# Homework Help: Blackbody radiation density

1. Apr 28, 2012

### glebovg

1. The problem statement, all variables and given/known data

A blackbody is radiating at a temperature of 2.50 x 103 K.

a) What is the total energy density of the radiation?
b) What fraction of the energy is emitted in the interval between 1.00 and 1.05 eV?
c) What fraction is emitted between 10.00 and 10.05 eV?

2. Relevant equations

$\frac{P}{A} = \sigma {T^4}$, where $\sigma$ is the Stefan–Boltzmann constant.

3. The attempt at a solution

a) $\frac{P}{A} = (5.67 \times {10^{ - 8}}\frac{W}{{{m^2}{K^4}}}){(2.50 \times {10^3}K)^4} = 2.21\frac{W}{{{m^2}}}$.

I am not sure how to approach part b and c.

2. Apr 29, 2012

### fluidistic

Hi.
For parts b and c I'd use Planck's law (http://en.wikipedia.org/wiki/Planck's_law). You can integrate $B_{\nu \text { or } \lambda } (T)$ from lambda or nu equal to 0 to infinity, this will equal to your result in part a) I think.
Then integrate either of them from a wavelength corresponding to 1 eV up to 1.05 eV. Etc.
Just to be sure, keep out the units in all calculations so that you can spot any clear mistake.

3. Apr 29, 2012

### glebovg

Are you sure this is the correct approach?

How do I find the wavelength? Using Planck's Postulate or the de Broglie relations?

Last edited: Apr 29, 2012
4. Apr 29, 2012

### fluidistic

I'm not sure there's a single approach but I'm almost sure this would work. (See https://www.physicsforums.com/showthread.php?t=427488&highlight=black+body for a similar problem).
As for your 2nd question, I'd use the formula $E=h\nu _1 =1eV$ and $E=h\nu _2 =1.05 eV$.
You should carry the units with the formula given in wikipedia because when you integrate it you might have either power/area or power/volume. There's a factor of c/4 or something like that if I remember well that you should take care of.

5. Apr 29, 2012

### glebovg

So, for part b and c I need to find the frequency and then subtract the spectral radiances?

6. Apr 29, 2012

### fluidistic

No.
If you integrate the spectral radiance from nu 1 to nu 2, you get the power/area or the power/volume (depending upon if there's the c/(4 pi) factor if I remember well) emitted between the frequencies nu 1 and nu 2.
From part a, you also know that if you integrate the spectral radiance over all frequencies possible, you'd get $2.21 \frac{W} {{m^2}}$. Well you'd get the result of part a). A simple analysis shows me that you made a mistake in the last step. You have something to the 3th power elevated to the 4th power so you should get something elevated to the 12th power. This, multiplied by something to the minus 8th power, your result should be close to something to the 4th power, i.e a result around 10⁴, not 10⁰.

7. Apr 29, 2012

### glebovg

It is 10^6 I think. I forgot to add that part in my original post. In part a I applied the Stefan Boltzmann law. Is it correct?

8. Apr 29, 2012

### fluidistic

I'm almost sure it is correct. However according to wikipedia it should be power/volume.

9. Apr 29, 2012

### glebovg

I think the approach you are proposing is incorrect. There must be something easier because Maple cannot even calculate such a small quantity.

10. Apr 29, 2012

### Steely Dan

Well then, instead of using Maple, try doing the integral by hand :)

11. Apr 29, 2012

### glebovg

The reason why I think your approach is incorrect is because we are interested in energy density, but what are the units for spectral radiance? Is it watts per steradian per square meter per hertz?

Last edited: Apr 29, 2012
12. Apr 29, 2012

### fluidistic

Does the following convinces you a bit: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/radfrac.html#c1?
Fraction of powers will result in fraction of energy densities because they are proportional I think.

13. Apr 29, 2012

### glebovg

14. Apr 29, 2012

### glebovg

Never mind. I do not think neither you nor I have a clue.

15. Apr 29, 2012

### Steely Dan

"Watts per hertz" is really just energy, so spectral radiance is simply energy per unit area per steradian per unit wavelength. Consequently, if you know the fraction of all the intensity in this spectral range compared to the total intensity, you also know the energy ratio.