`11xx10^(11)` photons are incident on a surface in 100s.These photons correspond to a wavelength of 10Å .If the surface area of the given surface is `0.01 m^(2)` ,the intensity of given radiations is ………(velocity of light is `3xx10^(8)ms^(-1),h=6.625x10^(-34)JS)`

Explanation

Number of photons incident in 10s=`11xx10^(11)`
`therefore` Number of photons incident in 1 s`=11xx10^(10)`
Now ,these photons being incident on area `0.01m^(2)` Number of photons being incident on `1m^(2)` in 1 s,
`n=(11xx10^(10))/(0.01)`
`=(11xx10^(10))/(10^(-2))=11xx10^(12)`
Energy associated with n photons,`=(11xx10^(10)xx6.6xx10^(-34)xx3xx10^(8))/(10xx10^(-10))` `therefore` Intensity of incident radiation
`=2.18xx10^(-3)Wm^(-2)`