-8}[/tex]W/m[tex]^{2}[/tex]K[tex]^{4}[/tex]

The Attempt at a Solution


I'm not sure if the blackbody radiation equation is relevant, but I'm not sure where to begin with this.


This formula (the Stefan-Boltzmann law) you gave gives the intensity radiated at all wavelength. so it's useless for your question. You need the function [itex] I(\lambda,T)[/itex] which you will have to integrate over the range of wavelength provided (and you could check that integrating from 0 to infinity would reproduce the SB law).

Hope this helps
 
  • #3
http://en.wikipedia.org/wiki/Planck's_law_of_black_body_radiation

Then you perform the integration between 350 and 700, then compare the result with the integration from 0 to infinity.

Hint: That last result is also know as a "theorem", what is it called and how does it look like?
 
  • #4
I'm working this same problem right now, and have done everything suggested (and then some), but even using maple to integrate I keep getting that the definite integral comes out to approx. 0 - 0 = 0. This is obviously wrong since more than zero energy gets radiated in the visible portion.

Converting the energy density function to 8pi*hc(kT/hc)^4*int(x^3/(e^x-1)) using
x = hc/lambda*kT, Maple gives me the antiderivative [of int(x^3/(e^x-1))] to be -1/4x^4 + x^3*ln(e^x-1) + 3x^2*polylog(2,e^x) - 6x*polylog(3,e^x) + 6*polylog(4,e^x). Converting 350 nm and 700 nm to values of x and evaluating gives the 0 answer.

Can anyone point out what retarded mistake I must be making? Also sorry about typing the formulas out like that but I haven't figured out how to format it yet.
 

1. What is the blackbody fraction of radiation?

The blackbody fraction of radiation refers to the percentage of emitted radiation that is absorbed by a blackbody. A blackbody is an idealized object that absorbs all incoming radiation and emits it back out at a specific temperature.

2. How is the blackbody fraction of radiation calculated?

The blackbody fraction of radiation is calculated using the Stefan-Boltzmann law, which states that the total emitted radiation (also known as the blackbody radiation) is proportional to the fourth power of the object's absolute temperature. This law can be used to calculate the blackbody fraction of radiation at a given temperature.

3. Why is the blackbody fraction of radiation important?

The blackbody fraction of radiation is important because it helps us understand how much radiation is being absorbed and emitted by an object, which is crucial in many scientific fields such as thermodynamics, astrophysics, and climate science. It also helps us make predictions about the behavior of objects at different temperatures.

4. How does the blackbody fraction of radiation relate to the color of an object?

The blackbody fraction of radiation is closely related to the color of an object. At low temperatures, most of the radiation emitted by an object is in the infrared region, making it appear red. As the temperature increases, the object will emit more radiation in the visible spectrum, causing it to appear orange, yellow, and eventually white at very high temperatures. This is known as blackbody radiation or thermal radiation.

5. Can the blackbody fraction of radiation be greater than 1?

No, the blackbody fraction of radiation cannot be greater than 1. This is because a blackbody absorbs all incoming radiation, so the maximum amount of emitted radiation is equal to the total absorbed radiation. Therefore, the blackbody fraction of radiation is always between 0 and 1.

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