[Vo]:Rossi H and Ni consumption
From: http://www.rossilivecat.com/ Quote: - - - - - - - - - - - - - - - - - - - - Andrea Rossi October 25th, 2011 at 4:59 PM Dear Thomas Blakeslee: Grams/Power for a 180 days charge Hydrogen: 18000 g Nickel: 1 g Warm Regards, A.R. - - - - - - - - - - - - - - - - - - - - End quote. At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an atoommic weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This means 10.48 atoms of H need be provided per 1 atom of Ni. Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms of H is consumed, thus 170.4 mols of H and a170.4 mols of Ni are consumed, maximum. This involves the obviously wrong assumption that all the Ni atoms are transmuted, not a more realistic 3 percent. There is also an outside possibility the H reacts with daughter products, giving the possibility of 10 subsequent daughter reactions per primary Ni+H reaction. Three such reactions is an outside possibility. One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is (6.241x10^24 eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms / mol) = 9.464x10^5 eV/(Ni atom). If there is one reaction per atom and all Ni is consumed by single reactions than that is 0.9464 MeV per Ni-H event. The gammas from this would be lethal at short range, even through 2 cm of lead. If it is assumed that 3% of the Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 MeV per reaction. If there are an average of 3 daughter reactions per primary reactions that is about 10 Mev per reaction. If 10 MeV gammas are produced then 5 cm of lead shielding will be of no use in protecting the operators. If near 1 MeV gammas are produced the lead shielding is inadequate. One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas, 6.24x10^18 gammas per second. using: I = I0 * exp(-mu * rho * L) where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/ cm^3, we have for 5 cm of lead: I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *(5 cm)) I = 2x10^18 free gammas per second. About half that, or 10^18 gammas/s would be directed toward the interior of the container housing the E-cats, and most of the 2x10^18 gammas per second would end up escaping the container. This is an approximate calculation. Even if it is off by an order of magnitude, this kind of 1 MeV gamma flux, even 1/32 of it from one E-cat, would be readily detected by a geiger counter at significant range. It does not seem credible the energy from a Ni-H reaction, at least in the form of one gamma per reaction, provides any explanation for 1 MW of heat, if that thermal power is in fact achieved. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
RE: EXTERNAL: [Vo]:Rossi H and Ni consumption
On Thurs Oct 27, 2011 Horace said [snip] It does not seem credible the energy from a Ni-H reaction, at least in the form of one gamma per reaction, provides any explanation for 1 MW of heat, if that thermal power is in fact achieved.[/snip] Horace, Assuming the thermal power is in fact achieved, and the reaction is not Ni-H, what do you feel is the next most credible theory ? Fran -Original Message- From: Horace Heffner [mailto:hheff...@mtaonline.net] Sent: Thursday, October 27, 2011 7:49 AM To: Vortex-L Subject: EXTERNAL: [Vo]:Rossi H and Ni consumption From: http://www.rossilivecat.com/ Quote: - - - - - - - - - - - - - - - - - - - - Andrea Rossi October 25th, 2011 at 4:59 PM Dear Thomas Blakeslee: Grams/Power for a 180 days charge Hydrogen: 18000 g Nickel: 1 g Warm Regards, A.R. - - - - - - - - - - - - - - - - - - - - End quote. At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an atoommic weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This means 10.48 atoms of H need be provided per 1 atom of Ni. Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms of H is consumed, thus 170.4 mols of H and a170.4 mols of Ni are consumed, maximum. This involves the obviously wrong assumption that all the Ni atoms are transmuted, not a more realistic 3 percent. There is also an outside possibility the H reacts with daughter products, giving the possibility of 10 subsequent daughter reactions per primary Ni+H reaction. Three such reactions is an outside possibility. One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is (6.241x10^24 eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms / mol) = 9.464x10^5 eV/(Ni atom). If there is one reaction per atom and all Ni is consumed by single reactions than that is 0.9464 MeV per Ni-H event. The gammas from this would be lethal at short range, even through 2 cm of lead. If it is assumed that 3% of the Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 MeV per reaction. If there are an average of 3 daughter reactions per primary reactions that is about 10 Mev per reaction. If 10 MeV gammas are produced then 5 cm of lead shielding will be of no use in protecting the operators. If near 1 MeV gammas are produced the lead shielding is inadequate. One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas, 6.24x10^18 gammas per second. using: I = I0 * exp(-mu * rho * L) where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/ cm^3, we have for 5 cm of lead: I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *(5 cm)) I = 2x10^18 free gammas per second. About half that, or 10^18 gammas/s would be directed toward the interior of the container housing the E-cats, and most of the 2x10^18 gammas per second would end up escaping the container. This is an approximate calculation. Even if it is off by an order of magnitude, this kind of 1 MeV gamma flux, even 1/32 of it from one E-cat, would be readily detected by a geiger counter at significant range. It does not seem credible the energy from a Ni-H reaction, at least in the form of one gamma per reaction, provides any explanation for 1 MW of heat, if that thermal power is in fact achieved. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi H and Ni consumption
