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Question
Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.
- Estimate no. of photons emitted by the bulb per second. [Assume no other losses]
- Will there be photoelectric emission?
- How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
- How many photons would atomic disk receive within time duration calculated in (iii) above?
- Can you explain how photoelectric effect was observed instantaneously?
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Solution
According to the problem, P = 20 W, λ = 5000 Å = 5000 × 10–10 m, distance (d) = 2 m, work function `phi_0` = 2 eV, radius r = 1.5 Å = 1.5 × 10–10 m
Now, Number of photon emitted by bulb per second, n' = `(dN)/(dt)`
i. Number of photon emitted by bulb per second is n’ = `(P)/((hc)/λ) = (Pλ)/(hc)`
= `(20 xx (5000 xx 10^-10))/((6.62 xx 10^-34) xx (3 xx 10^8))`
⇒ n' = 5 × 1019/sec
ii. Energy of the incident photon = `(hc)/λ`
= `((6.62 xx 10^-34) xx (3 xx 10^8))/(5000 xx 10^-10 xx 1.6 xx 10^-19)`
= 2.48 ev
As this energy is greater than 2 eV (i.e., a work function of the metal surface), hence photoelectric emission takes place.
iii. Let Δt be the time spent in getting the energy `phi` = (work function of metal).
Consider the figure, if P is the power of source then energy received by the atomic disc
`p/(4πd^2) xx pir^2Δt = phi_0`
⇒ Δt = `(4phi_0d^2)/(Pr^2)`
= `(4 xx (2 xx 1.6 xx 10^-19) xx 2^2)/(20 xx (1.5 xx 10^-10)^2`
= 2.84 s
iv. Number of photons received by the atomic disc in time Δt is
N = `(n^' xx pir^2)/(4pid^2) xx Δt`
= `(n^'r^2Δt)/(4d^2)`
= `((5 xx 10^19) xx (1.5 xx 10^-10)^2 xx 28.4)/(4 xx (2)^2`
= 2
Now let us discuss the last part in detail. As the time of emission of electrons is 11.04 s.
v. In photoelectric emission, there is a collision between the incident photon and free electron of the metal surface, which lasts for a very short interval of time (≈ 10–9 s), hence we say photoelectric emission is instantaneous.
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