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

Given , I = ` 10^(5) W//m^(2) , A = 2 cm ^(2) = 2 xx 10^(-4) " m"^(2) , phi _(0) = 2eV `
Let t be the time
The effective atomic area of Na = `0^(-20) m^(2)` nand it contains conduction electron per atom .
No. of conduction electrons in five layers `= (5 xx "Area of one laye")/("Effective atomic area ") = (5xx 2 xx 10^(-4))/(10^(-20) = 10^(17)`
We know that sodium has one free electron (or conduction electron ) per atom .
Incident power on the surface area of photocell
= Incident intensity `xx` Area on the surface area of photo cell
` 10^(-5) xx 2xx 10^(-4) = 2 xx 10^(-9) ` W
The electron present in all the 5 layers of sodium will share the incident energy equally
Energy absorbed per second per electron , E = ` ("Incident power ")/(" No. of electrons in 5 layers ")`
` = ( 2 xx 10^(-9))/(10^(17)) = 2 xx 10^(-26)` W
Time required for mission by each electron ,
`T = ("Energy required per electron ")/("Energy absorbed per second ") = (2 xx 1.6 xx 10^(-19))/(2 xx10^(-26)) = 1.6 xx10^(7) ` s , which is about 0.5 yr.
The answer obtained implies that the tme of emission of electron is very large and is not agreement with the observed time of emission of photoelectron .
Thus, it is implied that the wave theory cannot be applied in this experiment .