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Evaporation of liquid-metal alloys

  1. Nov 14, 2009 #1
    Hi everyone

    This is my first post, please be kind!

    I am part of a research group at Oxford University, UK. We are investigating ways to make multicrystalline Si solar cells more efficient. One of the characterisation techniques we use is electron beam induced conductivity (EBIC) measurement, which is a scanning electron microscopy technique.

    To use the EBIC technique, we have to make a Schottky contact on the top surface of the sample mc-Si wafer (surface exposed to electron beam), and an Ohmic contact on the back. Ohmic contact is made with high-purity liquid metal alloy of Indium and Gallium (InGa alloy) applied to wafer surface, then a layer of "silver-dag" is painted on and used to affix sample to copper stub that then goes into SEM. SEM vacuum is approximately 10^-7 Torr.

    Here is my problem: I want to know whether the InGa alloy, which is liquid at room temperature and atmospheric pressure, will evaporate at room temperature and 10^-7 Torr pressure. If it does, then we cannot use the technique, since we do not want the inside of the SEM to become coated with evaporated InGa!

    I cannot find data for enthalpy of vaporization of InGa alloy. This led me to consider in general terms, the problem of evaporation of liquid metal alloys. I cannot find any answers to the questions:

    1) Can 2-component liquid-metal alloys evaporate eg InGa (l) --> InGa (g)?

    2) If answer to 1 is "yes", then how does this occur?

    3) If answer to 1 is "no", do liquid-metal alloys remain liquid at low pressures? If not, then how do they evaporate?

    4) Can one calculate enthalpy of vaporization of alloy using enthalpies of vaporization of components and Gibbs energy of mixing, and if so, how?


    I have been looking at the literature all afternoon and am at a loss. Any help will be greatly appreciated.

    Thanks for taking the time to read my question!

    Johnny
     
  2. jcsd
  3. Nov 14, 2009 #2
    I suppose you could contact a company that does Physical Vacuum Deposition and get them to run a little experiment.

    Here's what I would do:

    Put the material in the vacuum chamber. These types of coating chambers have metal shutters that can open and close so you can choose when to start deposition. Make sure the shutter is closed so that it's directly above the alloy. Tape a glass slide to the shutter and then pump down the chamber to the desired pressure. "Soak" it at this pressure for a given amount of time and then vent the chamber. Then you can look at the glass under a microscope to determine if any alloy material was vaporized and deposited on the glass. **Note: you are not applying any voltage to the electron beam emitter, so you are not heating up the material at all, which is what these chambers are typically intended for** A technician could actually run your experiment while they're performing a leak-back test for a given chamber (since they wouldn't use the emitter during that either). Anyways, it's just an idea...
     
  4. Nov 15, 2009 #3

    Mapes

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    Here's an even more sensitive test: Put the material in an evaporator (which is actually used to heat samples by Joule heating or e-beam impingement and obtain a high vapor pressure for deposition) that has a quartz crystal to monitor deposition. Don't apply any heating, of course, just pump down to the desired pressure. (First off, if you can't reach that pressure until a later time than usual and the material is gone when you open the chamber again, your material is too volatile :smile:) The quartz sensor should be able to measure deposition down to Angstroms per second to let you know if there's any possibility of contaminating other chambers at similar pressures.

    EDIT: I should add, there's often a shutter that be closed over the sample. You'll need to zero the sensor before opening the shutter, since the sensor will pick up all kinds of outgassing as background noise.
     
  5. Nov 15, 2009 #4

    Mapes

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    I just took a look at the In-Ga phase diagram. Since the two elements don't form a compound (in fact, they form a eutectic, indicating that the In-Ga bond in the solid state is unfavorable), it seems unlikely that they would evaporate as InGa(g). It seems much more likely that the evaporated material would be dominated by the more volatile element, which is gallium. Since they're in the liquid state, perhaps you could assume ideal mixing and apply the relationships

    [tex]\mu=\mu_0+RT\ln a[/tex]

    [tex]\Delta G=\Delta H-T\Delta S[/tex]

    where [itex]\mu[/itex] is the chemical potential, or partial Gibbs free energy, and [itex]a[/itex] is the activity, equal to the concentration (or partial pressure) for ideal mixtures. The chemical potentials of liquid and vapor Ga are equal when the two states are in equilibrium:

    [tex]\mu_\mathrm{Ga,L,0}+RT\ln\,C_\mathrm{Ga,L}=\mu_\mathrm{Ga,L}=\mu_\mathrm{Ga,V}=\mu_\mathrm{Ga,V,0}+RT\ln\,P_\mathrm{Ga,V}[/tex]


    [tex]RT\ln\,P_\mathrm{Ga,V}=\mu_\mathrm{Ga,L}-\mu_\mathrm{Ga,V}+RT\ln\,C_\mathrm{Ga,L}=-\Delta G_\mathrm{Ga,L\rightarrow V}+RT\ln\,C_\mathrm{Ga,L}[/tex]

    [tex]P_\mathrm{Ga,V}=C_\mathrm{Ga,L}\exp(\Delta S_\mathrm{Ga,L\rightarrow V}/R)\exp(-\Delta H_\mathrm{Ga,L\rightarrow V}/RT)[/tex]

    which should give the vapor pressure of gallium. If my model is right, it's approximately just the concentration (85 at.% near the eutectic) multiplied by the vapor pressure of pure gallium. Does this help at all?
     
  6. Nov 15, 2009 #5
    Thanks guys. I was thinking about this today and I thought of a similar experiment to find out, using our thermal evaporator. I was going to use a sensitive electronic mass scale to weigh the sample and then pump it down and see if any evaporated. Nice to see other people think that's a do-able plan, great minds think alike and all that jazz...
    Problem is that the SEM reaches a higher vacuum than the evaporator so this might not be proof that the sample won't evaporate inside the SEM.

    Mapes, I think you're right, and the calculation (using X_Ga = 85%) and the vapour pressure of Ga = 9.31E-36Pa (@29.9°C) gives a vapour pressure of approx. 7E-38 Torr, so this won't coat the chamber. I think it's actually even more unlikely that evaporation occurs, because the separated In and Ga components are in the solid phase at room temperature. Mixing the two solids in powdered form results in the phase change to liquid alloy. To separate the components the Gibbs energy of mixing must be overcome. Or I'm just being ignorant, also likely. :)

    So thanks guys. I am going to do the experiment (measure mass in evaporator pump-down) anyway, just to be sure, since the instrument is too important to risk. I will see if there is an evaporator with higher vacuum available.

    Thanks again!

    Johnny
     
  7. Nov 15, 2009 #6

    Mapes

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    Upon further study, I see I'm mistaken to assume that gallium will have the higher vapor pressure because of its low melting temperature. Vapor pressure depends on the boiling temperature, not the melting temperature, and indium's boiling temperature is lower than gallium's by a couple hundred kelvins. However, it's still around 2300 K, so its vapor pressure is likely acceptably low, apparently <<10-20 Torr at room temperature.

    (Note that Google searches of vapor pressure should not be trusted; a value of 1017 Pa at 156°C for In appears in multiple places, which is incorrect; possible the reference is intended to be 10-17 Pa. Note to self: use a reputable scientific handbook!)
     
  8. Dec 17, 2009 #7
    I'm curious to know how your experiment came out. How did you end up going about it?
     
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