erblo wrote:Are you sure? It's been over two years since i took a course in nuclear physics, but I thought that one can simply look at the mass before and after the reaction: According to
WolframAlpha the mass of one Ni-62 plus one H is 0.0066u (or 6.15MeV) larger than that of one Cu-63. This is only a rounding error from the 6.12MeV they claim to get out of the reaction.
KitemanSA wrote:
Code: Select all
63Cu 62.9295975
62Ni 61.9283451
----------
delta 1.0012524
H 1.0078250
Looks like
SOME mass is lost which should mean SOME energy produced, no?
I regret to say that
I believe this is all a highly naive argument. Perhaps, even, the one that the researchers have gone along with, without thinking it through.
I'll run this from two different angles:
1) The binding energy of 62Ni and 63Cu are both around 8.7MeV/nucleon. If you throw in a proton into the mix (a single nucleon) and the total extra 'mass-energy' you are adding is 6.15MeV, then why do you think you'd suddenly be able to make a new atom where each of the nucleons has 8.7MeV worth of binding energy? It's a bit like a group of people going to an event paying $8.70 per ticket. An extra comes along and says to the cashier 'I've got an extra $6.15 to get in with this group'. Is the cashier likely to a) let you in, b) turn you away? That group is
bound by their contribution of $8.70 each to the party fund. An upstart proton that blags its way in with $6.15 isn't going to be welcome.
2) Let us say, just for the sake of argument, that the proton does manage to muscle its way into the 62Ni, endothermically or by physical processes not yet known [that run against thermodynamics!]. Then we'd have an excited 63Cu. Let us say it has an excess energy of 6.15MeV. ..... What does it do next?
Q1: Is this 63Cu 'hot'?
A: No. It was a reaction that occurred in the inertial frame of the collision, so it isn't going anywhere. It is a 'cold' excited atom that is undetectable by ''thermal' measurements, unless the energy is released from the atom, somehow. So... no heat.
{Q1a: Why does fusion generate heat?
A: Because there are fusion products that take that excess energy away as kinetic energy [which manifests itself as 'heat', in ensembles of such fast, kinetic, products].}
Q2: What would a 63Cu do with this extra 6.15MeV?...
A: Well, I am guessing that the
thermodynamic thing to do would be to chuck the proton out again. I do not know for sure, but I think you will find that the energy necessary for an excited 63Cu to shed a proton is 6.15MeV! [I say that, mainly because of the principle of (1) above, but don't have those exact figures to confirm that.] But let us just still hold on there with a glimmer of hope that this 63Cu does something else;
a) it can undergo a strong-force mediated outcome; this means that it will chuck out nucleons. As 62Ni is the bottom of the pile, we know immediately that if it chucks out anything other than that extra proton, then it is going to be an endothermic release of nucleons,
b) it can undergo a weak-force mediated reaction; this means that either a proton gives out a positron and turns into a neutron, becoming 63Ni, or a neutron gives up an electron giving 63Zn. Without even resorting to calculating those energy equations, I am sure those all work out endothermic [again... because 62Ni is already the bottom of the energy barrel].
c) it can undergo an electromagnetic-force mediated reaction; OK, so now we have something that ....er.... could be argued. A 6.15 MeV photon emission would [I presume, not really thought about it too hard] naively gives an energy balance. If so, then it will be very easy to detect 6.15MeV gammas....so...show me the Geiger tube readings while this thing is running...... oh, and also, if this is
the reaction, then there will be no heat.....
heat will merely demonstrate it is not an EM mediated reaction!
So, in summary; I see no thermodynamic reason for the proton to enter the 62Ni, and that even if it were there, there is no known nuclear mechanism by which heat would be released.