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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].
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Solution
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
ρNa = 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.
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