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

Here, `I=10^(-5)Wm^(-2), A=2xx10^(-4)m^(2), n=5, t=?, phi_(0)=2eV=2xx1.6xx10^(-19)J`
Sodium has one conduction electron per atom and effective atomic area `A'=10^(-20)m^(2)`
No. of conduction electrons in five layers `=(5xx"area one layer")/("effective atomic area")=(5xx2xx10^(-4))/(10^(-20))=10^(17)`
Incident power, P=Intensity x area `=10^(-15)xx2xx10^(-4)=2xx10^(-9)W`
According to wave picture, the incident power is uniformaly absorbed by all the electrons continously.
Hence, energy absorbed per second per electron `E = ("Incident power")/("no. of electrons in five layers") =(2xx10^(-9))/(10^(17))=2xx10^(-26)W`
`:.` Time required for photo-electric emission will be
`t = ("energy required per electron for ejection"(phi_(0)))/("energy absorbed per sec per atom"(E)) = (2xx1.6xx10^(-19))/(2xx10^(-26))=1.6xx10^(7)s`
Since the time for photoelectric emission is very large so wave picture cannot be applied to this experiment.