The energy flux of sunlight reaching the surface of the earth is `1.388xx10^(3)W//m^(2)` .How many photons (nearly) per square meter are incident on the Earth per second?Assume that the photons in the sunlight have an average wavelength of 550 nm.

Explanation

Energy flux of the sun =`1.388xx10^(3)W//m^(2)`
`lambda=550 nm=55xx10^(-8)m`
`h=6.63xx10^(-34)Js`
`c=3xx10^(8)m//s`
`implies` Energy of each photon,
`E=(hc)/(lambda)`
`therefore E=(6.63xx10^(-34xx3xx10^(8)))/(55xx10^(-8))`
`thereforeE=0.3616xx10^(-18) therefore E~~3.62xx10^(-19)J`
`implies` No. Of photon incident on surface of the earth per unit area per unit time
`=("Energy flux")/("Energy of each photon")`
`(1.388xx10^(3))/(3.62xx10^(-19))`
`=0.3834xx10^(22)`
`~~3.8xx10^(21)("photon")/(m^(2)xxs)`
Second Method:
If N number of photon incident per unit area the NE=P
`N=(P)/(E)=(P)/((hc)/(lambda))=(Plambda)/(hc)`
`=(1.388xx10^(3)xx5.5xx10^(-7))/(6.63xx10^(-34xx3xx10^(8)))=3.8xx10^(21)`