A small `10 W` source of wavelength `99 nm` is held at a distance `0.1 m` from a metal surface. The radius of an atom of the metal is approximately `0.05 nm`. Find (i) the average number of photons striking an atom per second. (ii) the number of photoelectrons emitted per unit area per second if the efficiency of liberation of photoelectrons is `1%`

Answer

`(5)/(16),(10^(20))/(80pi)`

Explanation

Energy incident/area of sphere /sec`=10/(4pi(0.1)^(2))` Energy incident on atom
`=10/(4pi (0.1)^(2))xxpixx(0.05xx10^(-9))^(2)=6.25xx10^(-19) J`
Energy of one photon `=(hc)/lambda=12.525 eV`
No. of photon striking atom/sec
`=(6.25xx10^(-19))=0.3125`
No. of photons /area.
`=10/(4pi (0.1)^(2))xx(1)/(12.525xx1.6xx10^(-19))=(10^(20)xx10xx10)/(80pi)`
since `n=1% n=10^(20)/(80 pi)`