There are some ifs and buts associated with this subject. It has been known for over a hundred years how that hydrogen will defuse through a hot metal enclosure. The rate of diffusion is subject to the temperature and pressure of the hydrogen, together with the exact kind, thickness, and temperature of the metal. These are all variables in the calculation of the diffusion rate. Furthermore, the presence of oxides and/or carbides on the surface of the metal can reduce the rate of diffusion of hydrogen by up to 5 orders of magnitude. We don’t know for sure what the accurate values of some of these variables are and additionally they would vary widely within an operational range throughout the operational lifetime of the E-Cat. However, since hydrogen is very slippery and notoriously hard to contain, a good guess can be made that most of the hydrogen consumed by the Rossi reactor would be lost through diffusion through the hot walls of the stainless steel reaction vessel. Because of all these large uncertainties, calculation of the nuclear reaction rates as a function of hydrogen consumption implying a clue to the nuclear processes going on inside the E-Cat reaction vessel cannot be made in my opinion. With best regards, Axil On Thu, Oct 27, 2011 at 7:48 AM, Horace Heffner hheff...@mtaonline.netwrote: From: http://www.rossilivecat.com/ Quote: - - - - - - - - - - - - - - - - - - - - Andrea Rossi October 25th, 2011 at 4:59 PM Dear Thomas Blakeslee: Grams/Power for a 180 days charge Hydrogen: 18000 g Nickel: 1 g Warm Regards, A.R. - - - - - - - - - - - - - - - - - - - - End quote. At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an atoommic weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This means 10.48 atoms of H need be provided per 1 atom of Ni. Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms of H is consumed, thus 170.4 mols of H and a170.4 mols of Ni are consumed, maximum. This involves the obviously wrong assumption that all the Ni atoms are transmuted, not a more realistic 3 percent. There is also an outside possibility the H reacts with daughter products, giving the possibility of 10 subsequent daughter reactions per primary Ni+H reaction. Three such reactions is an outside possibility. One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is (6.241x10^24 eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms /mol) = 9.464x10^5 eV/(Ni atom). If there is one reaction per atom and all Ni is consumed by single reactions than that is 0.9464 MeV per Ni-H event. The gammas from this would be lethal at short range, even through 2 cm of lead. If it is assumed that 3% of the Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 MeV per reaction. If there are an average of 3 daughter reactions per primary reactions that is about 10 Mev per reaction. If 10 MeV gammas are produced then 5 cm of lead shielding will be of no use in protecting the operators. If near 1 MeV gammas are produced the lead shielding is inadequate. One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas, 6.24x10^18 gammas per second. using: I = I0 * exp(-mu * rho * L) where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/cm^3, we have for 5 cm of lead: I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *(5 cm)) I = 2x10^18 free gammas per second. About half that, or 10^18 gammas/s would be directed toward the interior of the container housing the E-cats, and most of the 2x10^18 gammas per second would end up escaping the container. This is an approximate calculation. Even if it is off by an order of magnitude, this kind of 1 MeV gamma flux, even 1/32 of it from one E-cat, would be readily detected by a geiger counter at significant range. It does not seem credible the energy from a Ni-H reaction, at least in the form of one gamma per reaction, provides any explanation for 1 MW of heat, if that thermal power is in fact achieved. Best regards, Horace Heffner http://www.mtaonline.net/~**hheffner/http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi H and Ni consumption
