Question

Consider a thin target (10^–2 cm square, 10^–3 m thickness) of sodium, which produces a photocurrent of 100 µA when a light of intensity 100W/m² (λ = 660 nm) falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of Na = 0.97 kg/m³].

Answer 1

According to the problem, the area of the target A = 10^–2 cm² = 10^–4 m²

And thickness, d = 10^–3 m

Photocurrent, i = 100 × 10^–6 A= 10^–4 A

Intensity, I = 100 W/m²

⇒ λ = 660 nm = 660 × 10^–9 m

ρ_Na = 0.97 kg/m³

Avogadro number = 6 × 10²⁶ kg atom

Volume of sodium target = A × d

= 10^–4 × 10^–3

= 10^–7 m³

We know that 6 × 10²⁶ atoms of sodium weigh = 23 kg

Density of sodium = 0.97 kg/m³

Hence the volume of 6 × 10²⁶ sodium atoms = `23/0.97` m³

Volume occupied by one sodium atom = `23/(0.97 xx (6 xx 10^36))` = 3.95 × 10^–26 m³

Number of sodium atoms in target `(N_"sodium") =  10^-7/(3.95 xx 10^-26)` = 2.53 × 10¹⁸ 

Let m be the number of photons falling per second on the target.

Energy of each photon = `(hc)/A`

Total energy falling per second on target  = `(nhc)/λ = IA`

∴ `n = (IAλ)/(hc)`

= `(100 xx 10^-4 xx (660 xx 10^-9))/((6.62 xx 10^-34) xx (3 xx 10^8))` = 3.3 × 10¹⁶

Let P be the probability of emission per atom per photon. The number of photoelectrons emitted per second

`N = P xx n xx (N_"sodium")`

= `P xx (33 xx 10^16) xx (2.53 xx 10^18)`

Now, according to the question,

i = 100 µA = 100 × 10^–6 = 10^–4 A

Current, i = Ne

∴ `10^-4 xx P xx (3.3 xx 10^16) xx (2.53 xx 10^18) xx (1.6 xx 10^-19)`

⇒ `P = 10^-4/((3.3 xx 10^16) xx (2.53 xx 10^18) xx (1.6 xx 10^-19))`

= 7.48 × 10^–21

Then, the probability of photoemission by a single photon on a single atom is very much less than 1. Because the absorption of two photons by an atom is negligible.