- #1
AStaunton
- 105
- 1
Had a quantum assignment recently and one of the problems was:
What is the photon flux (photon cm^-2 s^-1) at a wavelength of 600nm at a distance of 1km from a blackbody soiurce at 2000k emitting 50W of radiation between 400nm-700nm.
My thinking was:
This involves Planck's radiation Equation:
[tex]I(\lambda)=\frac{2\pi hc^{2}}{\lambda^{5}(e^{hc/\lambda kT}-1)}[/tex]
we are interested in 600nm but have info about 400-700nm..Also know that [tex]I\propto P[/tex] so should find area under Plank's equation between limits of 400-700nm and then get area under plank's function from [tex]600nm[/tex] to [tex]600_{nm}+\Delta\lambda[/tex] as limit [tex]\Delta\lambda[/tex] tends to 0, express this latter area as a fraction of the former:
[tex]\frac{P_{600nm}}{P_{400-700nm}}[/tex]
now, find intensity for the 400-700nm range at a distance of 1km and then can find intensity for 600nm by:
[tex]I_{600nm}=I_{400-700nm}\frac{P_{600nm}}{P_{400-700nm}}[/tex]
now divide this intensity value by energy of 600nm photon: [tex]E=\frac{h}{\lambda}[/tex] and we have photon flux...
firstly I would be grateful if someone could confirm for me if I have gotten the concept right and this is the correct method to use...
I also have a technical question:
My ability with calculus is quite poor and so was not able to integrate Planck's function and so had to find area under curve between 400-700nm numerically, which was abviously less precise and quite time consuming! similar problem for area under curve at 600nm...although I think it can be done analytically by: [tex]600nm[/tex] to [tex]600_{nm}+\Delta\lambda[/tex] as limit [tex]\Delta\lambda[/tex] tends to 0, my calculus let me down and so I did it again numerically and found area between limits of 599.9 and 600.1.
So my question is, can what is the method to integrate plank's function and also is the idea of [tex]600nm[/tex] to [tex]600_{nm}+\Delta\lambda[/tex] as limit [tex]\Delta\lambda[/tex] tends to 0 a correct method to use.
Any advice appreciated.
What is the photon flux (photon cm^-2 s^-1) at a wavelength of 600nm at a distance of 1km from a blackbody soiurce at 2000k emitting 50W of radiation between 400nm-700nm.
My thinking was:
This involves Planck's radiation Equation:
[tex]I(\lambda)=\frac{2\pi hc^{2}}{\lambda^{5}(e^{hc/\lambda kT}-1)}[/tex]
we are interested in 600nm but have info about 400-700nm..Also know that [tex]I\propto P[/tex] so should find area under Plank's equation between limits of 400-700nm and then get area under plank's function from [tex]600nm[/tex] to [tex]600_{nm}+\Delta\lambda[/tex] as limit [tex]\Delta\lambda[/tex] tends to 0, express this latter area as a fraction of the former:
[tex]\frac{P_{600nm}}{P_{400-700nm}}[/tex]
now, find intensity for the 400-700nm range at a distance of 1km and then can find intensity for 600nm by:
[tex]I_{600nm}=I_{400-700nm}\frac{P_{600nm}}{P_{400-700nm}}[/tex]
now divide this intensity value by energy of 600nm photon: [tex]E=\frac{h}{\lambda}[/tex] and we have photon flux...
firstly I would be grateful if someone could confirm for me if I have gotten the concept right and this is the correct method to use...
I also have a technical question:
My ability with calculus is quite poor and so was not able to integrate Planck's function and so had to find area under curve between 400-700nm numerically, which was abviously less precise and quite time consuming! similar problem for area under curve at 600nm...although I think it can be done analytically by: [tex]600nm[/tex] to [tex]600_{nm}+\Delta\lambda[/tex] as limit [tex]\Delta\lambda[/tex] tends to 0, my calculus let me down and so I did it again numerically and found area between limits of 599.9 and 600.1.
So my question is, can what is the method to integrate plank's function and also is the idea of [tex]600nm[/tex] to [tex]600_{nm}+\Delta\lambda[/tex] as limit [tex]\Delta\lambda[/tex] tends to 0 a correct method to use.
Any advice appreciated.