Most Common Pu-239 Fission Chain Reaction?
Mike Fontenot
reaction of plutonium-239 is?
--
Mike Fontenot
glen herrmannsfeldt
> Anyone on this newsgroup know what the highest-probability
> fission chain reaction of plutonium-239 is?
You mean fission products?
I used to have a Chart of the Nuclides which had big pink
and blue areas where the U235 and Pu239 fission products were.
The range is pretty broad, but many have very short half-lives,
so the ones you actually see are the longer lived decay
products of the original fragments.
According to wikipedia, the exact products depenend on the
initiating neutron energy. See the Nuclear_fission_product
article.
-- glen
Mike Fontenot
> Mike Fontenot<mlf...@comcast.net> wrote:
>
>> Anyone on this newsgroup know what the highest-probability
>> fission chain reaction of plutonium-239 is?
>
> You mean fission products?
>
with the "yield tables" and "yield charts": they don't directly show the
most likely PAIR of nuclei that results from a single fission of a
single Pu-239 nucleus. They also aren't usually limited to the
IMMEDIATE products of a fission event ... I'm interested in determining
the most likely IMMEDIATE pair of nuclei produced in the fissioning of a
Pu-239 nucleus (together with the number of neutrons emitted, and
whether those neutrons have energies capable of producing further Pu_239
fissions).
I think the question, "What is the most probable immediate reaction when
Pu-239 absorbs a neutron, and splits into two large nuclei and a few
neutrons that are capable of producing a chain reaction?", is
well-defined, and should have a simple answer. I'm surprised, given the
importance of Pu-239 fission, that the answer is apparently not widely
known.
>
> According to wikipedia, the exact products depend on the
> initiating neutron energy.
Yes, I've heard that only relatively low-energy neutrons can initiate
fission of U-235 or Pu-239. But a missing capability, for me, is
determining IF the neutrons emitted in any proposed reaction are in the
required energy range.
For example, I came up with what I thought MIGHT be a likely reaction
for Pu-239, because it should result in a relatively large amount of
energy being released (and it produces an additional emitted neutron
beyond the minimum required for a chain reaction):
94_Pu_239 + 1n -> 58_Ce_148 + 36_Kr_89 + 3n.
(I arrived at the above possible reaction by starting with a similar
reaction for U-235 that I had seen on the internet somewhere, and then
"tweaked" it for Pu-239 instead of U-235, trying to maximize the
resulting energy release).
But what I don't know is, do those three emitted neutrons have
appropriate energies to produce a chain reaction or not. Also, I'm not
sure that it follows that the most likely reaction is necessarily the
reaction with the highest energy release.
Thanks for your response, Glen ... I appreciate it.
--
Mike Fontenot
glen herrmannsfeldt
> On 05/24/2011 09:37 PM, glen herrmannsfeldt wrote:
>> Mike Fontenot<mlf...@comcast.net> wrote:
>>> Anyone on this newsgroup know what the highest-probability
>>> fission chain reaction of plutonium-239 is?
>> You mean fission products?
> No, I want the specific REACTION. That's the problem I've encountered
> with the "yield tables" and "yield charts": they don't directly show the
> most likely PAIR of nuclei that results from a single fission of a
> single Pu-239 nucleus. They also aren't usually limited to the
> IMMEDIATE products of a fission event ... I'm interested in determining
> the most likely IMMEDIATE pair of nuclei produced in the fissioning of a
> Pu-239 nucleus (together with the number of neutrons emitted, and
> whether those neutrons have energies capable of producing further Pu_239
> fissions).
I believe the usual meaning of "fission products" is the immediate pair,
but in some cases it might also mean their decay products. If, for
example, one is interested in spent fuel rods, then decay products
are important.
> I think the question, "What is the most probable immediate reaction when
> Pu-239 absorbs a neutron, and splits into two large nuclei and a few
> neutrons that are capable of producing a chain reaction?", is
> well-defined, and should have a simple answer. I'm surprised, given the
> importance of Pu-239 fission, that the answer is apparently not widely
> known.
