A small potassium foil is placed (perpendicular to the direction of incidence of light) a distance `r( =0.5 m)` from a point light source whose output power `P_(0)` is `1.0W`. Assuming wave nature of light how long would it take for the foil to soak up enough energy `(=1.8eV)` from the beam to eject an electron? Assume that the ejected photoelectron collected its energy from a circular area of the foil whose radius eqals the radius of a potassium atom `(1.3xx10^(-10)m)`

Explanation

`I=(P_(0))/(4pir^(2))=(1.0W)/(4pi(0.5m)^(2))=0.32 W//m^(2)`
The target area `A` is `pi(1.3xx10^(-10)m)^(2)` or `5.3xx10^(-20)m^(2)`, so that the rate at which energy falls on the target is given by
`P=IA=(0.32W//m^(2))(5.3xx10^(-20)m^(2))`
`=1.7xx10^(-20)J//s`
If all this incoming energy is abosorbed, the time required to accumulate enough energy for the electron to escape is
`t=((1.8eV)/(1.7xx10^(-20)J//s))((1.6xx10^(-19))/(1eV))=17s`
Our selection of a radius for the effective target area was some-what arbitary, but no matter what reasonable area we choose, we should still calculate a ''soak-up time'' within the range of easy measurement. However, no time delay has ever been observed under any circumstances, the early experiments setting an upper limit of about `10^(-9)S` for such delays.