# franck-hertz-experiment - direction of acceleration

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?  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)...

• 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. – rfaenger Oct 2 '16 at 0:57
• 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? – rfaenger Oct 2 '16 at 1:09
• 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! – rfaenger Oct 2 '16 at 1:38