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?

Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. - Physics

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?

Long Answer

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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Experimental Study of Photoelectric Effect

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Chapter 11: Dual Nature Of Radiation And Matter - Exercises [Page 74]

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NCERT Exemplar Physics [English] Class 12

Chapter 11 Dual Nature Of Radiation And Matter
Exercises | Q 11.29 | Page 74

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