# Cold Fusion

### Bologna University

Recently a press conference was held at Bologna University to announce a cold fusion reactor. Complete details of the announcement are found in the video of the demonstration and related press conference on: the Energy Catalyzer at Bologna University

In this note we will attempt, using pretty elementary nuclear physics, to prove/disprove the claims made by Prof. Sergio Focardi & his associates. We start by obtaining the isotopic masses of Nickel and Copper isotopes from the periodic table of elements

The table shows that Nickel has the following isotopes: Ni58, Ni60, Ni61, Ni62 & Ni64. Where the numerical postfix indicates the total number of handrons (neutrons + protons) of the isotope. The table also shows that the isotopes of Copper are: Cu63 & Cu65. It is interesting to note that all Cu isotopes interleave with Ni isotopes. Namely. Ni62, Cu63, Ni64, Cu65. In other words the nucleic structures of these isotopes are close.

Next we try to determine which Ni isotopes could be involved in a transmutation to Cu isotope. The obvious candidates are Ni62=>Cu63 and Ni64=>Cu65. Both of these Ni isotopes are pretty rare with Ni62 present at a 3.634% level and Ni64 with .926%. Assuming that both are involved, they account for a little less than 5% of the total mass of nickel on hand.

Next we proceed to calculate the mass defect implied by such hypothetical reactions. The mass defect of the Ni62 to Cu63 transition is:

61.928349 + 1.0079 - 62.929601 = 0.006648

That of the Ni64 to Cu65 is:

63.927970 + 1.0079 - 64.927794 = 0.008076

Using the relative frequencies we get the following average mass defect:

(3.634 x .006648 + .926 x .008076)/(3.634 + .926) = (.024258 + .007478)/4.56= .031736/4.56 = .006959

To compute the maximum total mass converted we compute first the amount of Ni62 & Ni64 available in the Ni in the reactor. The briefing above states that the reactor was loaded with 1 gram of Ni. The percentage of this that would be Ni62 or Ni64 isotopes was calculated to be 4.56%. Therefore, 45.6 milligrams (mg) would be available, that is 4.56 x 10**-2. The average mass defect is .005959 per Ni atom involved. This corresponds to not more than .0059/63.9279 of the 45.6 mg. Thus m = 45.6 mg x .0059/63.9279 = .00421 = 4.21 nanograms (ng) potentially "burned". Next we calculate the amount of energy released by "burning" one nanogram of Ni and the average power level generated if such energy is released over a six months period.

A nanogram corresponds to:

E = mc**2 = 10**-6 x (3 x 10**8)**2 = 10**-6 x 9 x 10**16 = 9 x 10**10 = 90 GigaJoules = 9 x 10**10/3.6 x 10**6 = 2.5 x 10**4 = 25000 KWh = 25 MWh

Next we can compute the level of power that could be generated over a period of six months. Six month are:

6 x 30 x 24 x 3600 sec = 15,552,000 sec = 1.555 x 10**7 secs

Thus the average power level in Watts, from one nanogram of nuclear fuel is:

9.0 x 10**10/1.555 x 10**7 = 5.79 x 10**3 Watts = 5790 Watts = 5.8 KW

The amount of energy potentially generated is thus:

E = 25 x 4.21 = 105.25 MWh

That is, the 4.21 nanograms (ng) could get converted into 105.25 MegaWattHours!!! This is why producing energy from nuclear sources is so attractive. Especially if it can be done without releasing harmful radiation and requiring difficult to provide extreme temperatures & pressure in the reactor.

Next we can compute the level of power that could be potentially generated by the Rossi/Focardi reactor if this energy is released over six months.

The average power level in Watts is:

4.21 x 5.8 KW = 24.4 KW

In actuality a lower power output will be obtained since the efficiency of the assumed nuclear reaction would not be 100%

#### SRI

In this video Michael McKubre of SRI talks about his experiments, over many years, using Palladium (Pd) instead of Ni. It turns out that Pd & silver (Ag) are the next level of the table of elements in positions that correspond to respectively Ni & Cu, metals used by the Focardi Rossi team. Let us repeat similar calculations for the Pd/Ag case.

The isotopes of Pd are: Pd102, Pd104, Pd105, Pd106, Pd108 & Pd110. The isotopes of Ag are: Ag107 & Ag109. Therefore we have a very similar pattern as the Ni Cu case. Namely the two Ag isotopes interleave with the Pd106 & Pd108 palladium isotopes. The frequency of Pd106 is 27.33% and that of Pd108 is 26.46%. Thus both of them are present in significant amounts in palladium.

