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

Give , intensity , `I=10^(-5)W//m^2`
Area , A = 2 `cm^2`
Work function , `W_0=2eV`
Sodium has one conduction electron per atom
`implies` Effective atomic area `~10^(20)m^2` .
Number of atom per layer `=("Area of one layer ")/("Effective atomic area")`
`=2xx10^(-4)/(10^(-20))`
Number of electrons in 5 layers
`n=(5xx2xx10^(-4)m^2)/(10^(-20)m^2)=10^(17)`
Incident power , `P = I xxA = 10 ^(-5) W//m^2xx2xx10^(-4)m^2` .
`=2xx10^(-9)W`
In the wave picture , incident power is uniformly
absorbed by all the electrons continuously.
Consequently ,
Energy absorbed by an electron per second
Energy required by one electron to come out = Work function `(W_0)`
= 2eV
`=2xx1.6xx10^(-19)J`
Time required for photoelectron emission
`=W_0/("Energy absorbed per second per electron")`
`=(2xx1.6xx10^(-19))/(2xx10^(-26))`
`=1.6xx10^(7)` s = about 0.5 year
it can be implicated that experimentally photoelectron emission is observed nearly instantaneously (nearly about a nanosecond ). Thus the wave picture is in gross disagreement with experiment . While in the photon picture , energy of the radiation is not continuously shared by all the electrons in the top layers , it comes in discontinuous quanta. A photon is either absorbed or not absorbed by an electron nearly instantly.