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प्रश्न
It is observed in an experiment on the photoelectric effect that an increase in the intensity of the incident radiation does not change the maximum kinetic energy of the electrons. Where does the extra energy of the incident radiation go? Is it lost? State your answer with explanatory reasoning.
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उत्तर
Electrons are emitted when electromagnetic radiation with a frequency greater than the threshold frequency is incident on a metal surface. It has been noticed that not every incident photon is capable of realising an electron. In fact, the number of electrons emitted per second is far less than the number of photons incident per second. Photons that are ineffective in the liberation of electrons are reflected (or scattered) or absorbed, causing the temperature of the metal surface to rise. The maximum kinetic energy of a photoelectron is determined by the frequency of incident radiation and the metal's threshold frequency. It has nothing to do with the intensity of the incident radiation. As the intensity increases, so does the number of electrons emitted per second.
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संबंधित प्रश्न
Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.
The threshold wavelength of tungsten is 2.76 x 10-5 cm.
(a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 x 10-5 cm.
(b) What will be the maximum kinetic energy of electrons ejected in each of the following cases
(i) if ultraviolet radiation of wavelength λ = 1.80 × 10-5 cm and
(ii) radiation of frequency 4 x 1015 Hz is made incident on the tungsten surface?
As the intensity of incident light increases ______
Find the energy of photon which have momentum 2 × 10-16 gm-cm/sec.
Draw a neat labelled diagram of a schematic of the experimental setup for the photoelectric effect.
With the help of a circuit diagram describing an experiment to study the photoelectric effect.
The energy of a photon is 2 eV. Find its frequency and wavelength.
The following graph shows the stopping potential V0 versus frequency v for photoelectric emission from two metals A and B. The slope of each of the lines gives ______
The graph of stopping potential `"V"_"s"` against frequency v of incident radiation is plotted for two different metals P and Q as shown in the graph. ΦP and ΦQ are work-functions of P and Q respectively, then
When a light of wavelength 4000 Å falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, light of wavelength 6000 Å is sufficient for photo emission. The work functions of the two emitters are in the ratio of ____________.
The work function of a metal is 1.6 x 10-19 J. When the metal surface is illuminated by the light of wavelength 6400 Å, then the maximum kinetic energy of emitted photo-electrons will be (Planck's constant h = 6.4 x 10-34 Js) ____________.
The work function of a metallic surface is 5.01 eV. The photoelectrons are emitted when light of wavelength 2000 Å falls on it. The potential difference applied to stop the fastest photoelectrons is [h = 4.14 x 10-15 eV sec] ____________.
The threshold frequency for a certain photosensitive metal is v0. When it is illuminated by light of frequency v = 2v0, the maximum velocity of photoelectrons is v0. What will be the maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency
v = 5v0?
A metal surface is illuminated by photons of energy 5 eV and 2.5 eV respectively. The ratio of their wavelengths is ____________.
When wavelength of incident radiation on the metal surface is reduced from 'λ1' to 'λ2', the kinetic energy of emitted photoelectrons is tripled. The work function of the metal is ______.
(h = Planck's constant, c =velocity of light)
Light of different frequencies, whose photons have energies 3 eV and 18 eV respectively, successively illuminate a metal of work function 2 eV. The ratio of the maximum speeds of the emitted electrons will be ______.
Which one of the following statements ts INCORRECT for stopping potential in photoelectric emission?
When a metal with work function 0.6 eV is illuminated with light of energy 2 eV, the stopping potential will be ____________.
In experiment of photoelectric effect, the stopping potential for incident yellow light of wavelength 5890 Å is 4 volt. If the yellow light is replaced by blue light of wavelength 4000 Å, the stopping potential is ____________.
The radiation corresponding to the 3 → 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5 × 10-4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, and the work function of the metal is ______.
(Mass of electron = 9.1 × 10-31 kg)
In a photoelectric experiment, ultraviolet light of wavelength 280 nm is used with a lithium cathode having work function Φ = 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential.
(h = 6.63 × 10-34 Js, c = 3 × 108 ms-1)
When radiation of wavelength λ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3λ, the stopping potential is `"V"/4`. If the threshold wavelength for the metallic surface is nλ. then value of n will be ______.
When ultraviolet light of wavelength 100 nm is incident upon a sample of silver metal, a potential difference of 7.7 volt is required to stop the photoelectrons from reaching the collector plate. The potential required to stop photo electrons when light of wavelength 200 nm is incident upon silver is ______.
The maximum kinetic energy of the photoelectrons ejected will be ______ eV when the light of wavelength 350 nm is incident on a cesium surface. The work function of cesium = 1.9 eV.
If the electron in hydrogen atom jumps from second Bohr orbit to ground state and difference between energies of the two states is radiated in the form of photons. If the work function of the material is 4.2 eV, then stopping potential is ______.
[Energy of electron in nth orbit = `-13.6/"n"^2` eV ]