We can hypothesize two nuclear reactions. Namely:

Pd106 + proton => Ag107 & Pd108 + proton => Ag109

The total amount of "source" Pd is 27.33% + 26.46% = 53.79%

Next we proceed to calculate the mass defect implied by such hypothetical reactions. The mass defect of the Pd106 to Ag107 transition is:

105.903483 + 1.0079 - 106.905093 = 0.006829

That of the Pd108 to Ag109 is:

107.903894 + 1.0079 - 108.904756= 0.007038

To compute the maximum total mass converted we compute first the amount of Pd106 available in the Pd in the reactor. Assuming that the reactor was loaded with 1 gram of Pd. The percentage of this that would be Pd106 isotope is 27.33%. Therefore, 273 milligrams (mg) would be available, that is 2.73 x 10**-1. The mass defect is .006829 per Pd atom involved. This corresponds to not more than .006821/105.903483 of the 273mg. Thus m = 273 mg x .006829/105.903483 = .0176 mg= 17.6 ng "burned".

Similarly for the Pd108 we have 264.6 mg. The mass defect is .007038 and the total mass "burned" would be 264.6 x .007038/107.903894 = 17.0 ng. Thus the total mass potentially burned is: 34.6 nanograms

The amount of energy in MWh released is thus:

34.6 x 25 MWh = 865 MWh

The average power level in KW is:

5.8 x 34.6 = 200.7 KW

In actuality one would get a lower power output since the reaction efficiency will not be 100%

#### Platinum/Gold

On the next level of the periodic table of elements the platinum (Pt) gold (Au) pair corresponds to the Ni/Cu & Pd/Ag pairs. Therefore it is worth checking if a similar transmutation is energetically feasible.

In the case of Au there is a single isotope Au197 which therefore accounts for 100% of available gold. Its atomic weight is 196.966552. Therefore, we need to evaluate the energy balance for the transmutation:

Pt196 + proton => Au197

The mass defect then is:

195.964935 + 1.0079 - 196.066552 = 0.906283

Therefore this reaction is energetically feasible.

The total amount of "source" Pt is 25.3%

To compute the maximum total mass converted we compute first the amount of Pt196 available in the Pt in the reactor. Assuming that the reactor was loaded with 1 gram of Pt. The percentage of this that would be Pt196 isotope is 25.3%. Therefore, 253 milligrams (mg) would be available, that is 2.53 x 10**-1. The mass defect is .906283 per Pt atom involved. This corresponds to not more than .906283/195.964935 of the 253mg. Thus m = 253 mg x .906283/195.964935 = 1.17 mg "burned". The amount of energy potentially generated is thus:

E = 25 x 1.17 = 29.25 MWh

Next we can compute the level of power that could be potentially generated by the Pt/Au reactor. Assuming that the reactor needs to be refueled every six months or so. The average power level in KW is:

1.17 x 5.8 KW = 6.8 KWs

In actuality one would get a lower power output since the reaction efficiency will not be 100%

# Other Possible Transmutations

Both the Ni/Cu transmutation alleged for the Focardi/Rossi reactor and the Pt/Ag of the McKumbre reactor coincide with a tight interleaving of one or more isotopes of the source element with one isotope of the target element. The Isotope Map (click to enlarge) shows all nucleids coded with different colors depending upon their mechanism of nuclear decay. The stable nuclei are represented by black pixels The first observation is that the set of black pixels is very "thin" indeed. Secondly the type of interleaving observed in the Ni/Cu & Pt/Ag cases should be visibly apparent on this map of isotopes. In fact, it should correspond to the next handron being added being a proton (Changing Ni into Cu or alternatively Pt into Ag) and NOT a neutron. This would graphically show up as an additive black pixel to the right (# protons increasing by 1) and NOT in the up direction (# of neutrons not increasing). We can thus look at the Isotope Map and spot all these types of kinks or gaps & investigate their potential energy balance.

# First gap S Cl

S isotopes:

S32 31.9720707 95.02%

S33 32.9714585 0.75%

S34 33.9678668 4.21%

S3S 5.9670809 0.02%

Cl isotopes:

Cl35 34.96885271 75.77%

Cl37 36,96590260 24.23%

In this case the possibilities are S34=>Cl35 & S36=>Cl37; S34 is present only at the 4.21% and S36 at the 0.02%. Thus, especially the latter are not likely to be good source of transmutation energy.

# Second gap Ar K

Ar isotopes:

Ar36 35.9675463 0.337%

Ar37 36.966759 *

Ar38 37.9627322 0.063%

Ar39 38.964313 *

Ar40 39.962383123 99.6%

K isotopes:

K39 38.9637068 93.2581%

K40 39.9639987 0.011%

K41 40.9618260 6.7302%

Both A37 & A38 are available in trace quantities. Therefore the only potential transmutation is Ar40=>K41.

