Light of intensity `10^(-5)Wm^(-2)` falls on a sodium photocell of surface area `2cm^(2)`. Assuming that the top 5 layers of sodium absorb the incident energy, estimate the time required for photoelectric emission in the wave picture of radiation. The work function of the metal is given to be about 2eV. What is the implication of your answer? effective atomic area `=10^(-20)m^(2)`.

Explanation

Intensity of incident light, `I = 10^(-5)Wm^(-2)`
Surface area of a sodium photocell, `A = 2cm^(2) = 2xx10^(-4)m^(2)`
Incident power of the light, `P = IxxA`
`= 10^(-5)xx2xx10^(4)`
`= 2xx10^(-9)W`
Work function of the metal, `phi_(0) = 2eV`
`= 2xx1.6xx10^(-19)`
`= 3.2xx10^(19)J`
Number of layers of sodium that absorbs the incident energy, `n = 5`
We know that the effective atomic acrea of a sodium atom, `A_(e)` is `10^(-20) m^(2)`
Hence, the number of conduction electrons in n layers is given as:
`n' = nxx(A)/(A_( e))`
`= 5xx(2xx10^(-4))/(10^(-20)) = 10^(17)`
The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount of energy absorbed per second per electron is:
`E = (P)/(n')`
`= (2xx10^(-9))/(10^(17)) = 2xx10^(-26) J//s`
Time required for photoelectric emission:
`t = ((phi)_(0))/(E)`
`= (3.2xx10^(-19))/(2xx10^(-36)) = 1.6xx10^(7) s~~ 0.507` years
The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.