This is a nonsensical argument. The less hydrogen available for nuclear reactions the *more* the MeV per reaction that is required to make the 1 MW output, thus the less effective any shielding would be, and the *less credible* it is that the MW heat comes from nuclear reactions. On Oct 27, 2011, at 11:14 AM, Axil Axil wrote: There are some ifs and buts associated with this subject. It has been known for over a hundred years how that hydrogen will defuse through a hot metal enclosure. The rate of diffusion is subject to the temperature and pressure of the hydrogen, together with the exact kind, thickness, and temperature of the metal. These are all variables in the calculation of the diffusion rate. Furthermore, the presence of oxides and/or carbides on the surface of the metal can reduce the rate of diffusion of hydrogen by up to 5 orders of magnitude. We don’t know for sure what the accurate values of some of these variables are and additionally they would vary widely within an operational range throughout the operational lifetime of the E-Cat. However, since hydrogen is very slippery and notoriously hard to contain, a good guess can be made that most of the hydrogen consumed by the Rossi reactor would be lost through diffusion through the hot walls of the stainless steel reaction vessel. Because of all these large uncertainties, calculation of the nuclear reaction rates as a function of hydrogen consumption implying a clue to the nuclear processes going on inside the E-Cat reaction vessel cannot be made in my opinion. With best regards, Axil On Thu, Oct 27, 2011 at 7:48 AM, Horace Heffner hheff...@mtaonline.net wrote: From: http://www.rossilivecat.com/ Quote: - - - - - - - - - - - - - - - - - - - - Andrea Rossi October 25th, 2011 at 4:59 PM Dear Thomas Blakeslee: Grams/Power for a 180 days charge Hydrogen: 18000 g Nickel: 1 g Warm Regards, A.R. - - - - - - - - - - - - - - - - - - - - End quote. At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an atoommic weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This means 10.48 atoms of H need be provided per 1 atom of Ni. Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms of H is consumed, thus 170.4 mols of H and a170.4 mols of Ni are consumed, maximum. This involves the obviously wrong assumption that all the Ni atoms are transmuted, not a more realistic 3 percent. There is also an outside possibility the H reacts with daughter products, giving the possibility of 10 subsequent daughter reactions per primary Ni+H reaction. Three such reactions is an outside possibility. One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is (6.241x10^24 eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms / mol) = 9.464x10^5 eV/(Ni atom). If there is one reaction per atom and all Ni is consumed by single reactions than that is 0.9464 MeV per Ni-H event. The gammas from this would be lethal at short range, even through 2 cm of lead. If it is assumed that 3% of the Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 MeV per reaction. If there are an average of 3 daughter reactions per primary reactions that is about 10 Mev per reaction. If 10 MeV gammas are produced then 5 cm of lead shielding will be of no use in protecting the operators. If near 1 MeV gammas are produced the lead shielding is inadequate. One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas, 6.24x10^18 gammas per second. using: I = I0 * exp(-mu * rho * L) where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/ cm^3, we have for 5 cm of lead: I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *(5 cm)) I = 2x10^18 free gammas per second. About half that, or 10^18 gammas/s would be directed toward the interior of the container housing the E-cats, and most of the 2x10^18 gammas per second would end up escaping the container. This is an approximate calculation. Even if it is off by an order of magnitude, this kind of 1 MeV gamma flux, even 1/32 of it from one E-cat, would be readily detected by a geiger counter at significant range. It does not seem credible the energy from a Ni-H reaction, at least in the form of one gamma per reaction, provides any explanation for 1 MW of heat, if that thermal power is in fact achieved. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi H and Ni consumption
In the Miley presentation that he has recently released, Miley shows transmutation to 39 isotopes over possible contamination levels. The nuclear reactions and transmutation patterns that are going on inside the Rossi reactor are similar to what Miley documents as mentioned in Rossi’s original patent. The presence of a large amount of iron in the Miley results is interesting and similar iron contamination was found in the Rossi ash(10%) when they were analyzed by the swedes. The assumption that the nuclear reactions taking place in the Rossi reaction are exclusively restricted to copper transmutation is mistaken in my opinion. The possibility that the reactions going on are hydrogen only cannot be ignored with the production of copper as only one of many reactions going on. On Thu, Oct 27, 2011 at 3:21 PM, Horace Heffner hheff...