Please help me solve this waves and optics question!!! >"<?

blue01

2009-06-09 4:45 am

The average solar insolation (sunlight that falls on surface) on the earth in the absence of atmosphere is 342 W per m^2. The atmosphere reduce this to about 250W per m^2 on a clear day. Approximately 2% of the solar power is UV (average λ = 340nm), 47% is visible ( average λ = 580nm) and 51% is infrared (average λ = 1540 nm)

1) The surface area of New Zealand is 270, 530km^2. On a clear day, how much energy (in J) falls on New Zealand in average in 1 hour from the sun?

2) Solar photovoltaic generally collect ultraviolet and visible light from the sun and convert it to useful energy. How many visible photons fall on New Zealand in average in 1 hour on a clear day?

回答 (1)

[匿名]

2009-06-09 7:05 am

✔ 最佳答案

Energy = Intensity x Area x Time
Energy = ( 250 J s^-1 m^-2 ) x ( 270530 x 10^6 m^2 ) x ( 3600 s )
= 2.43477 x 10^17 J

Energy of visible light = 2.43477 x 10^17 x 47/100
= 1.1443419 x 10^17 J

frequency = velocity / wavelength
Average frequency of light = 3 x 10^8 ms^-1 / 580 x 10^-9 m
= 5.1724138 x 10^14 Hz

Energy of a photon = Plank constant x frequency
= 6.625 x 10^-34 Js x 5.1724138 x 10^14 Hz
= 3.426724 x 10^-19 J

Number of photons = total energy / energy of a photon
= 1.1443419 x 10^17 J / 3.426724 x 10^-19 J
= 3.339463 x 10^-35

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