Advertisements
Advertisements
प्रश्न
Answer the following question.
Why is the wave theory of electromagnetic radiation not able to explain the photoelectric effect? How does a photon picture resolve this problem?
Advertisements
उत्तर
There are three main drawbacks:
- Intensity: If we consider light as a wave then as the intensity of the light is increased the amplitude of the oscillation of the electron will increase. Thus, as the intensity of the incident light is increased the maximum kinetic energy of the emitted electron will also increase. But it was observed that the kinetic energy of the emitted electrons does not depend on the intensity whereas the magnitude of the photoelectric current increases with the frequency.
- Frequency: If we consider the light as a wave then the photoelectric emission should happen on any frequency, but it was observed that the electrons are emitted after a particular frequency. If the frequency of the incident light is lesser than this frequency there is no photoelectric emission observed.
- Time Delay: According to the wave theory the energy is uniformly distributed over the wavefront. As the light falls on the metallic surface, it will take some time for the electron to gain sufficient energy to get emitted. But experimentally it was observed that the electrons are emitted instantaneously as the light falls on the metallic surface.
How the photon theory can explain the photoelectric effect:
- According to photon theory increasing the intensity means increasing the number of photons that do not change the maximum kinetic energy but changes the number of ejected electrons.
- The energy of a photon is given as E = hf that explains the dependence of the energy on the frequency, after a particular frequency of a photon that is threshold frequency there is photoelectric emission.
- As soon as a photon falls on the metallic surface it is absorbed, hence the electron is ejected instantaneously. Hence, all these are in accordance with the experimental observations.
संबंधित प्रश्न
Define the term "cut off frequency" in photoelectric emission. The threshod frequency of a metal is f. When the light of frequency 2f is incident on the metal plate, the maximum velocity of photo-electrons is v1. When the frequency of the incident radiation is increased to 5f, the maximum velocity of phto-electrons is v2. Find the ratio v1 : v2.
Two neutral particles are kept 1 m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
In an experiment on photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W. The potential of the collector plate is made sufficiently positive with respect to the emitter, so that the current reaches its saturation value. Assuming that on average, one out of every 106 photons is able to eject a photoelectron, find the photocurrent in the circuit.
A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. (a) A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal. (b) Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal.
In the arrangement shown in the figure, y = 1.0 mm, d = 0.24 mm and D = 1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
In an experiment on photoelectric effect, the emitter and the collector plates are placed at a separation of 10 cm and are connected through an ammeter without any cell. A magnetic field B exists parallel to the plates. The work function of the emitter is 2.39 eV and the light incident on it has wavelengths between 400 nm and 600 nm. Find the minimum value of B for which the current registered by the ammeter is zero. Neglect any effect of space charge.
Plot a graph to show the variation of stopping potential with frequency of incident radiation in relation to photoelectric effect.
Work function of aluminium is 4.2 eV. If two photons each of energy 2.5 eV are incident on its surface, will the emission of electrons take place? Justify your answer.
In Photoelectric effect ______.
For a given frequency of light and a positive plate potential in the set up below, If the intensity of light is increased then ______.