>> According to wikipedia, the exact products depend on the
>> initiating neutron energy.
> Yes, I've heard that only relatively low-energy neutrons can initiate
> fission of U-235 or Pu-239. But a missing capability, for me, is
> determining IF the neutrons emitted in any proposed reaction are in the
> required energy range.
No, high energy neutrons also initiate fission. (That is important
in bombs.) In reactors, the moderator is needed to reduce the
absorption of neutrons by U238.
> For example, I came up with what I thought MIGHT be a likely reaction
> for Pu-239, because it should result in a relatively large amount of
> energy being released (and it produces an additional emitted neutron
> beyond the minimum required for a chain reaction):
> 94_Pu_239 + 1n -> 58_Ce_148 + 36_Kr_89 + 3n.
>From the graph in wikipedia Nuclear_fission_products, it looks
like the peaks are at A=103 and 134. The curves are complicated
by the preference for even Z and N.
> (I arrived at the above possible reaction by starting with a similar
> reaction for U-235 that I had seen on the internet somewhere, and then
> "tweaked" it for Pu-239 instead of U-235, trying to maximize the
> resulting energy release).
It seems that the peaks move closer for either higher incident
neutron energy or higher mass of fissioning nuclide.
> But what I don't know is, do those three emitted neutrons have
> appropriate energies to produce a chain reaction or not. Also, I'm not
> sure that it follows that the most likely reaction is necessarily the
> reaction with the highest energy release.
It takes no energy from the incoming neutron. Or, to put it another
way, there is an attractive (nuclear) force on the neutron.
One of the more interesting (especially in power reactors) fission
poducts (or fission product decay products) is Xe-135. (It has its
own wikipedia page.) Xe-135 has a neutron cross section of about
2 million barns, or about 2 million times that of U-238.
That is, if a neutron gets even close to an Xe-135 nucleus it
can get sucked in by the nuclear force. That is the same way
that neutrons start fission.
It is also an important part of the cause of the Chernobyl exposion.
-- glen
Mike Fontenot
>
> No, high energy neutrons also initiate fission. (That is important
> in bombs.)
>
Yes, you are right ... I had a fundamental misunderstanding about that.
I finally found a wiki webpage (on neutron moderators) that clears up a
lot of my misconceptions. Although it IS true that a high-energy
neutron, for a GIVEN encounter with a Pu-239 nucleus, is MUCH less
likely (than a low-energy neutron) to be absorbed (and thereby cause a
fission), it is NOT true that high-energy neutrons are useless as
producers of fission. Early in the Manhattan project, it was concluded
that in bombs, high-energy neutrons actually result in a faster (and
thus more complete) chain reaction. The plutonium pit is surrounded by
beryllium metal, which reflects the high-energy neutrons. So the
neutrons bounce back and forth through the pit, and even though the
probability that they cause fission on ANY GIVEN encounter with a Pu-239
nucleus is very low, they have a huge number of encounters in a very
short time. The result is that the chain reaction goes much faster than
it would with low-energy neutrons.
It was also clear from that webpage that essentially ALL neutrons that
are emitted by a fissioning Pu-239 nucleus are high-energy neutrons, no
matter what the particular reaction happens to be. So I don't NEED to
determine the energy of the emitted neutrons in my prospective reaction,
in order to determine if that reaction is capable of a chain reaction.
--
Mike Fontenot
Mike Fontenot
> From the graph in wikipedia Nuclear_fission_products, it looks
> like the peaks are at A=103 and 134. The curves are complicated
> by the preference for even Z and N.
