Photons per Unit Volume (Quantum)

  • #1

Homework Statement


[itex]\rho(\lambda)d\lambda=\frac{8(\pi)hc}{\lambda^{5}}\frac{d\lambda}{e^{hc/(\lambda)KT}-1}[/itex]

What does this represent?
Calculate the number of photons per unit volume in a blackbody from [itex]\lambda[/itex] to [itex]d\lambda[/itex]


Homework Equations


N/A


The Attempt at a Solution


I believe this represents the energy per unit volume in wavelength [itex]\lambda[/itex] to [itex]d\lambda[/itex]. To calculate the number of photons per unit volume, would I just calculate

[itex]\int^{d\lambda}_{\lambda}\frac{8(\pi)hc}{\lambda^{5}}\frac{d\lambda}{e^{hc/(\lambda)KT}-1}[/itex]? It seems pretty messy
 

Answers and Replies

  • #2
cepheid
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If rho is the energy per unit volume (and per unit frequency interval), then to get the number of photons per unit volume and per unit frequency interval, you'd have to divide by the energy per photon.

Also, you're thinking about the integral wrong. It doesn't make sense to have lambda and d(lambda) as limits in an integral, since the whole point of d(lambda) is that it represents an "infinitesimal" wavelength interval (it's not a number). You integrate once you want to get the change over some finite wavelength interval, like between lambda1 and lambda2. Conceptually, at least, this operation corresponds to adding up all the changes in all the little infinitesimal intervals within that finite wavelength range.
 
  • #3
Using your hint I believe I need to integrate from 0→∞

[itex]\frac{8(\pi)hc}{\lambda^{5}}\frac{d\lambda}{e^{hc/(\lambda)KT}-1}\frac{\lambda}{hc}[/itex]

I used [itex]E=\frac{hc}{\lambda}[/itex] for the energy of a photon. My final integral appears to be

[itex]8\pi\int^{\infty}_{0}\frac{1}{\lambda^{4}}\frac{d\lambda}{e^{hc/(\lambda)KT}-1}[/itex]

It appears to be a pretty messy integral. Am I on the right track?

Thank you!
 
  • #4
cepheid
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Science Advisor
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Using your hint I believe I need to integrate from 0→∞

[itex]\frac{8(\pi)hc}{\lambda^{5}}\frac{d\lambda}{e^{hc/(\lambda)KT}-1}\frac{\lambda}{hc}[/itex]

I used [itex]E=\frac{hc}{\lambda}[/itex] for the energy of a photon. My final integral appears to be

[itex]8\pi\int^{\infty}_{0}\frac{1}{\lambda^{4}}\frac{d\lambda}{e^{hc/(\lambda)KT}-1}[/itex]

It appears to be a pretty messy integral. Am I on the right track?

Thank you!

I don't think you need to integrate at all. It only asks for the number of photons per volume in infinitesimal wavelength interval dλ, which is precisely what this function is defined to be. ρ(λ) is the energy per unit volume in the infinitesimal wavelength interval dλ.

If you integrated from 0 to ∞, you'd be finding the amount of energy (or number of photons) per unit volume due to photons at ALL POSSIBLE WAVELENGTHS, which is not what the question is asking.
 
  • #5
I see what you're saying. Then the number of photons per volume in a black body from [itex]\lambda[/itex] to [itex]d\lambda[/itex] should be simply

[itex]\frac{8\pi}{\lambda^{4}}\frac{d\lambda}{e^{hc/(\lambda)KT}-1}[/itex]?

It seems too easy though
 

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