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प्रश्न
Plot a graph showing the variation of photoelectric current with collector plate potential at a given frequency but for two different intensities I1 and I2, where I2 > I1.
उत्तर
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संबंधित प्रश्न
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 109 eV)
Briefly explain the three observed features which can be explained by Einstein’s photoelectric equation.
The frequency and intensity of a light source are doubled. Consider the following statements.
(A) The saturation photocurrent remains almost the same.
(B) The maximum kinetic energy of the photoelectrons is doubled.
The electric field at a point associated with a light wave is `E = (100 "Vm"^-1) sin [(3.0 xx 10^15 "s"^-1)t] sin [(6.0 xx 10^15 "s"^-1)t]`.If this light falls on a metal surface with a work function of 2.0 eV, what will be the maximum kinetic energy of the photoelectrons?
(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 monochromatic light source of intensity 5 mW emits 8 × 1015 photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of 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)
In a photoelectric experiment, the collector plate is at 2.0 V with respect to the emitter plate made of copper (φ = 4.5 eV). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm. Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector.
Consider the situation of the previous problem. Consider the faster electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s photoelectric equation?
- In the explanation of photo electric effect, we assume one photon of frequency ν collides with an electron and transfers its energy. This leads to the equation for the maximum energy Emax of the emitted electron as Emax = hν – φ0 where φ0 is the work function of the metal. If an electron absorbs 2 photons (each of frequency ν) what will be the maximum energy for the emitted electron?
- Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?
A photon of wavelength 663 nm is incident on a metal surface. The work function of the metal is 1.50 eV. The maximum kinetic energy of the emitted photoelectrons is ______.
