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प्रश्न
Photoelectric effect supports quantum nature of light because
(a) there is a minimum frequency below which no photoelectrons are emitted
(b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity
(c) even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately
(d) electric charge of the photoelectrons is quantised
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उत्तर
(a) there is a minimum frequency below which no photoelectrons are emitted
(b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity
(c) even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately
Photoelectric effect can be explained on the basis of quantum nature of light. According to the quantum nature of light, energy in light is not uniformly spread. It is contained in packets or quanta known as photons.
Energy of a photon, E = hv, where h is Planck's constant and v is the frequency of light.
Above a particular frequency, called threshold frequency, energy of a photon is sufficient to emit an electron from the metal surface and below which, no photoelectron is emitted, as the energy of the photon is low. Hence, option (a) supports the quantum nature of light.
Now, kinetic energy of an electron,
`K = hv_0 - varphi`
Thus, kinetic energy of a photoelectron depends only on the frequency of light (or energy). This shows that if the intensity of light is increased, it only increases the number of photons and not the energy of photons. Kinetic energy of photons can be increased by increasing the frequency of light or by increasing the energy of photon, which supports E = hv and, hence, the quantum nature of light. Hence, option (b) also supports the quantum nature of light.
Photoelectrons are emitted from a metal surface even if the metal surface is faintly illuminated; it means that less photons will interact with the electrons. However, few electrons absorb energy from the incident photons and come out from the metal. This shows the quantum nature of light. Hence, (c) also supports the quantum nature of light.
Electric charge of the photoelectrons is quantised; but this statement does not support the quantum nature of light.
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संबंधित प्रश्न
The work function for the following metals is given:
Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV
Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
Can a photon be deflected by an electric field? Or by a magnetic field?
The threshold wavelength of a metal is λ0. Light of wavelength slightly less than λ0 is incident on an insulated plate made of this metal. It is found that photoelectrons are emitted for some time and after that the emission stops. Explain.
Planck's constant has the same dimensions as
If the frequency of light in a photoelectric experiment is doubled, the stopping potential will ______.
The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. A light source is put on and a saturation photocurrent is recorded. An electric field is switched on that has a vertically downward direction.
A parallel beam of monochromatic light of wavelength 663 nm is incident on a totally reflecting plane mirror. The angle of incidence is 60° and the number of photons striking the mirror per second is 1.0 × 1019. Calculate the force exerted by the light beam on the mirror.
(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 work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is 2.5 V.
(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 electric field associated with a light wave is given by `E = E_0 sin [(1.57 xx 10^7 "m"^-1)(x - ct)]`. Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV.
(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)
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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.
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- Can you explain how photoelectric effect was observed instantaneously?
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What type of potential can the battery create between the plates?
Which plate is kept illuminated by monochromatic radiation?
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What is used to change light frequency and study its effect on emission and electron energy?
Why does greater light intensity produce more emitted electrons?
