The energy flux of sunlight reaching the surface of Earth is `1.388 xx10^(3)Wm^(-2)` . How many photons per square meter are incident on Earth per second? Assume the wavelength of photon at an average of 550 nm.

Options

A. `2xx10^(10)`
B. `5xx10^(7)`
C. `2.25xx10^(17)`
D. `3.85xx10^(21)`

Answer

D

Explanation

Given `lambda=550 nm=550 xx10^(-9)m`
Energy flux `phi=1.388xx10^(3)Wm^(-2)`
Also, `h=6.626xx10^(-34)Js`
and `c=3xx10^(8) ms^(-1)`
`:.` energy of each photon `E=(hc)/(lambda)`
`=(6.626xx10^(-34)xx3xx10^(8))/(550xx10^(-9))`
`=6.31xx10^(-19)J`
`:.` no. of photons per square metere of earth
`=(phi)/(E)=(1.388xx10^(3))/(3.61xx10^(-19))=3.85xx10^(21)`