#### Question

Light of intensity 10^{−5} W m^{−2} falls on a sodium photo-cell of surface area 2 cm^{2}. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?

#### Solution

Intensity of incident light, *I* = 10^{−5} W m^{−2}

Surface area of a sodium photocell, *A* = 2 cm^{2} = 2 × 10^{−4} m^{2}

Incident power of the light, *P = I × A*

= 10^{−5} × 2 × 10^{−4}

= 2 × 10^{−9} W

Work function of the metal,`phi_0` = 2 eV

= 2 × 1.6 × 10^{−19}

= 3.2 × 10^{−19} J

Number of layers of sodium that absorbs the incident energy, *n* = 5

We know that the effective atomic area of a sodium atom, *A*_{e} is 10^{−20} m^{2}_{.}

Hence, the number of conduction electrons in *n* layers is given as:

`n' = n xx A/A_e`

`= 5 xx (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 =2 xx 10^(-26) "J/s"`

Time required for photoelectric emission:

`t = phi_0/E`

`= (3.2 xx 10^(-19))/(2xx10^(-26)) = 1.6 xx 10^7 s ~~ 0.507 "year"`

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.