Calculating Radiation Power of a Ball at 10000K in the Range of 400nm-800nm

In summary, the student is trying to calculate the radiation power of a ball in a 400 nm-800 nm range. They use Planck's equation and integrate it. They find that they don't know how to integrate and ask for help. They use a site to find that they get -4.31119 W/m2/sr.
  • #1
kubajed
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Homework Statement


I need to calculate radiation power of ball in range 400nm-800nm.
T=10000K
d=1um.

Homework Equations


I think I need to use Planck equation and integrate that.
My equation in link: http://imgur.com/XbAUwJ8

The Attempt at a Solution


I look for equations and laws and try to do my equation (above). If this is correct, I ask for help with calculate that.
 
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  • #3
Thanks. Problem is that I don't know how to calculate integrate.
 
  • #4
If you have not learned how to integrate you should consider reading up or watching some explanatory videos on YouTube. I suppose that if the question expects to you integrate for your solution you should have been taught it at some point before.

But if you do know how to integrate, and your only problem is that you are not confident if you did it right, we can help you here if you show us how you did it so we can correct your mistakes (and you can learn from it too!).
 
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  • #5
I don't know how to integrate. That isn't a homework for all. It's an extra exercise for me. I hope that you will help me with it.
I do calculations in Mathematica. I put that:
Integrate[((2*Pi*6.63*10^-34*(3*10^9)^2)/x^5)*(1/(E^(6.63*10^-34*3*10^9/x*1.38*10^-23*10000))-1)*1.256*10^-11, {x,4*10^-7,8*10^-7}].
I use my equation just multiplied by c/4 and area of ball. Result is -4.31119.
This site: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/radfrac.html claim that result is 0.00269. Where I am wrong?
 
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  • #6
kubajed said:
I don't know how to integrate. That isn't a homework for all. It's an extra exercise for me. I hope that you will help me with it.
I do calculations in Mathematica. I put that:
Integrate[((2*Pi*6.63*10^-34*(3*10^9)^2)/x^5)*(1/(E^(6.63*10^-34*3*10^9/x*1.38*10^-23*10000))-1)*1.256*10^-11, {x,4*10^-7,8*10^-7}].
I use my equation just multiplied by c/4 and area of ball. Result is -4.31119.
This site: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/radfrac.html claim that result is 0.00269. Where I am wrong?
This link shows the integral over a finite range http://www.spectralcalc.com/blackbody/inband_radiance.html
 
  • #7
Which one of them is that what I am looking for? On this site is calculator too. It claims that result is 6.77561e+07 W/m2/sr. My site: 2.1447e+08 W/m2.
 
  • #8
I have other question: how much photons (in visible range) will be registered in detector located 10cm far with diameter 1mm in 100ps?
 
  • #9
kubajed said:
Which one of them is that what I am looking for? On this site is calculator too. It claims that result is 6.77561e+07 W/m2/sr. My site: 2.1447e+08 W/m2.
Those seem about right for the radiance. It asks for the power. The emitter is a tiny ball. Using https://astrogeology.usgs.gov/tools/thermal-radiance-calculator/, I get more like 2.2E-4W.

(There's something I'm not understanding in this topic. I see radiance quoted as W·sr−1·m−2. Suppose the ball's surface has a radiance R and radius r. It has a surface area 4πr2, so that satisfies the m−2. What solid angle should I use to satisfy the sr−1? 2π, on the basis that each area element is emitting into a half space? Doesn't really make sense to me.
I note that one of your results quotes sr−1 but the other doesn't, and the ratio is about π.)
 

FAQ: Calculating Radiation Power of a Ball at 10000K in the Range of 400nm-800nm

1. How is radiation power of a ball at 10000K calculated?

The radiation power of a ball can be calculated using the Stefan-Boltzmann law, which states that the radiation power is proportional to the fourth power of the temperature. The formula for calculating radiation power is P = σAT^4, where P is the power, σ is the Stefan-Boltzmann constant, A is the surface area of the ball, and T is the temperature in Kelvin.

2. What is the significance of a ball's temperature being 10000K?

A temperature of 10000K is extremely high and indicates that the ball is emitting a large amount of thermal radiation. This temperature is often used in scientific calculations and models to simulate extreme heat sources, such as stars or nuclear reactions.

3. What is the range of wavelengths considered in the calculation of radiation power?

The range of wavelengths considered in this calculation is 400nm to 800nm, which falls within the visible light spectrum. This range is often used when studying objects that emit thermal radiation, as it covers the wavelengths that are visible to the human eye.

4. How does the surface area of the ball affect the radiation power?

The surface area of the ball directly affects the radiation power, as seen in the formula P = σAT^4. A larger surface area means more space for thermal radiation to be emitted from, resulting in a higher radiation power. Similarly, a smaller surface area would result in a lower radiation power.

5. Can the Stefan-Boltzmann law be used to calculate radiation power for objects other than a ball?

Yes, the Stefan-Boltzmann law can be used to calculate the radiation power for any object with a known temperature and surface area. This law is applicable to all objects that emit thermal radiation, including stars, planets, and even human bodies.

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