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प्रश्न
Why it is the frequency and not the intensity of the light source that determines whether the emission of photoelectrons will occur or not? Explain.
उत्तर
The frequency, not the intensity, of the light source, determines the emission because if any incident light of a specific (i.e., threshold or more) frequency is incident, it results in emission regardless of its intensity.
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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?
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10−9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
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Let nr and nb be the number of photons emitted by a red bulb and a blue bulb, respectively, of equal power in a given time.
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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)
In the case of photoelectric effect experiment, explain the following facts, giving reasons.
The photoelectric current increases with increase of intensity of incident light.
If photons of ultraviolet light of energy 12 eV are incident on a metal surface of work function of 4 eV, then the stopping potential (in eV) will be :
- Assertion (A): For the radiation of a frequency greater than the threshold frequency, the photoelectric current is proportional to the intensity of the radiation.
- Reason (R): Greater the number of energy quanta available, the greater the number of electrons absorbing the energy quanta and the greater the number of electrons coming out of the metal.
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