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(D) Given: Energy flux $I = 1.388 	imes 10^{3} \; W/m^{2}$, Wavelength $\lambda = 550 \; nm = 550 	imes 10^{-9} \; m$.The energy of a single photon

Vedclass · Accepted Answer

$4 	imes 10^{21} \; 	ext{photons}/m^{2} \cdot s$

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(D) Given: Energy flux $I = 1.388 	imes 10^{3} \; W/m^{2}$, Wavelength $\lambda = 550 \; nm = 550 	imes 10^{-9} \; m$.The energy of a single photon is given by $E = \frac{hc}{\lambda}$.Using $h = 6.626 	imes 10^{-34} \; J \cdot s$ and $c = 3 	imes 10^{8} \; m/s$:$E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^{8}}{550 	imes 10^{-9}} \approx 3.614 	imes 10^{-19} \; J$.The number of photons $n$ incident per unit area per second is given by $n = \frac{I}{E}$.$n = \frac{1.388 	imes 10^{3}}{3.614 	imes 10^{-19}} \approx 3.84 	imes 10^{21} \; 	ext{photons}/m^{2} \cdot s$.Rounding to the nearest value, we get $n \approx 4 	imes 10^{21} \; 	ext{photons}/m^{2} \cdot s$.

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

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