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Franck-Hertz curve

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    I've just done a practical to plot a franck-hertz curve for mercury. There are a few post-practical questions, and i'm not sure of 2 of them:
    1) estimate the contact potential
    2) estimate the mean free path of an electron in mercury vapour at a temperature of 160C and vapour pressure 11.28mmHg.


    2. Relevant equations

    1) Contact potential = (U1+U2) - Delta U2
    (Delta U2 average distance between peaks, U1 = driving voltage (constant), U2 = acclerating voltage for first peak)
    2) v = (3KbT/m)^1/2

    3. The attempt at a solution
    1)Peaks at: 7.8V, 12.5V,17.5V, 22.4V, 27.6V. U1 = 1.9V
    Average ΔU2 = [(12.5-7.8) + (17.5-12.5) + (22.4-17.5) + (27.6-22.4)]/4 eV =4.95 V

    Error in ΔU2 = [±4(0.1+0.1)]/4 = ±0.2 V

    First excitation potential of mercury = 5.0±0.2V
    (Excitation energy of mercury = 5.0±0.2eV)

    However, the first peak appears at 7.8V. This is as a result of contact potential:

    Kinetic energy of the electrons at grid G2 = U1 + U2 = (7.8±0.1 + 1.9±0.1)eV = 9.7±0.2 eV

    Contact potential = KE - ΔU2 = 9.7±0.2 – 5.0±0.2 = 4.7±0.4V

    2) v = (3*1.38 × 10^-23 m2 kg s-2 K-1* 433.15K/9.109 × 10-31 kg)^1/2
    =140308m/s
    I'm not sure where to go from here, as i don't know how to use the vapour pressure to find the mean free path. Any help at all would be greatly appreciated!
     
  2. jcsd
  3. Nov 14, 2008 #2

    Redbelly98

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    I'll take a stab at helping, since nobody has replied in 1.5 days.

    Can you explain what is meant by U1, the driving voltage? I have done this experiment myself, but that was 27 years ago. Also, what is in the circuit? I presume a power source and the mercury tube, of course. And a voltmeter is connected across the tube electrodes? Anything else? There must be something to limit the current, perhaps that is built into the power supply itself.

    From what I remember, mean-free-path is dependent primarily on the number-density of atoms, and the diameter of the atoms. I would expect a search at google or wikipedia could provide a clue.
     
  4. Nov 17, 2008 #3
    In the experiment, a heated cathode emits electrons which are sent by a driving voltage U1 towards a grid control electrode and then accelerated by the field produced by voltage U2 towards a positively charged acceleration grid electrode. Those electrons with sufficient kinetic energy to pass through the field generated by the potential U3 are collected at an anode and produce current as measured by an ammeter. This all takes place inside a previously evacuated tube now filled with mercury. This is connected to a supply unit which is used to set values for U1 and U3, to vary U2 and to measure how the current changes with U2.

    My problem with the mean free path is the values that i'm given in the question. You're right, most equations i've found depend on the density of the atoms or the diameter but none factor in pressure as a variable without introducing the diameter. I did find a value for the diameter of a mercury atom and plugged it into the equation
    mean free path = [RT]/[sqrt(2)*pi*diameter^2*Pressure*Avagadro's number]
    This came out at 0.39 x 10^-6 m. However, this particular question appeared on an exam paper last year without the diameter given, so i was wondering if there was another method just using the T and P given and any standard constants?
     
  5. Nov 17, 2008 #4

    Redbelly98

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    I'm not sure what U3 is in your setup, but here are my thoughts.

    You got the correct contact potential, but there is an issue with how you wrote out the analysis. When the 1st peak appears and electrons have just enough kinetic energy to excite mercury at grid G2, those electrons have a kinetic energy of 5.0±0.2 eV. I.e., they have the minimum kinetic energy required to excite mercury.

    If we could neglect the contact potential, the electron KE would be simply U1+U2. However, the actual KE is less than U1+U2 because the electrons had to overcome the contact potential in order to leave the electrode. If you write out the equation
    KE=5.0±0.2eV = ______, you can solve for contact potential given U1+U2 value where excitation starts to occur.

    Mean free path: you're given temperature and pressure of the atoms, so density n/V can be calculated from a well-known equation.

    Not sure how this is possible without the diameter given. In the limit d→0, i.e. the atoms are infinitesimally small, the mfp will approach ∞.
     
  6. Nov 17, 2008 #5

    Redbelly98

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    Oh, by the way, you can do a ballpark check of your mean free path calculation. It should be comparable to the thickness of those excitation disks you observed in the mercury tube.
     
  7. Nov 17, 2008 #6
    Okay, so the contact potential is U1+U2 - KE = 9.7 - 5.0 = 4.7eV.
    I got the density from the ideal gas equation, and i'm just going to use the literature value for the diameter to get the mean free path.
    That's cleared a lot up. Thanks for your help!
     
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