I am studying physics and I am reading a guide to the franck-hertz-experiment and I am a little bit confused about the +/- notation of the potential $U_A$ and $U_B$.

The experiment is divided in two parts. In the first one we want to measure the current $I_A$ (electrons, who arrive at the anode). These electrons are accerlated by $U_B$.

First question: To accelerate the electrons, $U_B$ has to be positive ($U_B > 0$) (corresponding to the picture which I've added), right?

To break the electrons, $U_A$ is also $>0$ because the +/- in the picture is switched. Correct?

Now where the confusion began:

To measure the ionisation energy of Hg the manual want me to put a constant negative voltage on $U_A$ (so $U_A < 0$) to keep the electrons away from the anode and provide that all possibly created ions arrive at the anode. But using a negative voltage $U_A < 0$ would mean that the electric field lines would show in the different directions, right?

With $U_B > 0$ and $U_A < 0$ there would be just a complete acceleration in y-direction, wouldn't it?

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I couldn't find any answer to solve my confusion. I hope someone can help me out. Maybe its a notation error in the manual or I didn't understand the experiment.

What I thought: To measure the ionisation-energy, we increase $U_B$ until we measure a current $I_A$ (caused by positive ions)...


There should be no confusion. The circuit diagram is correct. The grid has a positive voltage with respect to the cathode accelerating the electrons. The anode, where the small anode current is measured, is (weakly) negatively biased with respect to the grid so that only electrons traversing the grid with a certain kinetic threshold energy can reach the anode and contribute to the anode current. When the ionization energy is reached by electrons in the space between the cathode and the grid ionization of Hg atoms occurs and the electrons lose kinetic energy so that they cannot reach the anode and the anode current falls. Positive ion currents don't play a significant role in the explanation of the Franck-Hertz experiment. The electron current reaching the anode is very small as compared to the total current at the cathode.

  • $\begingroup$ Thank you for answer. Lets say we do $U_B = +10V$. The electrons will be accelerated from the kathode to the grid. When we put $U_A = +5V$ the electrons will slowed down moving from the grid to the anode, right? So when we do $U_A = -5V$ they should be accelerated? The manual has a graph, where the ionisation current starts at a certain $U_B = ~10V$. With $U_A = -30 V$. Only after ionisation there is current. $U_B < ~10V$ there is no measured current at all. Well - this is getting out of hand I guess. Thank you anyway. $\endgroup$ – rfaenger Oct 2 '16 at 0:57
  • $\begingroup$ Yes, the electrons are accelerated by the positive grid voltage and some of them will traverse the grid and run against the decelerating field between the anode and the grid caused by the small negative anode voltage (typically around -1V) and reach the anode because of their high kinetic energy. When the ionize Hg atoms the number of energetic electron that can still manage to reach the anode will decrease. $\endgroup$ – freecharly Oct 2 '16 at 1:02
  • $\begingroup$ I added another picture. It says: the current is at zero level until $U_B$ reaches the ionisation-voltage ($U_{ion} ~ 10V$). If $U_A = -30V$ it should be ALSO (same as $U_B$) accelerating the electrons in the direction of the anode and there should be a current at all time. What am I missing? Why has $U_B$ to be increased so much until we measure a current? What particles cause the current? negative electrons or positive ions? $\endgroup$ – rfaenger Oct 2 '16 at 1:09
  • $\begingroup$ This is indeed confusing. Maybe in your apparatus it is the current of the positive ions that are created upon impact ionization that is measured at the anode. This would explain the very large negative anode voltage used of -30V. In this situation, no electron from the cathode traversing the grid would be able to reach the anode. But then I don't understand that they are talking about electrons accelerated towards the anode. Mabe you have a look at this link en.wikipedia.org/wiki/Franck–Hertz_experiment $\endgroup$ – freecharly Oct 2 '16 at 1:22
  • $\begingroup$ Okay, thank you very much for taking your time to think about it! So $U_A = -30V$ is in respect to the anode to block off the electrons and it is indeed the current of the positive ions. But the change of sign (+ to -) doesnt really suit here; because $U_A$ still has the same "task" -> decelerating the electrons! $\endgroup$ – rfaenger Oct 2 '16 at 1:38

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