Therefore, we need to evaluate the energy balance for the transmutation:

Ar40 + proton =>K41

The mass defect then is:

39.962383 + 1.0079 - 40.961826= 0.008457

The total amount of "source" Ar is 99.6%

To compute the maximum total mass converted we compute first the amount of Ar40 available in the Ar in the reactor. Assuming that the reactor was loaded with 1 gram of Ar. The percentage of this that would be Ar40 isotope is 99.6%. Therefore, 996 milligrams (mg) would be available, that is 9.96 x 10**-1. The mass defect is .008457 per Ar atom involved. This corresponds to not more than .008457/39.962383 of the 996 mg. Thus m = 996 mg x .008457/39.962383 = 0.21 mg "burned".

The amount of energy in KWH potentially generated is thus:

E = 25 x 210 = 5250 MWh

Next we can compute the level of power that could be potentially generated by the Ar/K reactor. Assuming that the reactor needs to be refueled every six months or so.

210 x 5.8 KW = 1218 KW

In actuality one would get a lower power output since the reaction efficiency will not be 100%

# Third gap Ga Ge

Ga isotopes:

Ga69 68.925581 60.108%

Ga71 70.924705 39.892%

Ge isotopes:

Ge70 69.924250 21.23%

Ge72 71.922078 27.66%

Ge73 72.923459 7.73

Ge74 73.921178 35.94

Ge76 75.921403 7.44

Both the Ga69=>Ge70 & Ga71=>Ge72 are possible since both Ga isotopes are available at the 60% & 40% level and other than the possible scarcity of Ga the supply odds appear favorable.

Next we proceed to calculate the mass defect implied by such hypothetical reactions. The mass defect of the Ga69 to Ge70 transition is:

68.925581 + 1.0079 - 69.924250 = 0.009231

That of the Ga71 to Ge72 is:

70.924705 + 1.0079 - 71.922078= 0.010527

To compute the maximum total mass converted we compute first the amount of Ga69 available in the Ga in the reactor. Assuming that the reactor was loaded with 1 gram of Ga. The percentage of this that would be Ga69 isotope is 60.11%. Therefore, 601 milligrams (mg) would be available, that is 6.01 x 10**-1. The mass defect is .009231 per Ga atom involved. This corresponds to not more than .009231/68.925581 of the 601 mg. Thus m = 601 mg x .009231/68.925581 = .080 mg= 80 ng "burned".

Similarly for the Ga71=>Ge72 transmutation we have:

Total mass converted: 399 mg of Ga71 available. The converted amount is:

399 x .01527/70.924705 mg = .086 mg = 86 ng "burned"

Total potentially transmuted is 166 ng.

E = 25 MWh x 166 = 4150 MWh

Next we can compute the level of power that could be potentially generated by a Ga/Ge reactor. Assuming that the reactor needs to be refueled every six months or so.

166 x 5.8 KW = 962.8 KW

In actuality one would get a lower power output since the reaction efficiency will not be 100%

# Fourth gap Sb Te

Sb isotopes:

121Sb 120.903818 57.36%

123Sb 122.904216 42.64%

Te isotopes:

120Te 119.90402 0.096%

122Te 121.903047 2.603%

123Te 122.904273 0.908%

124Te 123.902819 4.816%

125Te 124.904425 7.139%

126Te 125.903306 18.95%

128Te 127.904461 31.69%

130Te 129.906223 33.80%

Both the 121Sb=>122Te & 123Sb=>124Te are possible since both Sb isotopes are available at the 57% & 43% level and other than the possible scarcity of Sb the supply odds appear favorable.

Next we proceed to calculate the mass defect implied by such hypothetical reactions. The mass defect of the 121Sb to 122Te transition is:

120.903818 + 1.0079 - 121.903047 = 0.008671

That of the 123Sb to 124Te is:

122.904216 + 1.0079 - 123.902819= 0.009297

To compute the maximum total mass converted we compute first the amount of 121Sb available in the Sb in the reactor. Assuming that the reactor was loaded with 1 gram of Sb. The percentage of this that would be 121Sb isotope is 57.36%. Therefore, 574 milligrams (mg) would be available, that is 5.74 x 10**-1. The mass defect is .008671 per Sb atom involved. This corresponds to not more than .008671/120.903818 of the 574 mg. Thus m = 574 mg x .008671/120.903818 = .041 mg= 41 ng "burned".

Similarly for the 123Sb=>124Te transmutation we have:

Total mass converted: 426 mg of 123Sb available. The converted amount is:

426 x .009297/122.904216 mg = .032 mg = 32 ng "burned"

Total potentially transmuted is 73 ng.

