{"id":473,"date":"2011-04-04T06:07:25","date_gmt":"2011-04-04T11:07:25","guid":{"rendered":"http:\/\/www.journal-of-nuclear-physics.com\/?p=473"},"modified":"2011-04-06T02:47:16","modified_gmt":"2011-04-06T07:47:16","slug":"how-can-30-of-nickel-in-rossi%e2%80%99s-reactor-be-transmuted-into-copper","status":"publish","type":"post","link":"https:\/\/www.journal-of-nuclear-physics.com\/?p=473","title":{"rendered":"How can 30% of nickel in Rossi\u2019s reactor be transmuted into copper?"},"content":{"rendered":"

by Dott. Giuliano Bettini
\nRetired. Earlier: Selenia SpA, Rome and IDS SpA, Pisa
\nAlso Adjunct Professor at the University of Pisa
\n<\/em>Adjunct Professor at Naval Academy, Leghorn (Italian Navy) <\/em><\/p>\n

Abstract<\/strong><\/div>\n
In the present article I would like to answer a question posed by L. Kowalsky in a recent paper: how can 30% of nickel in Rossi\u2019s reactor be transmuted into copper? “Everything should be made as simple as possible, but not simpler”, says a guy. I apologizes if I am too simplistic here.<\/div>\n

Introduction<\/strong>
\nThe interest on Andrea Rossi’s Nickel-Hydrogen Cold Fusion technology is accelerating [1]. However, Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. Kowalski [2]. says that \u201cthis seems to be impossible because the produced copper isotopes rapidly decay into Ni\u201d. But how it works?<\/p>\n

<\/p>\n

How it works<\/strong>
\nFollowing Focardi Rossi [3]. a Ni58 nucleus produces a Copper nucleus according to the reaction<\/p>\n

Ni58 + p \u2192 Cu59<\/p>\n

Copper nucleus Cu59 decays with positron (e+) and neutrino (\u03bd) emission in Ni59 nucleus according to<\/p>\n

Cu59 \u2192 Ni59 + \u03bd + e+<\/p>\n

Then (e+) annichilates with (e-) in two gamma-rays<\/p>\n

e- + e+ \u2192 \u03b3 + \u03b3<\/p>\n

Starting [3] from Ni58 which is the more abundant isotope, we can obtain as described in the two above processes Copper formation and its successive decay in Nickel, producing Ni59, Ni60, Ni61 and Ni62. Because Cu63, which can be formed starting by Ni62, is stable and does not decay in Ni63, the chain stops at Ni62 (i.e. Cu63). Each process means some MeV.<\/p>\n

Of course how can a proton p gets captured by the Ni58 nucleus? (and subsequent Ni59, Ni60, Ni61 and Ni62). Following Stremmenos [4]. a neutron-like particle, an electron proton pair, a mini-atom, a proton masked as a neutron, gets captured by the Ni58.<\/p>\n

\"\"<\/p>\n

\n

If the masked proton becomes a neutron the result is Ni59.
\nIn order to have Cu59 (increase of atomic number from 28 to 29) the electron (of the masked proton) gets ejected from the nucleus. The masked proton becomes a proton.<\/p>\n

\"\"<\/p>\n

\n

The same process holds for all the subsequent transformations, until Cu63.
\nIt remains to be understood the issue of the gamma radiation in the MeV range.<\/p>\n

