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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
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
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)
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 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)
The minimum energy required to remove an electron is called ______.
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.
A student performs an experiment on photoelectric effect, using two materials A and B. A plot of Vstop vs ν is given in Figure.
- Which material A or B has a higher work function?
- Given the electric charge of an electron = 1.6 × 10–19 C, find the value of h obtained from the experiment for both A and B.
Comment on whether it is consistent with Einstein’s theory:
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 ______.
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:
