Number of modes in Cubic Cavity

  • #1

Homework Statement


Calculate the number of modes in a cubic cavity of length a=2.5 cm in the wavelength interval (λ1,λ2) where λ1=500 nm and λ2=501 nm. What's the total energy which radiates from the cavity if it's kept at a constant temperature of T=1500 K.

Homework Equations


I imagine these would be rather relevant: Number of modes in a cavity N= (8π/3)*(a/λ)^3 and the average energy for each mode which would be either k*T if we work with Rayleigh-Jeans or hν/e^(hν/kT)-1 if we work with Planck's derivation.

The Attempt at a Solution


I thought of simply finding the number of modes as N1-N2 where N1 is the number for λ1 and N2 for λ2, and then multiplying by the average energy (Which, could someone please double-check that formula? I might have misinterpreted it as average energy of a mode and it might be something different.) Of course, if we use Planck's formula I would find the frequency as being c/λ so no biggie there. I just want to know if it's the correct way to go about it.
 

Answers and Replies

  • #2
nrqed
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Homework Statement


Calculate the number of modes in a cubic cavity of length a=2.5 cm in the wavelength interval (λ1,λ2) where λ1=500 nm and λ2=501 nm. What's the total energy which radiates from the cavity if it's kept at a constant temperature of T=1500 K.

Homework Equations


I imagine these would be rather relevant: Number of modes in a cavity N= (8π/3)*(a/λ)^3 and the average energy for each mode which would be either k*T if we work with Rayleigh-Jeans or hν/e^(hν/kT)-1 if we work with Planck's derivation.

The Attempt at a Solution


I thought of simply finding the number of modes as N1-N2 where N1 is the number for λ1 and N2 for λ2, and then multiplying by the average energy (Which, could someone please double-check that formula? I might have misinterpreted it as average energy of a mode and it might be something different.) Of course, if we use Planck's formula I would find the frequency as being c/λ so no biggie there. I just want to know if it's the correct way to go about it.
Your formula are correct. And yes you may calculate N1-N2 that way. In this case, since the difference of wavelength is small compared to the wavelengths themselves, one could also take the derivative to find dN in terms of ##d \lambda## and use that to get dN directly.
 
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  • #3
Thank you! As for the radiant energy, would mere multiplication with N give me the right answer? I see no reason why it should not, just making sure.
 

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