>
But I don't think that actually implies that, for a particular common
REACTION, that the two large nuclei produced have mass numbers of 103
and 134. And even if it DID imply that, knowledge of the two mass
numbers doesn't uniquely specify the particular isotopes (even for the
case when there are no immediate proton conversions (conservation of
Z)). That's the problem: the yield curves don't specify REACTIONS ...
there is no direct correlation information between the two peaks ... no
direct information about the two specific products produced in a given
reaction.
If the curve really concerns ONLY immediate products, then I suppose
that it may indicate that the two products in the most likely reactions
DO usually have mass numbers in the vicinity of 103 and 134. If so,
that means that my prospective reaction is probably NOT very likely ...
i.e., that there are other factors that determine the likelihood of a
given reaction, beyond just the amount of energy produced.
Some of the yield curves I've seen were labeled as "after 1 year", or
"after 2 years", etc. I HAVE seen the term "prompt products" used as a
label for a yield curve, but I'm still not sure that "prompt" really
means "immediate".
--
Mike Fontenot
glen herrmannsfeldt
(snip, I wrote)
>> From the graph in wikipedia Nuclear_fission_products, it looks
>> like the peaks are at A=103 and 134. The curves are complicated
>> by the preference for even Z and N.
> But I don't think that actually implies that, for a particular common
> REACTION, that the two large nuclei produced have mass numbers of 103
> and 134. And even if it DID imply that, knowledge of the two mass
> numbers doesn't uniquely specify the particular isotopes (even for the
> case when there are no immediate proton conversions (conservation of
> Z)). That's the problem: the yield curves don't specify REACTIONS ...
> there is no direct correlation information between the two peaks ... no
> direct information about the two specific products produced in a given
> reaction.
I am not an expert, or even close, in this. It does seem that I
am the only one replying. I haven't thought about this for a while,
but I believe that it really is mostly or only mass that counts.
To first order, the force between nucleons is the same, protons
or neutrons.
I do agree that it doesn't mean that the two peaks necessarily
come out together, but it does seem likely.
> If the curve really concerns ONLY immediate products, then I suppose
> that it may indicate that the two products in the most likely reactions
> DO usually have mass numbers in the vicinity of 103 and 134. If so,
> that means that my prospective reaction is probably NOT very likely ...
> i.e., that there are other factors that determine the likelihood of a
> given reaction, beyond just the amount of energy produced.
> Some of the yield curves I've seen were labeled as "after 1 year", or
> "after 2 years", etc. I HAVE seen the term "prompt products" used as a
> label for a yield curve, but I'm still not sure that "prompt" really
> means "immediate".
In fission, prompt neutrons are those used for bombs. Some of the
fission fragments also give delayed neutrons, useful in reactors
but not bombs. Besides delayed neutrons, beta decay is the usual
decay mode for fission fragments, which doesn't change the mass (Z+N).
-- glen
glen herrmannsfeldt
(snip)
> I finally found a wiki webpage (on neutron moderators) that clears up a
> lot of my misconceptions. Although it IS true that a high-energy
> neutron, for a GIVEN encounter with a Pu-239 nucleus, is MUCH less
> likely (than a low-energy neutron) to be absorbed (and thereby cause a
> fission), it is NOT true that high-energy neutrons are useless as
> producers of fission.
Well, it is pretty much the wave nature of quantum mechanics.
Without that, the cross section would be pretty much the cross
sectional area of an appropriate sized sphere. For U238, which
isn't especially interested in neutrons, the neutron absorption
cross section is pretty much the physical size of the nucleus.
The thermal (low energy) neutron cross section of U238 is 2.683 barns,
so about the size of the nucleus. For Xe135 is is 2665000 barns,
many times the size of the nucleus. That happens as a thermal
neutron has a long wavelength, and so also large uncertainty in
position. And Xe135 very much wants another neutron.
-- glen
glen herrmannsfeldt
> Anyone on this newsgroup know what the highest-probability
> fission chain reaction of plutonium-239 is?
OK, if you look at http://www.nndc.bnl.gov/chart/chartNuc.jsp
is it all there.
