Calculating Quanta Emitted by a Blackbody at 500 C

AI Thread Summary
To calculate the number of quanta emitted by a blackbody at 500 C (773.15 K) with an intensity of 2.0 W/m^2, the peak wavelength can be determined using Wien's displacement law, resulting in a λmax of 3750 nm. The intensity provided is related to the energy emitted per unit area, which can be integrated over all wavelengths to find total energy. The relationship J=∫^{∞}_{0}J_{λ}(T)dλ is crucial for calculating the total energy radiated. Further calculations will involve converting the energy to the number of quanta using the energy of a single photon. This approach will yield the desired number of quanta emitted per second per square meter.
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Homework Statement



The temperature of a blackbody is 500 C. If the intensity of the emitted radiation, 2.0 W/m^2, were due entirely to the most intense frequency component, how many quanta of radiation would be emiteed per second per square meter?

Homework Equations



λmaxT = 2.90 x 10^-3 m K


The Attempt at a Solution



I'm not sure where to begin or how to use intensity.
500 C = 773.15 K

I calculated λmax = 2.90 x 10^-3 m K / 773.15 K = 3750 nm or 3.75 x 10 ^-6 m. I'm not sure where to go from here. Please help. Thank you!
 
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A useful relationship to consider is this one:

J=∫^{∞}_{0}J_{λ}(T)dλ

This tells you the total energy radiated per unit time per unit surface area of a black body.
 
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Thank you!
 
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