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Franck-Hertz (my head) Experiment

  1. Apr 10, 2008 #1
    My questions are about the classic Franck-Hertz experiment where electrons are given energy by an accelerating voltage. These collide with mercury atoms, raising their internal energy. The electrons are then left with insufficient energy to overcome the retarding voltage of a collector plate and produce a drop in current. These drops occur at regular intervals which demonstrates the discrete nature of the atom.

    Parts of the experiment confuse me, my thought process is scattered at the best of times, so I’d appreciate it if you could help set me straight. Thank you for your time.

    1. I have read that if the mercury vapour is too rare, more than one energy level transition may be involved. Why is this?
    I thought that the accelerating voltage ( and therefore the energy of the electrons ) would control which energy levels were involved, not the density of the vapour.
    I can understand that if the vapour is too dense, too many collisions will occur and the current will remain low, but I have a hard time trying to rationalise why a rare vapour would produce such an effect. The current is measured when
    (i) no transitions take place and
    (ii) when the ground-lowest transition is present.

    Ideally we want the drop in current to be due to this ground-lowest energy level transition only. These electrons have been involved in a collision and now don’t have enough energy to overcome the retarding voltage of the collector plate.

    What we don’t want to measure is a lack of an electron due to a different energy transition, unless this electron goes on to cause the lowest energy transition too.

    If the vapour is very rare, there’s less chance that this second collision will occur and the drop in current may be due to energy transitions we’re not interested in?

    2. The electrons pass through 2 grids, the first controls how many electrons there are, but I ask myself, why do we want to do this and how is it done?

    My guess would be that we want the number of electrons entering the chamber at any time to be constant. Fair enough, but I would’ve thought this would be dependant on the number leaving the heated filament, so how can we control it?

    How would a negatively charged grid affect the number of electrons passing through? I would have thought it would just repel electrons or slow them down, unless it filtered out those below a certain energy? The energy of the electrons depend on the energy with which they leave the filament. Energy from heat supplied by current minus the ionising energy of the filament maybe?

    2.b The idea that the accelerating voltage imparts an energy of V [eV] to each electron is used. At what distance from the grid can you say an electron has this energy? That’s kind of a useless question but my point is that even if we assume it’s travelling from rest and calculate the answer, how do we know, the collisions haven’t taken place before the electrons reach this energy?

    3. Admittedly, I’m mostly lost when it comes to electronics. This is my rewording of a lab manual, except the bit in inverted commas which is a direct quote. All I have to say when it comes to this is, what are ya talking about, ya mad yoke?!?

    The electrometer uses a current sensing resistor as a feedback element. This is done to reduce the input resistance to RF/A so as to minimise the voltage-burden problem. This problem is that the “voltage-burden developed across the current sensing resistor adversely affects the experiment” by altering the p.d. between the second grid and the collector plate.

    Right, from what little “understanding” I do have of electronics, I can grasp minimising input resistance, minimise voltage developed across it. Good man Ohm! That‘s fine with me. A battery is keeping the collector plate negative with respect to the grid so this voltage must be going in the opposite direction because it “adversely affects the experiment”.
    After that, I’m lost with what’s happening here. Where does this gain, A, come into play?
  2. jcsd
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