Calculating Quanta Emitted by a Blackbody at 500 C

In summary, the temperature of a blackbody is 500 C or 773.15 K. If the intensity of the emitted radiation is 2.0 W/m^2, and this is due entirely to the most intense frequency component, it can be calculated that 3750 nm or 3.75 x 10^-6 m is the maximum wavelength. The total energy radiated per unit time per unit surface area can be determined using the relationship J=∫^{∞}_{0}J_{λ}(T)dλ.
  • #1
Aoiumi
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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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  • #2
A useful relationship to consider is this one:

[itex]J=∫^{∞}_{0}J_{λ}(T)dλ[/itex]

This tells you the total energy radiated per unit time per unit surface area of a black body.
 
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  • #3
Thank you!
 
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