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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
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A student performs an experiment on photoelectric effect, using two materials A and B. A plot of Vstop vs ν is given in Figure.
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Comment on whether it is consistent with Einstein’s theory:
Radiation of frequency 1015 Hz is incident on three photosensitive surfaces A, B and C. Following observations are recorded:
Surface A: no photoemission occurs
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Using Einstein’s photo-electric equation, explain the three observations.
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 ______.
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