In reply to Eric Walker's message of Wed, 9 Sep 2015 23:15:35 -0500: Hi, [snip] >On Wed, Sep 9, 2015 at 10:25 PM, <[email protected]> wrote: > >The formula could probably be adapted, if I knew the appropriate function >> for the nuclear force. IOW what I know is only qualitative, not >> quantitative. > > >To restate what you've said: > > - The strong force drops off by the sixth power. > - The electromagnetic force drops off by the second power. > - Holding the two nuclei near one another for any amount of time goes a > long way. > >I was wondering whether the system could be modeled at a basic level in the >following way. First, assume there is an applied voltage that is drawing >the ionized 7Li towards the nickel nucleus. (This is the electric arc >thing I always go back to.) I recall seeing a formula that will tell you >how close the 7Li will approach for a given initial kinetic energy against >the Coulomb repulsion of the nickel nucleus.
For an arc, this would depend on the average number of electrons that found themselves between the two nuclei. You might also get positive ions going the other way that neutralize the force exerted by the electrons. >I imagine it can be adapted >to make use of an applied voltage in place of kinetic energy. The adapted >formula would give you the separation distance. I think current would be more appropriate than voltage. Note that as the distance between the nuclei decreased, so would the number of electrons between them, given a fixed current density. In short as they get closer together the attractive force would decrease, and the repulsive force increase. >I assume the separation >distance would remain constant for however long the voltage is applied. You would probably reach a point where there were just enough electrons to neutralize the charges on the nuclei. Note that this is usually called an "atom", and we already know that the distance between ordinary atoms is too large for nuclear reactions at any significant level. > >Now model the strong force using two spheres of large diameter that are >characterized by a density function that drops off with the sixth power. >Multiply the values of the points where the two spheres overlap with one >another. Now integrate the multiples over the two volumes and through time >to get a value for the strength of the interaction at a given distance and >period of time. Call this value S. I'm not sure that an integral over the volume is appropriate as we are essentially looking at a reaction along a straight line between the nuclei. > >Now calibrate the model by placing the 7Li and the nickel nucleus at the >distance that would lie between them if the 7Li approached with 10 MeV, and >hold them there for the brief period of time they would be in proximity. That distance is at least 12 fm. >Whatever the model says "S" is a that point is calibrated to 1 barn. It's not quite that simple, since cross sections also take into account tunneling from distances larger than the closest approach distance. You really need to integrate the chance at any given distance over the entire separation distance from the closest approach to infinity to get the total chance. (I think time may need to be in there too somewhere. ;) > >Would this model capture the qualitative dimensions, or is it missing >something important? > >Eric Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
