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प्रश्न
How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s photoelectric equation?
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उत्तर १
Einstein’s Photoelectric equation is
KEmax = hv - hv0
eV0 = h(v - v0)
From above equation
Case - I:
If v > v0 [Emission with K.E.]
If v = v0 [Just emission]
If v < v0 [No emission]
उत्तर २
Einstein's photoelectric equation is, K. E = hv - Φ or hv = Φ + K. E.
This equation explains that when a photon of certain energy is incident on a photosensitive surface, a particular amount of energy gets used as a work function to eject electrons from their shells and the rest of the energy is acquired by ejected electrons as their Kinetic energy. The emission of electrons from a photosensitive surface takes place only if the incident energy of photons is greater than the work function.
संबंधित प्रश्न
point out any two characteristic properties of photons on which Einstein’s photoelectric equation is based ?
Define the terms (i) ‘cut-off voltage’ and (ii) ‘threshold frequency’ in relation to the phenomenon of photoelectric effect.
Using Einstein’s photoelectric equation shows how the cut-off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable plot/graph.
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.
A non-monochromatic light is used in an experiment on photoelectric effect. The stopping potential
The minimum energy required to remove an electron is called ______.
- 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?
There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength.
Radiation of frequency 1015 Hz is incident on three photosensitive surfaces A, B and C. Following observations are recorded:
Surface A: no photoemission occurs
Surface B: photoemission occurs but the photoelectrons have zero kinetic energy.
Surface C: photo emission occurs and photoelectrons have some kinetic energy.
Using Einstein’s photo-electric equation, explain the three observations.
The photon emitted during the de-excitation from the first excited level to the ground state of a hydrogen atom is used to irradiate a photocathode in which the stopping potential is 5 V. Calculate the work function of the cathode used.
If c is the velocity of light in free space, the correct statements about photon among the following are:
- The energy of a photon is E = hv.
- The velocity of a photon is c.
- The momentum of a photon, ρ = `(h v)/c`.
- In a photon-electron collision, both total energy and total momentum are conserved.
- Photon possesses positive charge.
Choose the correct answer from the options given below:
