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At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average power of approximately 4.0...Asked by Deaven
At noon on a clear day, sunlight reaches the earth's surface at Madison, Wisconsin, with an average power of approximately 4.00 kJ·s–1·m–2. If the sunlight consists of photons with an average wavelength of 510.0 nm, how many photons strike a 6.60 cm2 area per second?
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Answered by
DrBob222
E of 510 nm photon = hc/wavelength
Substitute and solve for E in J and I get about 4E-19 J but you need to do that more accurately.
We need to change the energy hitting Madison to cm^2 and not m^2 so
4000 J/sec*m^2 x (1 m/100 cm) x (1 m/100 cm) = about 0.4 J/sec*cm^2 and that x 6.60 cm^2 = about 2.6 J/sec*cm^2. Again that's only an estimate.
Then 4E-19 J/photon x #photons = 2.6 J.
Solve for # photons.
Substitute and solve for E in J and I get about 4E-19 J but you need to do that more accurately.
We need to change the energy hitting Madison to cm^2 and not m^2 so
4000 J/sec*m^2 x (1 m/100 cm) x (1 m/100 cm) = about 0.4 J/sec*cm^2 and that x 6.60 cm^2 = about 2.6 J/sec*cm^2. Again that's only an estimate.
Then 4E-19 J/photon x #photons = 2.6 J.
Solve for # photons.
Answered by
Darb
DrBob your explanations are incredibly unclear.
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