Assuming the sun is directly overhead, the surface of the earth receives sunlight at 1.4 × 10^3 W m^ -2 . Treat the light as monochromatic with an average wavelength of 500 nm , and assume no light is absorbed between

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Correct answer: B. 9.9 × 10^ 44

Intensity of sunlight at the earth's surface,

I = 1.4 \times 10^3 \text{ W m}^{-2}

Distance from the sun to the earth,

R = 1.5 \times 10^{11} \text{ m}

Total power emitted by the sun,

P = I \times 4 \pi R^2

P = 1.4 \times 10^3 \times 4 \times 3.14 \times (1.5 \times 10^{11})^2 = 3.96 \times 10^{26} \text{ W}

Energy of a single photon,

E = \dfrac{hc}{\lambda}

E = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 3.98 \times 10^{-19} \text{ J}

Number of photons emitted per second,

N = \dfrac{P}{E}

N = \dfrac{3.96 \times 10^{26}}{3.98 \times 10^{-19}} \approx 9.9 \times 10^{44}

Assuming the sun is directly overhead, the surface of the earth...

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