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प्रश्न
On the basis of the graphs shown in the figure, answer the following questions :
(a) Which physical parameter is kept constant for the three curves?
(b) Which is the highest frequency among v1, v2, and v3?
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उत्तर
(a) Intensity is kept constant for all three curves. The given graph shows the variation of photocurrent with collector plate potential for the same intensity of light radiation at various frequencies.
(b) The stopping potential is found to be changing linearly with the frequency of incident light. The stopping potential is more negative for higher frequencies of incident radiation. As an increase in the frequency of the incident light increases the kinetic energy of the emitted electrons, so the greater retarding potential is required to stop them completely.
Hence higher the negative potential, the more the frequency of incident electrons.
As V1 is more negative than V2 and V2 is more negative than V3, hence, V1 > V2 > V3
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संबंधित प्रश्न
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
λ1 = 3650 Å, λ2 = 4047 Å, λ3 = 4358 Å, λ4 = 5461 Å, λ5 = 6907 Å,
The stopping voltages, respectively, were measured to be:
V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 V
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[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10−19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Light of intensity 10−5 W m−2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
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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)
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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)
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- Estimate no. of photons emitted by the bulb per second. [Assume no other losses]
- Will there be photoelectric emission?
- How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
- How many photons would atomic disk receive within time duration calculated in (iii) above?
- Can you explain how photoelectric effect was observed instantaneously?
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 :
How would the stopping potential for a given photosensitive surface change if the intensity of incident radiation was decreased? Justify your answer.
The figure shows a plot of stopping potential (V0) versus `1/lambda`, where λ is the wavelength of the radiation causing photoelectric emission from a surface. The slope of the line is equal to ______.