The energy released is thus:

25 MWh x 73 = 1825 MWh

Next we can compute the level of power that could be potentially generated by a Sb/Te reactor. Assuming that the reactor needs to be refueled every six months or so.

The average power level in KW is:

73 x 5.8 KW = 423.4 KW

In actuality one would get a lower power output since the reaction efficiency will not be 100%

# Fifth gap Sm Eu

Sm isotopes:

144Sm 143.911995 3.1%

147Sm 146.914893 15.0%

148Sm 147.914818 11.3%

149Sm 148.917180 13.8%

150Sm 149.917271 7.4%

152Sm 151.919728 26.7%

154Sm 153.922205 22.7%

Eu isotopes:

151Eu 150.919846 47.8%

153Eu 152.921226 52.2%

Both the 150Sm=>151Eu & 152Sm=>153Eu are possible since both Sm isotopes are available at the 7.4% & 26.7% level and other than the possible scarcity of Sm the supply odds appear favorable.

Next we proceed to calculate the mass defect implied by such hypothetical reactions. The mass defect of the 150Sm to 151Eu transition is:

149.917271 + 1.0079 - 150.919846 = 0.005325

That of the 152Sm to 153Eu is:

151.919728 + 1.0079 - 152,921226= 0.006402

To compute the maximum total mass converted we compute first the amount of 150Sm available in the Sm in the reactor. Assuming that the reactor was loaded with 1 gram of Sm. The percentage of this that would be 150Sm isotope is 7.4%. Therefore, 74 milligrams (mg) would be available, that is 0.74 x 10**-1. The mass defect is .005325 per Sm atom involved. This corresponds to not more than .005325/149.917271 of the 74 mg. Thus m = 74 mg x .005325/149.917271 = .0026 mg= 2.6 ng "burned".

Similarly for the 152Sm=>153Eu transmutation we have:

Total mass converted: 267 mg of 152Sm available. The converted amount is:

267 x .006402/151.919728 mg = .011 mg = 11 ng "burned"

Total potentially transmuted is 13.6 ng.

E = 25 MWH x 13.6 = 340 MWh

Next we can compute the level of power that could be potentially generated by a Sm/Eu reactor. Assuming that the reactor needs to be refueled every six months or so.

The average power level in KW is:

13.6 x 5.8 KW = 78.88

In actuality one would get a lower power output since the reaction efficiency will not be 100%

## Conclusions

The results above are summarized in the following table:

Pair_______Burned ng_____Energy MWh______Power KW

Units__________1.0__________25.0____________5.8

Ni/Cu_________4.2_________105.2___________24.4

Pd/Ag________34.6_________865.0__________200.7

Pt/Au_________1.2__________29.2____________6.8

S/Cl___________*_____________*______________*

Ar/K________210.0________5250.0_________1218.0

Ga/Ge_______166.0________4150.0__________962.8

Sb/Te________73.0________1825.0__________423.4

Sm/Eu________13.6_________340.0___________78.9

The calculations above do NOT prove that the Focardi/Rossi or the McKumbre reactors work. They prove that IF the hypothesized nuclear reactions take place they would release very significant amounts of energy that could easily exceed the electrical energy supplied to the reactors. The proof of the positive & significant energy yield rests, as pointed out, on simple controlled calorimetric experiments & the chemical detection of nuclear "ashes" such as Cu & Ag in the two reactors fuel. Both kinds of empirical analyses can be performed by chemist & do NOT require the involvement of nuclear physicists.

There is no denying that adding a proton to either a Ni or Pd nucleus is not easy since nuclear forces act over extremely short distances while the electromagnetic forces causing the proton to proton repulsion act over much larger distances. This is the basic reason why fusion is normally observed at very high energy & pressure levels (interiors of stars, hydrogen bombs). However the strong isotope interleaving we have pointed out may hint to the possibility of an atypical "resonance" effect. Another possibility is that the wave function of the incoming proton is sufficiently diffused to encompass the target nucleus at room temperature/pressure conditions. In such case one could expect results that would be repeatable only in the aggregate.

It seems to us that given: a) the huge benefits that could be obtained by confirming/rebutting in a definitive way these findings; b) the very minor costs for doing so (million to low billion dollars); c) the obvious bias and conflict of interest of the groups involved in doing the debunking of the Fleischman & Pons experiments twenty one years ago; it would be wise to look at these possibilities on the basis of several laboratories in several countries involving neither people getting huge R&D funding for "Hot fusion" or dominant energy companies. After all. twenty one years later, the world is in a far more desperate energy balance situation.