@mtaonline.netwrote: This is a nonsensical argument. The less hydrogen available for nuclear reactions the *more* the MeV per reaction that is required to make the 1 MW output, thus the less effective any shielding would be, and the *less credible* it is that the MW heat comes from nuclear reactions. On Oct 27, 2011, at 11:14 AM, Axil Axil wrote: There are some ifs and buts associated with this subject. It has been known for over a hundred years how that hydrogen will defuse through a hot metal enclosure. The rate of diffusion is subject to the temperature and pressure of the hydrogen, together with the exact kind, thickness, and temperature of the metal. These are all variables in the calculation of the diffusion rate. Furthermore, the presence of oxides and/or carbides on the surface of the metal can reduce the rate of diffusion of hydrogen by up to 5 orders of magnitude. We don’t know for sure what the accurate values of some of these variables are and additionally they would vary widely within an operational range throughout the operational lifetime of the E-Cat. However, since hydrogen is very slippery and notoriously hard to contain, a good guess can be made that most of the hydrogen consumed by the Rossi reactor would be lost through diffusion through the hot walls of the stainless steel reaction vessel. Because of all these large uncertainties, calculation of the nuclear reaction rates as a function of hydrogen consumption implying a clue to the nuclear processes going on inside the E-Cat reaction vessel cannot be made in my opinion. With best regards, Axil On Thu, Oct 27, 2011 at 7:48 AM, Horace Heffner hheff...@mtaonline.netwrote: From: http://www.rossilivecat.com/ Quote: - - - - - - - - - - - - - - - - - - - - Andrea Rossi October 25th, 2011 at 4:59 PM Dear Thomas Blakeslee: Grams/Power for a 180 days charge Hydrogen: 18000 g Nickel: 1 g Warm Regards, A.R. - - - - - - - - - - - - - - - - - - - - End quote. At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an atoommic weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This means 10.48 atoms of H need be provided per 1 atom of Ni. Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms of H is consumed, thus 170.4 mols of H and a170.4 mols of Ni are consumed, maximum. This involves the obviously wrong assumption that all the Ni atoms are transmuted, not a more realistic 3 percent. There is also an outside possibility the H reacts with daughter products, giving the possibility of 10 subsequent daughter reactions per primary Ni+H reaction. Three such reactions is an outside possibility. One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is (6.241x10^24 eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms /mol) = 9.464x10^5 eV/(Ni atom). If there is one reaction per atom and all Ni is consumed by single reactions than that is 0.9464 MeV per Ni-H event. The gammas from this would be lethal at short range, even through 2 cm of lead. If it is assumed that 3% of the Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 MeV per reaction. If there are an average of 3 daughter reactions per primary reactions that is about 10 Mev per reaction. If 10 MeV gammas are produced then 5 cm of lead shielding will be of no use in protecting the operators. If near 1 MeV gammas are produced the lead shielding is inadequate. One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas, 6.24x10^18 gammas per second. using: I = I0 * exp(-mu * rho * L) where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/cm^3, we have for 5 cm of lead: I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *(5 cm)) I = 2x10^18 free gammas per second. About half that, or 10^18 gammas/s would be directed toward the interior of the container housing the E-cats, and most of the 2x10^18 gammas per second would end up escaping the container. This is an approximate calculation. Even if it is off by an order of magnitude, this kind of 1 MeV gamma flux,
Re: EXTERNAL: [Vo]:Rossi H and Ni consumption
On Oct 27, 2011, at 4:49 AM, Roarty, Francis X wrote: On Thurs Oct 27, 2011 Horace said [snip] It does not seem credible the energy from a Ni-H reaction, at least in the form of one gamma per reaction, provides any explanation for 1 MW of heat, if that thermal power is in fact achieved.[/snip] Horace, Assuming the thermal power is in fact achieved, and the reaction is not Ni-H, what do you feel is the next most credible theory ? Fran A Ni-H or even p-e-p nuclear interaction catalyzed by a Ni nucleus is not ruled out given there is a mechanism to disperse the nuclear energy in small increments and avoid radioactive products. I think the reaction begins with a Ni electron being momentarily delayed in the Ni nucleus in a deflated state interaction with a proton or quark, as defined here: http://www.mta online.net/~hheffner/FusionUpQuark.pdf http://mtaonline.net/~hheffner/DeflateP1.pdf This provides the Ni nucleus with a very large magnetic moment, and magnetic gradient, which permits it to be the target of tunneling of deflated state hydrogen from the lattice. This results in multiple hydrogen nuclei present in the Ni nucleus, and a highly de-energized Ni-H deflated nucleus cluster, with multiple trapped electrons which then radiate energy or transfer it directly to k-shell electrons via near field interactions. Various apparently non-radioactive products are thereby made feasible. Non-radioactive products are the branches nature prefers because they are the least energy products. It is notable that no nuclear reaction may result from a given Ni-H deflated cluster, and yet nuclear heat, in the form of zero point energy, is released and then replenished by the zero point field after the cluster breaks up. See: http://mta online.net/~hheffner/NuclearZPEtapping.pdf Discussion of this could be very academic if there is in fact no excess heat from the Rossi experiments. I am hoping to write a FAQ on deflation fusion, but have not had the time. I will be happy to discuss this at a later time. Best regards, Horace Heffner http://www.mta online.net/~hheffner/
