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Intensity and Frequency of Light
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by
on 2007-06-25 08:43:51 |
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Hi, I was doing my revision and I had a question that I happen to come about which often troubles me. This is the question... What is the effect on the photocurrent and the stopping potential when... a) Intensity of incident radiation is increased. b) Frequency of incident radiation is increased. Hope that anyone can help me with this question. (Actually, many gave me all kinds of answer... it's rather confusing that way.) Thanks a lot.  |
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Re: Intensity and Frequency of Light
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on 2007-06-25 10:45:17 |
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Photocurrent is a measure of the current that flows through a photosensitive device, in this case, due to the photoelectric effect. Stopping potential is the voltage required to reduce the photocurrent to zero. a) Intensity is a measure of the number of photons received per second. The intensity and current (photocurrent) are directly proportional, because the current is proportional to the passing charge - which is to say the number of electrons, each ejected by one matching photon - over time. The voltage (stopping potential) is independent of the intensity because the voltage of each ejected electron depends only on the frequency of light (incident radiation), and the work function, and not on the intensity. b) Frequency is a measure of the energy of each photon received. The frequency and voltage (stopping potential) are directly proportional for the same reasons given in a). The current is independent of the intensity, for the same reasons given in a), assuming zero internal resistance in the device. 
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Re: Intensity and Frequency of Light
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by
on 2007-06-25 11:01:43 (edited 2007-06-25 11:05:38)
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So can I say that increase in intensity (but the frequency remains the same) will increase the value of the photocurrent and have no change in stopping voltage values. Increase in frequency (but intensity remained constant) will cause the increase in stopping potential but decrease in photcurrent values? There's a graph to this topic. From your statement... I clearly understand the effect of intensity of to the photocurrent values as the stopping voltage remained the same (because the intensity is increased but the frequency is constant). And yes the frequency part is just as it was told in the book. But... This is the strange part. Incording to the text book where it shows a graph of Photocurrent versus Stopping voltage... there's two light. Light A has a higher frequency than light B but both light A and B have the same intensity. From the graph plotted... Light A have a higher stopping voltage value than light B but it's photocurrent value is lower than light B. Why is that so? PS: Sorry to trouble you with this topic. |
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Re: Intensity and Frequency of Light
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on 2007-06-27 16:51:49 |
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At the THRESHOLD FREQUENCY, photoelectrons are JUST released from the a metal surface (due to photoelectric effect~Millikan's Experiment). The photoelectrons released have ZERO kinetic energy. [INTENSITY of Light = the number of photoelectrons emitted per second] **Photoelectric emission occurs ONLY when sufficient energy is delivered to release photoelectrons by light of a CERTAIN fixed FREQUENCY the electrons could receive (photon theory~the frequency has to MATCH the exact amount of energy the electrons can accept) **Only when you increase the INTENSITY of light ABOVE THE CRITICAL/THRESHOLD FREQUENCY that the number of photoelectrons emitted are also increased ________________________________________________________________ As for your question about the graph, Light A, with the higher frequency, has a larger Stopping potential (which is more POSITIVE than the lower stopping potential for Light B ) On your graph of photocurrent against (potential difference)voltage, the side on the right of the y-axis is positive, and towards the left, is negative *By applying a POSITIVE potential to the target metal, the electrons are being slowed down. This prevents the electrons that had been emitted with low values of kinetic energy from reaching the electrode, therefore the CURRENT through the electrometer is DECREASED. In other words, Light A has a more positive, higher Stopping potential, so the current is decreased (explanation above)...Light B has a less positive stopping potential, so that the electrons do not slow down as much as those of Light A : current is greater that that of Light A ^this is how I interpret the situation.. i hope you understand my wording..it took me a while to put it in words lolz..if u get it at all  |
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Re: Intensity and Frequency of Light
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by
on 2007-06-28 10:24:52 |
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Well, I think you didn't really interpret the graph correctly. When the frequency of that incident radiation increases, the value of the stopping potential becomes more negative. From what I can understand the graph is that the reason fo it to be negative is probably because it represent the voltage that prevents that electron to be emitted thus with enough energy to overcome that stopping potental, the electron can be emitted. Incording to the graph, the higher the frequency, the more negative is the stopping postental but decreases the photocurrent value. From frequency light A bigger than light B's frequency example, light A has a higher (more negative stopping potential) than light B and that's an understandable fact but it's photocurrent value is lover than light B which is something that I don't understand. How do I know that the more higher is the frequency the more negative is the stopping potential value (more higher in stopping potential value) because I did some past year papers of the real exams and happened to find the same similiar question. Then why did I ask this question because I don't understand that only concept in changes I frequency and I want to know it in depth. About the situation where light A and B has the same frequency but different intensity which causes only the difference in photocurrent values but not the stopping potential values. I think it's wrong of you to interpret it that way but thanks for your efforts. ^^ |