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प्रश्न
How would the stopping potential for a given photosensitive surface change if the intensity of incident radiation was decreased? Justify your answer.
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उत्तर
A decrease in incident radiation intensity would have no effect on the stopping potential of a certain photosensitive surface. This is because the stopping potential is exclusively determined by frequency rather than intensity of incident radiation. The kinetic energy of photoelectrons is determined only by the frequency of incident radiation, but the number of photoelectrons emitted from the surface is inversely proportional to the intensity of the received light. As a result, decreasing the intensity of the incident radiation would result in fewer photons being produced, but it would have no influence on their kinetic energy or stopping potential.
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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
Determine the value of Planck’s constant h, the threshold frequency and work function for the material.
[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?
It is found that photosynthesis starts in certain plants when exposed to sunlight, but it does not start if the plants are exposed only to infrared light. Explain.
Two photons of
A beam of white light is incident normally on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10 W of power, find the force exerted by it on the surface.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photo current. Find the threshold wavelength for 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)
In an experiment on photoelectric effect, the stopping potential is measured for monochromatic light beams corresponding to different wavelengths. The data collected are as follows:-
Wavelength (nm): 350 400 450 500 550
Stopping potential (V): 1.45 1.00 0.66 0.38 0.16
Plot the stopping potential against inverse of wavelength (1/λ) on a graph paper and find (a) Planck's constant (b) the work function of the emitter and (c) the threshold wavelength.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
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.
Which part is the photosensitive metal plate?
When are photoelectrons emitted in relation to the threshold frequency?