Re: [Vo]:Rossi H and Ni consumption
You are off on a tangent. My point is that Rossi's claims are in conflict with the observed results. I will no longer respond for now. On Oct 27, 2011, at 12:15 PM, Axil Axil wrote: In the Miley presentation that he has recently released, Miley shows transmutation to 39 isotopes over possible contamination levels. The nuclear reactions and transmutation patterns that are going on inside the Rossi reactor are similar to what Miley documents as mentioned in Rossi’s original patent. The presence of a large amount of iron in the Miley results is interesting and similar iron contamination was found in the Rossi ash(10%) when they were analyzed by the swedes. The assumption that the nuclear reactions taking place in the Rossi reaction are exclusively restricted to copper transmutation is mistaken in my opinion. The possibility that the reactions going on are hydrogen only cannot be ignored with the production of copper as only one of many reactions going on. On Thu, Oct 27, 2011 at 3:21 PM, Horace Heffner hheff...@mtaonline.net wrote: This is a nonsensical argument. The less hydrogen available for nuclear reactions the *more* the MeV per reaction that is required to make the 1 MW output, thus the less effective any shielding would be, and the *less credible* it is that the MW heat comes from nuclear reactions. On Oct 27, 2011, at 11:14 AM, Axil Axil wrote: There are some ifs and buts associated with this subject. It has been known for over a hundred years how that hydrogen will defuse through a hot metal enclosure. The rate of diffusion is subject to the temperature and pressure of the hydrogen, together with the exact kind, thickness, and temperature of the metal. These are all variables in the calculation of the diffusion rate. Furthermore, the presence of oxides and/or carbides on the surface of the metal can reduce the rate of diffusion of hydrogen by up to 5 orders of magnitude. We don’t know for sure what the accurate values of some of these variables are and additionally they would vary widely within an operational range throughout the operational lifetime of the E-Cat. However, since hydrogen is very slippery and notoriously hard to contain, a good guess can be made that most of the hydrogen consumed by the Rossi reactor would be lost through diffusion through the hot walls of the stainless steel reaction vessel. Because of all these large uncertainties, calculation of the nuclear reaction rates as a function of hydrogen consumption implying a clue to the nuclear processes going on inside the E- Cat reaction vessel cannot be made in my opinion. With best regards, Axil On Thu, Oct 27, 2011 at 7:48 AM, Horace Heffner hheff...@mtaonline.net wrote: From: http://www.rossilivecat.com/ Quote: - - - - - - - - - - - - - - - - - - - - Andrea Rossi October 25th, 2011 at 4:59 PM Dear Thomas Blakeslee: Grams/Power for a 180 days charge Hydrogen: 18000 g Nickel: 1 g Warm Regards, A.R. - - - - - - - - - - - - - - - - - - - - End quote. At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an atoommic weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This means 10.48 atoms of H need be provided per 1 atom of Ni. Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms of H is consumed, thus 170.4 mols of H and a170.4 mols of Ni are consumed, maximum. This involves the obviously wrong assumption that all the Ni atoms are transmuted, not a more realistic 3 percent. There is also an outside possibility the H reacts with daughter products, giving the possibility of 10 subsequent daughter reactions per primary Ni+H reaction. Three such reactions is an outside possibility. One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is (6.241x10^24 eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms /mol) = 9.464x10^5 eV/(Ni atom). If there is one reaction per atom and all Ni is consumed by single reactions than that is 0.9464 MeV per Ni-H event. The gammas from this would be lethal at short range, even through 2 cm of lead. If it is assumed that 3% of the Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 MeV per reaction. If there are an average of 3 daughter reactions per primary reactions that is about 10 Mev per reaction. If 10 MeV gammas are produced then 5 cm of lead shielding will be of no use in protecting the operators. If near 1 MeV gammas are produced the lead shielding is inadequate. One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas, 6.24x10^18 gammas per second. using: I = I0 * exp(-mu * rho * L) where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/ cm^3, we have for 5 cm of lead: I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) * (5 cm)) I = 2x10^18 free gammas per second. About half that, or 10^18 gammas/s would be directed toward the interior of the container
