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प्रश्न
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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उत्तर
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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(a) Identify the variable X on the horizontal axis.
(b) What does the point A on the horizontal axis represent?
(c) Draw this graph for three different values of frequencies of incident radiation v1, v2 and v3 (v1 > v2 > v3) for same intensity.
(d) Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I1 > I2 > I3) having same frequency.
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Photoelectric effect supports quantum nature of light because
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(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
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When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photo current. Find the threshold wavelength for the metal.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
A small piece of cesium metal (φ = 1.9 eV) is kept at a distance of 20 cm from a large metal plate with a charge density of 1.0 × 10−9 C m−2 on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
The figure is the plot of stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find (a) the ratio h/e and (b) the work function.
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On the basis of the graphs shown in the figure, answer the following questions :
(a) Which physical parameter is kept constant for the three curves?
(b) Which is the highest frequency among v1, v2, and v3?
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Show the graphical variation in the above two cases.
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