Question

Light of intensity 10⁻⁵ W m⁻² falls on a sodium photo-cell of surface area 2 cm². 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?

Answer 1

Intensity of incident light, I = 10⁻⁵ W m⁻²

Surface area of a sodium photocell, A = 2 cm² = 2 × 10⁻⁴ m²

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 m²_.

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

`"n'" = "n" xx "A"/"A"_"e"`

= `5 xx (2xx10^(-4))/10^(-20)`

= 10¹⁷

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 × 10⁻²⁶ J/s

Time required for photoelectric emission:

`"t" = phi_0/"E"`

= `(3.2 xx 10^(-19))/(2xx10^(-26))`

= 1.6 × 10⁷ 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.