If the output power of the light source is P = 0.1 Watt, the intensity of the light at the location of the potassium foil (d = 0.5 meters away) will be

`I = P/(4pid^2) = 0.1/(4pi(0.5)^2) = 0.032 W/m^2`

This means the power (or the rate of energy) at which the circular area of foil with the given radius `r = 1.3*10^(-10)` m will absorb the light will be

`P_a = I*(pir^2) = 0.032 * pi(1.3*10^(-10))^2 = 1.69*10^(-21)` W. (We can neglect the fact that the points on this piece of foil are possibly at different distances from the light source because the radius is so much smaller than the distance from the light source.)

So the energy rate at which the given area absorbs the light is

`1.69 * 10^(-21) W` . Assuming the rate is uniform, the energy absorbed during time t will be

`E = P_a*t ` . The energy required to eject and electron is

`E = 1.8 eV = 1.8 * 1.6*10^(-19) = 2.88*10^(-19) J` .

Thus, the time it will take to absorb this energy is

`t = E/(P_a) = (2.88*10^(-19))/(1.69*10^(-21)) = 170.4 s`

**It would take the foil 170.4 seconds to saturate enough energy to absorb an electron**.

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