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प्रश्न
What is the effect of threshold frequency and stopping potential on increasing the frequency of the incident beam of light? Justify your answer.
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उत्तर
Threshold frequency: The threshold frequency is the lowest frequency of incident radiation that can cause an electron to be ejected from a metal. At frequencies below the threshold, there is no photoelectric emission.
The frequency of light that will produce an emission of electrons from the metal's surface is referred to as the threshold frequency.
If ν signifies the frequency of the incident photon and νth signifies threshold frequency, then;
- If ν < νth, then this denotes that no ejection of photoelectrons will occur.
- If ν = νth, then this denotes that photoelectrons are just ejected from the surface of the metal, however, the kinetic energy of the electron is equal to zero.
The minimal negative voltage that must be provided to the anode to halt the photocurrent is known as stopping potential. When expressed in electron volts, the stopping voltage corresponds to the electrons' maximum kinetic energy.
Stopping potential, `eV_0 = hν_"incident" - phi`
Where ν is the frequency of the incident radiation and Φ is the metal surface's work function. As a result, stopping potential rises as incident radiation frequency rises.
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संबंधित प्रश्न
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
The threshold wavelength of a metal is λ0. Light of wavelength slightly less than λ0 is incident on an insulated plate made of this metal. It is found that photoelectrons are emitted for some time and after that the emission stops. Explain.
The work function of a metal is hv0. Light of frequency v falls on this metal. Photoelectric effect will take place only if
Photoelectric effect supports quantum nature of light because
(a) there is a minimum frequency below which no photoelectrons are emitted
(b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity
(c) even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately
(d) electric charge of the photoelectrons is quantised
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The electric field associated with a light wave is given by `E = E_0 sin [(1.57 xx 10^7 "m"^-1)(x - ct)]`. Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV.
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(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
Consider a thin target (10–2 cm square, 10–3 m thickness) of sodium, which produces a photocurrent of 100 µA when a light of intensity 100W/m2 (λ = 660 nm) falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of Na = 0.97 kg/m3].
Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.
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- Can you explain how photoelectric effect was observed instantaneously?
