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Quantum physics - Frankhertz experiment

  1. Sep 16, 2009 #1
    I'm doing the frank hertz experiment and in preparation I'm trying to get a few questions answered.

    any help would be greatly appreciated,

    8.
    Considering that the energy of the 1st excited state of the mercury atom is ~4.8eV above that of the ground state, what is the maximum amount of energy that an electron with 4.0 eV of kinetic energy can loose to a mercury atom with which it collides? What about for a 6.0 eV electron (neglect the recoil at mercury atom)?


    I haven't really done quantum physics study before(not yet in a few weeks we start), BUT I do know from memory that,
    1eV = (1.6*10^-19C)*(1V) = 1.6*10^-19 Joules

    so I'm guessing that the Electron upon collision with the mercury atom, would transfer most of it's energy to the mercury atom,

    I'm going to say the whole 4eV = 4*(1.6*10^-19J) = 6.4*10^-19 Joules

    Not sure though, Do I have to use momentum and conservation of momentum?
    Is it an in-ellastic collision?

    How would I answer for the energy of a 6eV electron coliding with the mercury atom?


    7.
    Why is the collecting anode made negative with respect to the grid?

    I can actually answer this question, but would like some one to check

    The cathode emits electrons to pass through the grid and be collected at the anode,

    so the anode is kept at a lower potential than the grid to stop the electrons from getting extra kinetic energy


    9.
    Why must the Franck‐Hertz tube be operated at an elevated temperature? What is the consequence of going to a temperature even higher than recommended?

    Well I thought that the tube gets hot because of the cathode being heated up,

    the only way for the cathode to emit electrons would be to heat it to give it enough energy for electrons to be emitted

    consequences of heating the tube higher would be a higher resistance in the circuit perhaps? giving innacurate results?

    is this a valid answer to the question?
     
  2. jcsd
  3. Sep 16, 2009 #2
    I've thought about it a little bit

    and i've decided that all the kinetic energy of the electron is used in exciting the mercury atom,

    this is because if you think about it, it's an inelastic collison, because the mass of the electron is MUCH MUCH smaller than the mass of the mercury nucleus,
    so almost all of the energy on collision is transfered

    just like throwing a small ball of mud at a brick building, it's an inelastic collision because of the mass differences,

    so 4*(1.6*10^-19) = the energy the 4eV electron would lose

    similarly for the 6eV
    after gaining the first excitation level at around 4.8eV mercury would use the following 1.2eV to get to the next level of excitation

    or would it?
     
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