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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
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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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(b) photoelectric emission may or may not take place
(c) the stopping potential will increase
(d) the stopping potential will decrease
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(a) Which physical parameter is kept constant for the three curves?
(b) Which is the highest frequency among v1, v2, and v3?
The graph shows the variation of photocurrent for a photosensitive metal
- What does X and A on the horizontal axis represent?
- Draw this graph for three different values of frequencies of incident radiation ʋ1, ʋ2 and ʋ3 (ʋ3 > ʋ2 > ʋ1) for the same intensity.
- Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I3 > I2 > I1) having the same frequency.
