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Apr 23

The difference between cold fusion and cold fusion

by mitch | Categories: nuclear chemistry | (11024 Views)

In a recent article in C&EN, Steve Ritter writes about a cold fusion presentation at the recent ACS Chicago meeting. The article can potentially be read as lumping the loony-toony crack-pot conspiracy theorists cold fusion with the very real and valid field of low-energy nuclear reactions dealing with cold fusion. So, lets talk about the differences…

Article: http://pubs.acs.org/isubscribe/journals/cen/85/i17/html/8517sci2.html

Low Energy Nuclear Reactions
In this branch of nuclear chemistry there is a subfield called cold fusion. In “cold fusion” a heavy nucleus (Z>2) is accelerated and bombarded against a target composed entirely of Lead(Z=82), or Bismuth(Z=83), or other near neighbor. The term cold fusion is applied for these reactions because when these two nuclei come together they have an excitation energy of ~10-15MeV which is very small when you compare it to other types of nuclear fusion. Since the excitation energy is so low, the newly formed element is stable to fission and thus tends to stick around long enough to be detected by “conventional” means. Some examples of cold fusion reactions exploited in this way are shown below.

  • 209Bi + 58Fe -> 266Mt + 1n <~~Reaction lead to the discovery of Mt
  • 208Pb + 58Fe -> 265Hs + 1n <~~Reaction lead to the discovery of Hassium

Loony-Toony Cold Fusion
Loony-Toony Cold Fusion is the fusion of two light nuclei at room temperatures. The following 4 reactions are the typical cold fusion reactions investigated by the crack pots.

  • 2H + 2H -> 3He + 1n(2.45 MeV) (eqn.1)
  • 2H + 2H -> 3H + 1H(3.0 MeV) (eqn.2)
  • 2H + 2H -> 4He + gamma-ray(23.8 MeV) (eqn.3)
  • 1H + 2H -> 3He + gamma-ray(5.5 MeV) (eqn.4)

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Here is how the non-nuclear chemists test for cold fusion. They stick D2O in an electrochemical cell, turn it on, measure the heat coming out and compare it to the amount of energy that went in. Sounds like a very reasonable scientific thing to do; If the
heat out is greater than the energy in, than a fusion event has to of occurred to account for this mythical-ether like heat output. However, this is a horrible prehistoric way to look for fusion; It would be like calculating an area of an integral by cutting it out from a sheet of graph-paper and weighing it.

Reasons Why Loony-Toony Cold Fusion is bull…

  • We’ve looked for the heat and can’t find it. There is no excess heat generated from these kinds of reactions that can not be accounted for. See Henderson et al.’s paper: http://dx.doi.org/10.1007/BF01588282
  • Lets just do the freshman Physics and see if it makes sense with our knowledge of physical chemistry.
    In order to fuse, two nuclei need to overcome the coulomb force and touch each other, so lets calculate the coulomb energy for this configuration.

    So, for the case of a deuteron on deuteron…
    k = 8.9876×109 N m2 C-2, q1=q2=1.602×10-19 C, r = 2.8 x 10-15 m.
    Which by my calculation equals to 8.2 x 10-14 J

    Now we know the energy necessary to put two +1 charges close enough in order to undergo fusion. How fast must the two nuclei travel in order to have enough kinetic energy to overcome this 8.2 x 10-14 J barrier?
    The Total Kinetic Energy(KE):
    KE = .5(mass of deuteron)(velocity)2 + .5(mass of deuteron)(velocity)2 = (mass of deuteron)(velocity)2

    velocity = [KE / (mass of deuteron)]1/2
    velocity = [ 8.2 x 10-14 J / (3.3 x 10-27 Kg) ]1/2
    velocity = 4,984,825 meters per second!

    To get atoms moving that fast will require a high temperature, higher than can be achieved with water for sure. We’ve all taken Physical Chemistry so lets see what Temperature it will take to have gas molecules traveling that fast (notice how I already assume that fusion in the liquid phase is phony-bologne). If I assume the deuteron behaves as a monatomic gas then…

    Translational kinetic energy = (3/2)kT, where k is the Boltzmann constant and T is temperature.

    T = (2/3)KE / k = (2/3)(8.2 x 10-14 J) / (1.381×10-23 J / kelvin) = 3,958,484,190 Kelvins

    This temperature can not be reached by an electrochemical cell! This is all just complete rubbish.

    (Notes: (1) It would of been proper to use the energy in the center of mass instead of the lab-frame. (2)The correct coulomb barrier for real nuclei is approximately half of what was calculated using the formula above, which is only meant for rigid hard charge spheres.)

  • In (eqn.3) and (eqn.4) above, a very high mono-energetic photon(gamma-ray) is emitted. It is ridiculously simple to put a gamma detector in front of the electrochemical cell and look for the very specific photon that comes out with an energy of 23.8MeV or 5.5MeV respectively. From (eqn.1) an energetic neutron is emitted; it is also very simple to put a scintillating liquid around the electrochemical cell and look for the stereotypical slow-rise time of a neutron signal.

In conclusion, giving coverage to this fringe science only helps perpetuate the false belief that there exists any viability in cold fusion. The C&EN readership would be well served if more coverage of valid nuclear chemistry research was reported too. Hint-hint. Wink

Link to article:

http://pubs.acs.org/isubscribe/journals/cen/85/i17/html/8517sci2.html

Mitch

6 comments

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  1. Janique

    But would it be allowed? Got yah.

  2. Sili

    It would be like calculating an area of an integral by cutting it out from a sheet of graph-paper and weighing it.

    I’m pretty sure I’ve done that at least once …

  3. cliff52

    I noticed no one has offered a challenge to your math or chemistry or physics. We all want energy from cold fusion, but science reminds us that wanting is not enough. Thank you for a serious, brief review of the factual basis of science as we know it today.

  4. ttch

    Excess heat is not mentioned in the abstract of Henderson et al.’s paper; are you sure you have the right reference?

  5. Behnam

    My understanding is that many practitioners of “cold fusion” of the type you ridicule are not sure that what is actually happening is fusion. Furthermore, they more than happily acknowledge the fact that it is an anomalous phenomenon, meaning that a theory has not been developed to explain it.

    What they have done is to identify an anomalous, measurable, and repeatable phenomenon. The question is how the phenomenon is to be explained. They don’t claim to have an explanation that they know is true for sure. But nobody else seems to have such an explanation either.

    To me, all this is a sign that some exciting science could be done here. Something valuable is to be learned by probing this phenomenon and theorizing about it. The outcome is unlikely to alter the laws of physics, but it will nevertheless teach us about the physical world.

  6. Ben

    I’m not familiar with cold fusion. However, when you calculated the kinetic energy which is 3/2kt, it is a statistically averaging result and it is a result that people usually observed from experiment.
    The fusion people might are thinking about the fluctuation.

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