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The kinetic energy of an electron in a photoelectric tube increases with increase in the applied voltage across the plates of the tube, thus the velocity of the electrons also increases. Accordingly the time taken by the electron to reach the opposite plate should be less and so the current in the external circuit should increase as $i=\frac{dp}{dt}$ and so current is inversely proportional to time.But this is not the case as the value of saturation current remains the same.Why is the value not depending on the time taken by the electron to reach the opposite plate?

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  • $\begingroup$ Vacuum or gas-filled? $\endgroup$ – Jon Custer Sep 18 '15 at 17:07
  • $\begingroup$ Vacuum.But I would also like to know the difference in the result if the vacuum was replaces with air. $\endgroup$ – user2215860 Sep 18 '15 at 17:16
  • $\begingroup$ Beyond saturation the tube generates a constant amount of charge per photon. Why would the total current increase if the charge doesn't? At most the waveform generated gets a little faster, which makes the peak current increase. Are you measuring the time resolved pulses or just the average current? The average can't change because the charge is always the same. $\endgroup$ – CuriousOne Sep 18 '15 at 17:26
  • $\begingroup$ I am measuring the average current.Although the number of charge does not increase but the speed of the electron increases so the rate of flow of electrons should also increase and as current is the rate if flow of charge so the current should increase $\endgroup$ – user2215860 Sep 18 '15 at 17:31
  • $\begingroup$ Yep, that increases the peak current, but not the average. You are not measuring the peak and the average stays the same. Get yourself a really fast oscilloscope and look at the individual pulses. $\endgroup$ – CuriousOne Sep 18 '15 at 17:37
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The current in any circuit is given by i= dq/dt. q being the charge of the carriers not dp/dt. Increasing the voltage will only increase the kinetic energy of the electrons coming out but not the current because the number of electrons available is the same(unless you are varying the intensity of the light which is creating the effect).

But if their is air, it should resist the electron flow a little. Hence you might see a voltage dependence. But that possibly wouldn't be ohmic.

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  • $\begingroup$ Current depends on charge as well as time.Although the charge remains the same but the time taken by the electrons to flow decreases so the rate of flow of electrons should increase thereby increasing the current. $\endgroup$ – user2215860 Sep 18 '15 at 17:37
  • $\begingroup$ Please tell me where i am getting it wrong. $\endgroup$ – user2215860 Sep 18 '15 at 17:37
  • $\begingroup$ Yes it depends on time. But please understand that the travelling time is not important. The question is "How much charge is reaching anode per unit TIME?" Got it? $\endgroup$ – Ari Sep 18 '15 at 17:39
  • $\begingroup$ Lets take an example suppose 10 electrons are emmited from emitter plate and they reach the anode in 5 secs so the current will be 10/5=2 and let another setup emit 10 electrons which take 10 secs to reach the anode, the average current will be then 10/10=1.So how is it the same? $\endgroup$ – user2215860 Sep 18 '15 at 17:43
  • $\begingroup$ No. If 10 electrons are emitted and they reach the anode at the same time the current would be a sharp spike and then zero. It has nothing to do with time of flight. it's same in both cases. $\endgroup$ – Ari Sep 18 '15 at 17:52
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The saturation current is the same because the RATE of charge flow per unit time is the same. Shorter time, smaller charge flow. The charge is not fixed. The amount of photoelectrons emitted is maxed. So all that are emitted are being collected

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