\n
Numbers<\/strong><\/div>\n
I am an electronic engineer, so I need easy numbers in order to understand.<\/div>\n
However “Everything should be made as simple as possible, but not simpler”, says a guy. Maybe I am too simple here.<\/div>\n
Let\u2019s calculate.<\/div>\n<\/div>\n
\n
\u00a0<\/strong><\/div>\n
MeV for each Ni transformation<\/strong><\/div>\n
I read that starting from Ni58<\/strong> we can obtain Copper formation and its successive decay in Nickel, producing Ni59<\/strong>, Ni60<\/strong>, and Ni62<\/strong>. The chain stops at Cu63 <\/strong>stable.<\/div>\n
For simplicity I assume all the Nickel in the reactor in the form Ni58<\/strong>.<\/div>\n
For simplicity I suppose for each Ni58<\/strong> the whole sequence of events from Ni58<\/strong> to Cu63<\/strong> and as a rough estimate I calculate the mass defect between (Ni58<\/strong> plus 5 nucleons) and the final state Cu63<\/strong>.<\/div>\n
Ni58<\/strong> mass is calculated to be 57.95380\u00b1 15 amu<\/strong><\/div>\n
The actual mass of a copper-Cu63<\/strong> nucleus is 62.91367 amu<\/strong><\/div>\n
Mass of Ni58<\/strong> plus 5 nucleons is \u00a057.95380+5=62.95380 amu<\/strong><\/div>\n
Mass defect is 62.95380-62.91367=0.04013 amu<\/strong><\/div>\n
1 amu = 931 MeV is used as a standard conversion<\/div>\n
0.04013×931 MeV=37.36 MeV<\/div>\n
So each transformation of Ni58<\/strong> into Cu63<\/strong> releases 37.36MeV of nuclear energy.<\/div>\n<\/div>\n
\n
\u00a0<\/strong><\/div>\n
\u00a0<\/strong><\/div>\n
Nickel consumption<\/strong><\/div>\n
According to many blogs in the Internet \u201cOne hundred grams of nickel powder can power a 10 kW unit for a minimum of six months\u201d.<\/div>\n
How much of Ni58<\/strong> should be transformed, in six months of continuous operation, in order to generate 10 kW?<\/div>\n
I follow a procedure outlined in [2].<\/div>\n
10 kW is thermal or electrical (?) power. The nuclear power must be larger. Assume a nuclear power twice:<\/div>\n
20 kW = 20,000 J\/s = 1.25 x 10**17 MeV\/s.<\/div>\n
Each transformation of Ni58<\/strong> into Cu63<\/strong> releases 37.36MeV of nuclear energy.<\/div>\n
The number of Ni58<\/strong> transformations should thus be equal to (1.25 x 10**17)\/37.36 = 3.346 x 10**15 per second.<\/div>\n
Multiplying by the number of seconds in six months (1.55 x 10**7) the total number of transformed Ni58<\/strong> nuclei is 5.186 x 10**22.<\/div>\n
This means 5 grams.<\/div>\n
The order of magnitude is not exactly the same but seems to be plausible. This means also 5 grams of Nickel in Rossi\u2019s reactor transmuted into (stable) Copper after six months of continuous operation at the rate of 10 kW.<\/div>\n<\/div>\n
\n
\u00a0<\/strong><\/div>\n
Conclusions<\/strong><\/div>\n
Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. At first glance this seems to agree with calculations based on simple assumptions.<\/div>\n
\u00a0<\/strong><\/div>\n
References<\/strong><\/div>\n
[1] Link, \u00a0Cold Fusion “Andrea Rossi” Method<\/a><\/div>\n
[2] L. Kowalski, \u201cRossi’s Reactors Reality or Fiction?\u201d<\/a>, March 2011<\/div>\n
[3] A.Rossi, S. Focardi, https:\/\/www.journal-of-nuclear-physics.com<\/a><\/div>\n
[4] E. Stremmenos, \u201cHydrogen\/Nickel cold fusion probable mechanism\u201d<\/a>, March 2011<\/div>\n<\/div>\n

\n

\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

by Dott. Giuliano Bettini Retired. Earlier: Selenia SpA, Rome and IDS SpA, Pisa Also Adjunct Professor at the University of Pisa Adjunct Professor at Naval Academy, Leghorn (Italian Navy) <\/p>\n

Abstract In the present article I would like to answer a question posed by L. Kowalsky in a recent paper: how can 30% […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[3],"tags":[],"_links":{"self":[{"href":"https:\/\/www.journal-of-nuclear-physics.com\/index.php?rest_route=\/wp\/v2\/posts\/473"}],"collection":[{"href":"https:\/\/www.journal-of-nuclear-physics.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.journal-of-nuclear-physics.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.journal-of-nuclear-physics.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.journal-of-nuclear-physics.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=473"}],"version-history":[{"count":13,"href":"https:\/\/www.journal-of-nuclear-physics.com\/index.php?rest_route=\/wp\/v2\/posts\/473\/revisions"}],"predecessor-version":[{"id":484,"href":"https:\/\/www.journal-of-nuclear-physics.com\/index.php?rest_route=\/wp\/v2\/posts\/473\/revisions\/484"}],"wp:attachment":[{"href":"https:\/\/www.journal-of-nuclear-physics.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=473"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.journal-of-nuclear-physics.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=473"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.journal-of-nuclear-physics.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=473"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}