What is the energy density of blackbody radiation at 2.50 x 10^3 K?

[glebovg]

Homework Statement

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?

Homework Equations

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

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.

 

[fluidistic]

Homework Statement

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?

Homework Equations

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

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.

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.

 

[glebovg]

Are you sure this is the correct approach?

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

 

[fluidistic]

Are you sure this is the correct approach?

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

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.

 

[glebovg]

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

 

[fluidistic]

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

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.

 

[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?

 

[fluidistic]

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?

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

Planck's law can also be written in terms of the spectral energy density (u) by multiplying B by 4π/c:[8]

u_i(T) = \frac{4\pi}{c} B_i(T)

These distributions have units of energy per volume per spectral unit.

 

[glebovg]

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

 

[Steely Dan]

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

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

 

[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?

 

[fluidistic]

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?

Does the following convinces you a bit: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/radfrac.html#c1?

The total power radiated by a blackbody is given by the Stefan-Boltzmann equation, but it is often interesting to know the fraction of power which is emitted in the visible or some other wavelength range. [...] Finding the power radiated within a given wavelength range requires integration of the Planck radiation formula over that range. The radiated power per unit area is the Planck energy density multiplied by c/4. It can be approximated numerically by taking a sum of values of the Planck radiation density times a wavelength interval.

Fraction of powers will result in fraction of energy densities because they are proportional I think.

 

[glebovg]

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

That was very helpful.

 

[glebovg]

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

 

[Steely Dan]

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?

"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.