Saved Bookmarks
| 1. |
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. |
|
Answer» Solution :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)` |
|