KitemanSA wrote:93143 wrote: Dan, Kiteman - you're both wrong. But Dan is wronger.
When you add a proton to nickel-62, it is true that the proton's binding energy decreases dramatically. It is also true that the binding energy of every other nucleon involved increases slightly, which offsets the former much more than the graph seems to imply, due to the 62:1 number ratio.
Thank you for the detail, you are correct in intent. Please note however that we were talking approximations.
93143 wrote: ...and Kiteman? You made a math error in your example. 62*8.95 is 554.9, not 554.28. This gives a larger reaction energy, by 0.62 MeV (notice that it is the binding energy difference times the product nucleon count...).
...oops... I think it may have been a transcription error. I am even worse at typoing than at math!
I admit some of my reasoning has been ambigous at best and missleading at worse.
http://en.wikipedia.org/wiki/Nickel-62
http://www.chem.purdue.edu/gchelp/howto ... energy.htm
...However, nuclear binding energy is often expressed as kJ/mol of nuclei or as MeV/nucleon
.
Using energy per mole may be enlightening. It is more direct, without having to consider the energy per nucleon.
Finally (once again), I'll let a link do the talking as I am not doing so well myself.
http://en.wikipedia.org/wiki/Nuclear_binding_energy
Also, Cu63 can release a little more energy than Ni62 if it is broken down completely. It can even release a little excess energy when Ni62 is fused to Cu63. I'll concede that point. BUT, this ignores the energy required to add the proton to the nucleus. This is the Coulomb repulsion that has to be overcome in order for the proton to reach the nucleus against the increasingly strong electromagnetic repulsion from all of the protons already present. This barrier for light elements are in the range of 10s to 100s of KeV for reasonable fusion rates. By the time you reach iron or nickel the Coulomb barrier is in the millions of eVs
This factor has to be figured into the final energy balance. The change in energy has to be greater than this 'injection' energy for there to be a net gain (exothermic)
This injection energy might be decreased through some catalytic effect so that net gains might be possible, but the catalyst would need to be some synthetic material. As I argued, there are just to many possible natural situations where non synthetic material arrangements could occur and be easily observed (if you survived). If there is some catalytic action in typical LENR reactions of light elements, the effectiveness of the catalyst in the Rossi device would need to be perhaps ~ 10,000 times greater- the Coulomb barrier is ~ 100 times greater and the output is ~ 100 times greater.
So, I fall back to the position that one- a truly phenomenal catalyst must be discovered only now after over 20 years of intense LENR research by many labs. Rossi would have to be extremely lucky or smart. And, this is dependent on the LENR reactions working at all- it is still not resolved.
Secondly, the the radiation from several thousands of watts of energy output would not be blocked by a thin layer of lead if a significant portion of it is in the form of gamma rays .
Of course other arguments about very sloppy and partially hidden experimental design leaves several obvious paths for fraud, in addition to a moderately wide range in the accuracy of the measurements. All of the secrets seems to be contained in the can, so allowing a third party to design a setup to test the effects which is much more transparent and reliable would go a very long way towards validating the device. He could still insist on the absence of gamma ray diagnostics.
I don't know what the radiation output would be. Assuming most of the energy was in the form of Beta emissions, and gammas only carried away ~ 10% of the energy would mean gammas of ~ 600KeV. A centimeter of lead would only reduce this radiation by a factor of perhaps 10-100X. This means that you would be exposed to ~ 10-100 Watts of gamma radiation for hours, days or even months. Consider that a chest X-ray would be photons at perhaps 50KeV, and ~ 50 milliamps for perhaps 0.1 seconds. This would translate into ~ 250 Watts. This means you would be exposed to the equivalent of ~ 1 chest x-ray every 2 to 20 seconds . . Using the lower number, this would be ~ three chest X-rays per minute. or ~ 200 per hour. Each chest X-ray might represent ~ 20-50 mREM of radiation exposure (this is an older setup where exposure intensifiers are not used). So you would be exposed to ~ 20mRem * 200 exposures / hr, or ~ 4 REM per hour. . If you hung around for one day you would likely develop radiation poisoning, and in several more days an increasing chance of dieing in the short term.
Even if you reduce this with distance, lower energy gammas and/ or fewer gammas, the improvement might only be a hundred fold. That might give a daily exposure of ~ 40 mREM per hour, or almost 1 REM per day. Not something you would want in your home . Even a nuclear worker would reach his yearly allowable dose in only 5 days.
Of course this is supposition under some assumed conditions, but it illustrates how minimal the gamma contribution has to be before this feeble 1 cm lead shielding would suffice. It raises further skepticism that this is actually a fusion process, at least by conventional understanding of neucleosynthsis.
Is there a nuclear chemist that can calculate the gamma output? Is it even possible to calculate a likely range given the vague descriptions provided?
Ps: the energy difference between Ni62 and CU63 in the form of nuclear binding energy would be:
Ni62 has 8.7945 MeV per nucleon * 62 nucleons = 545.259 MeV
Cu63 has 8.758 MeV per nucleon * 63 nucleons = 551.754 MeV
The difference is 551.754- 545.259 = 6.4495 MeV.
From this I am guesstimating that the average energy needed to force a proton past the Coulomb barrier would in this example be ~ 6.5 MeV- thus it is endothermic by this reasoning. So the new physics would be a unheard of catalytic reaction that lowered the Coulomb barrier resistance to ~ 0.04 eV (400 degrees C)..
If this was the case then ~ 30 grams of accumulated copper over ~ 6 months would represent ~ 1/2 mole of copper or ~ 3 *10^23 fusions/ transmutations. Each delivering ~ 6.5 MeV. of excess energy. This would represent a total of ~ 2 *10^31 MeV of energy.
For comparison ~ 1 milliwatt of power comes from ~ 10^9 DD fusions/ sec. and with an assumption of a generous 10 MeV of excess energy per DD fusion , this would result in ~ 10^16 MeV per milliwatt or ~ 10^19 MeV per Watt. A claimed Rossi device output of ~ 10,000 Watts would represent ~ 10^23 MeV output per second. This divided into the above number would result in ~ 10^ 8 seconds of operation before 30 grams of copper was produced.
10^8 seconds is ~ 1200 days, or ~ 3 years.
There is about 6 times too much copper present. What does that mean? It means nothing. Without a control which shows the amount of natual copper isotopes contained in the virgin mixture, you cannot say how much additional copper was created, except that you have a minimal amount of ~ 5 grams. Of course there would need to be reasonable assurance that the samples provided are representative, and not artificially doctored.
Dan Tibbets
To error is human... and I'm very human.
). 
The amount of energy required to 'unbind' a nucleon from 62NI is the MOST amount of energy per nucleon for any nucleus, whereas protons clearly do/can not need to be 'unbound' any further.