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Q)

The energy flux of sunlight reaching the surface of the earth is $ 1.388 \times 10^3 \: W/m^2$. How many photons per square meter are incident on earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm.

$\begin {array} {1 1} (a)\;3.84 \times 10^{20}photons/m^2/s & \quad (b)\;38.4 \times 10^{21}photons/m^2/s \\ (c)\;0.384 \times 10^{20}photons/m^2/s & \quad (d)\;3.84 \times 10^{21}photons/m^2/s \end {array}$

 

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A)
Energy flux of sunlight, $ \phi =1.388 \times 10^3 W/m^2$
Hence, power of sunlight, $ P/m^2 = 1.388 \times 10^3W$
Average wavelength of photons in sunlight, $ \lambda = 550 nm = 550 \times 10^{-9}m$
No. of photons $ / m^2/s$ incident on earth = $n$
$P=nE$
$ \Rightarrow n=P/E=P\lambda / hc$
$ = ( 1.388 \times 10^3 \times 550 \times 10^{-9} ) / (6.626 \times 10^{-34} \times 3 \times 10^8)$
$ = 3.84 \times 10^{21} photons / m^2 / s$
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