"Carey Sublette" <careysub@earthling.net> wrote in message news:<7Lw7b.5876$Yt.5199@newsread4.news.pas.earthlink.net>...

Can anyone provide a reasonable (or at least plausible) explanation for the
fact that even-even nuclides (e.g. U-238) have higher rates for spontaneous
fission than even-odd nuclides (U-235)? For quantum-mechanical reasons, the
rate SF rate must mean a lower threshold barrier to this process. But
even-even nuclides have higher binding energies, and higher required
excitation energies for normal fission.

So, why the higher SF rate?

Carey Sublette

I looked at the nuclides for 20 years and the rate of spontaneous
fission in relation to the even-odd relation of the chart is a
stability in relation to the chart.

Meaning the applied nucleon theory of the atom as the chart, is
independent of the stability theory.

Making the hydrogen spontaneous fission rate a means to resolve
your question. And so a search for the prediction of a stability
would then begin.

And so it is one of the tough theory concept ideas to ponder.

strech, stretch......

Meaning a means to alter the spontaneous rate is obvious given the
lack of theory and a beautiful chart symmetry. And so to cause
the atom stability for self fission, as the abstract chart appears
to allow, is a simple conclusion?

So make the abstract nuclide chart an atom in the particular,
and the meaning of a mass alteration appears the sole cause.

Just add some even/odd mass to the hydrogen atom to alter the rate!!!!

Implying a crazy idea of figuring out the even/odd mass in the abstract
also.

What is it? A nucleon alteration of a special kind. So ponder
a while.

And so the, "this stinks" answer, appears again in need of a quantum
theory.

Alteration of the quanta by mass addition, causes another state's
alteration, and not the one undergoing spontaneous fission.

So take the hydrogen atom and fuse on many masses and make
a new chart!! All experiment until the same relation of
the even/odd is found!!!

I have to work on it for a while. Another posting for my answer I guess.

Douglas Eagleson
Gaithersburg, MD USA

"Carey Sublette" <careysub@earthling.net> wrote in message

Can anyone provide a reasonable (or at least plausible) explanation for

fact that even-even nuclides (e.g. U-238) have higher rates for

fission than even-odd nuclides (U-235)? For quantum-mechanical reasons,

rate SF rate must mean a lower threshold barrier to this process. But
even-even nuclides have higher binding energies, and higher required
excitation energies for normal fission.

Well, note that when you do "normal" fission case, you're transfering

AND A NEUTRON into the nucleus, so the energy to cross the fission barrier
comes from BOTH the energy of the incident neutron, and the difference in
binding energy with the neutron added (exothermic for odd -> even,

for even->odd). You then have to compare this to the threshold for the

nucleus, which has opposite parity.

So consider the hypothesis that even nuclei have higher SF rates due to

barriers. What does this predict for induced fission?

gs = groundstate
E = energy of incident neutron
e = binding energy change due to pairing

neutron (E) + even (gs) --> odd (E - e) high barrier
neutron (E) + odd (gs) --> even (E + e) low barrier

so the odd nucleus gets an extra energy "kick" from the spin pairing, plus
forms a configuration with a lower barrier. This allows the nucleus to
fision substantially easier. So we expect odd nuclei to have

lower fision thresholds (which they do).

So the even/odd barrier difference does seem to explain everything you
mentioned. The next question is, why is the even barrier lower than the
odd barrier, when the even nucleus is more stable?

First, I'll explain why one does not expect the stability alone to
substantially influence the barrier height. Since the reaction is highly
exothermic, the barrier height should not depend strongly on the binding
energy, since it has what a chemist would refer to as an "early"

state. In other words, the the higher stability of Pu-240 vs Pu-239 due
to spin pairing also substantially stabilizes the configuration at the top
of the barrier, causing the barrier height and width to be similar. So I
don't see that there is any reason based on binding energy to expect

to have a *lower* SF rate than Pu-239. The energy release for fission is
just too substantial for relatively minor energy differences in the

states to substantially affect the reaction profile.

On the other hand, I don't know of a good reason for it to have a *higher*
SF rate either, as we know it does. It may be that the stabilization due

spin pairing is even more important for the transition state than it is

the ground state nuclei, leading to a (somewhat) lower fision barrier for
even nuclei relative to odds ones.

But I'm speculating.

"Carey Sublette" <careysub@earthling.net> wrote in message news:<7Lw7b.5876$Yt.5199@newsread4.news.pas.earthlink.net>...

Can anyone provide a reasonable (or at least plausible) explanation for the
fact that even-even nuclides (e.g. U-238) have higher rates for spontaneous
fission than even-odd nuclides (U-235)? For quantum-mechanical reasons, the
rate SF rate must mean a lower threshold barrier to this process. But
even-even nuclides have higher binding energies, and higher required
excitation energies for normal fission.

Well, note that when you do "normal" fission case, you're transfering energy
AND A NEUTRON into the nucleus, so the energy to cross the fission barrier
comes from BOTH the energy of the incident neutron, and the difference in
binding energy with the neutron added (exothermic for odd -> even, endothermic
for even->odd). You then have to compare this to the threshold for the NEW
nucleus, which has opposite parity.

So consider the hypothesis that even nuclei have higher SF rates due to lower
barriers. What does this predict for induced fission?

gs = groundstate
E = energy of incident neutron
e = binding energy change due to pairing

neutron (E) + even (gs) --> odd (E - e) high barrier
neutron (E) + odd (gs) --> even (E + e) low barrier

so the odd nucleus gets an extra energy "kick" from the spin pairing, plus
forms a configuration with a lower barrier. This allows the nucleus to
fision substantially easier. So we expect odd nuclei to have substantially
lower fision thresholds (which they do).

So the even/odd barrier difference does seem to explain everything you
mentioned. The next question is, why is the even barrier lower than the
odd barrier, when the even nucleus is more stable?

First, I'll explain why one does not expect the stability alone to
substantially influence the barrier height. Since the reaction is highly
exothermic, the barrier height should not depend strongly on the binding
energy, since it has what a chemist would refer to as an "early" transition
state. In other words, the the higher stability of Pu-240 vs Pu-239 due
to spin pairing also substantially stabilizes the configuration at the top
of the barrier, causing the barrier height and width to be similar. So I
don't see that there is any reason based on binding energy to expect Pu-240
to have a *lower* SF rate than Pu-239. The energy release for fission is
just too substantial for relatively minor energy differences in the initial
states to substantially affect the reaction profile.

On the other hand, I don't know of a good reason for it to have a *higher*
SF rate either, as we know it does. It may be that the stabilization due to
spin pairing is even more important for the transition state than it is for
the ground state nuclei, leading to a (somewhat) lower fision barrier for
even nuclei relative to odds ones.

But I'm speculating.