Number density of photons

In summary, the number density of photons in the sun's core for the region of wavelengths between 599 nm and 601 nm is given by Planck's blackbody formula and is constant. However, to get the number of photons per unit volume with these wavelengths, you need to approximate the energy of a photon and then divide by hc/v.
  • #1

Homework Statement

In Problem 1 find the number density of photons in the sun’s core for the region of wavelengths between 599 nm and 601 nm. In problem 1 temperature is given.

Homework Equations

Planck blackbody formula and number density= constant times T^3 where T is temperature.

The Attempt at a Solution

Everyone is telling me to use the constant times T^3 formula but that makes no sense in this problem. The constant times T^3 is only when all frequencies are integrated. However only a slice of frequency is considered. Is this question an error or something. A lot of questions and slides had errors in this class so far.
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  • #2
You are right, you can't use that formula. You need to use Planck's distribution function for the energy per unit volume for radiation that has a wavelength between λ and λ + dλ. See equation 5 here:

Then think about how you would use this to get the number of photons per unit volume with wavelengths in the specified range.
  • #3
I know I got to integrate the wavelengths to get the energy density. However I'm having a problem finding a way to get from energy density to photon density. I know the energy of each photon is hc/(wavelength).
  • #4
You won't need to integrate. Note that you are considering a range of wavelengths Δλ that is very small. So, to a very good approximation you can let dλ ≈ Δλ.

To go from energy density to photon number density, consider the approximate energy of anyone photon in the range of wavelengths that your are dealing with.
  • #5
So once I apply the approximation and plug in 599nm with my dλ ≈ Δλ equaling 2nm I divide by hc/599nm
  • #6
Yes, that should do it. (You could also use 600 nm as a good average wavelength in the interval you are working with, but it won't make much difference.)
  • #7
Thanks a lot. I think what got me was that I never had a expression with df in it before where I did not integrate. I thought a infinitesimal was so small that you can't put a number on it. I knew I had to find the energy density and then divide by something. However because I thought I was was integrating from 601 nm to 599 nm and the wavelength was changing I didn't really have a hc/v to divide by and that got me hooked up for hours today.

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