# Dependence of saturation current in photoelectric tube on the time taken by the electron to reach the opposite plate?

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?

• Vacuum or gas-filled? – Jon Custer Sep 18 '15 at 17:07
• Vacuum.But I would also like to know the difference in the result if the vacuum was replaces with air. – user2215860 Sep 18 '15 at 17:16
• 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. – CuriousOne Sep 18 '15 at 17:26
• 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 – user2215860 Sep 18 '15 at 17:31
• 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. – CuriousOne Sep 18 '15 at 17:37