At the top, click on the 239Pu FY, which is fission yeild for Pu239.
The color then indicates visually the fission yield.
Dark red are the higher yield nuclides. If you click on one, then
it will show the data at the bottom, including the yield fraction.
You can also zoom, and move. When you get to zoom number 1, the
numbers appear, including, if selected above, Pu239 FY.
I135 is 0.043, which is pretty high.
-- glen
Mike Fontenot
That IS a good chart! Thanks.
It looks like my prospective reaction (yielding 58_Ce_48 and 36_Kr_89)
is not very likely (around 0.009 in yield). 52_Te_134 had about the
highest yield I saw (about 0.44, as I recall).
It would STILL be nice, though, if the chart DIRECTLY showed the most
likely OTHER nuclide, i.e., if the chart directly gave information about
the PAIR of nuclei produced by a particular REACTION, together with the
likelihood of that happening. It's probably possible to INFER that the
most likely other nucleus produced with 52_Te_134 would perhaps be
42_Mo_104, plus two neutrons ... its yield is similar to 52_Te_134.
42_Mo_103 would give three neutrons, but has quite a bit lower yield.
I AM starting to understand that the fraction of the nucleons that are
protons, in each of the nuclide products, is actually not very important
in the nuclear properties (as opposed to the chemical or radioactive
properties). So, once you specify the mass number A1 for one of the two
products, the most important additional parameter is perhaps the number
of neutrons emitted: once you specify that, then A2 is fixed, and what
the two Z numbers of the products are is perhaps of secondary importance
(except in issues of the stability and corrosiveness of the products, of
course). So maybe what would be nice to see would be a chart giving the
(say) three most likely A1 values (where A1 is the mass number of the
heavier of the two nuclides produced), and for each of them, show the
corresponding A2 values (together with their likelihoods, and total
energies released), for zero, one, two, or three emitted neutrons.
I think one BIG outage, in my understanding, is why the emitted neutrons
are always high-energy neutrons. It's pretty understandable why the two
nuclides would be flying apart, given the electrostatic repulsion of
those two closely-spaced clumps of protons, but not so obvious for
neutrons. I think my next step is to spend a lot more quality time with
one of my nuclear physics books.
Thanks again for your help.
--
Mike Fontenot
Mike Fontenot
In my last posting, I left off a zero in the 52_Te_134 yield: it SHOULD
have been 0.044, rather than 0.44 ... sorry for the error.
I SHOULD have said
"It looks like my prospective reaction (yielding 58_Ce_48 and
36_Kr_89) is not very likely (around 0.009 in yield). 52_Te_134 had
about the highest yield I saw (about 0.44, as I recall)."
--
Mike Fontenot
Mike Fontenot
I'm piling errors on top of errors ... some days are like that! My
correction SHOULD have said:
"It looks like my prospective reaction (yielding 58_Ce_48 and
36_Kr_89) is not very likely (around 0.009 in yield). 52_Te_134 had
about the highest yield I saw (about 0.044, as I recall)."
(Note to Charles Francis: Feel free to correct my original error, if
you'd like, and delete these subsequent correction postings, if you get
them all in time).
--
Mike Fontenot
brad
Try Googling this..." Nucleonics, April, 1958, Fission-product Yields
from U,Th,and Pu"
Or go to a library for that issue. Find graph inside for U, Th, Pu
fission yields.
Brad
glen herrmannsfeldt
(snip)
> It would STILL be nice, though, if the chart DIRECTLY showed the most
> likely OTHER nuclide, i.e., if the chart directly gave information about
> the PAIR of nuclei produced by a particular REACTION, together with the
> likelihood of that happening. It's probably possible to INFER that the
> most likely other nucleus produced with 52_Te_134 would perhaps be
> 42_Mo_104, plus two neutrons ... its yield is similar to 52_Te_134.
> 42_Mo_103 would give three neutrons, but has quite a bit lower yield.
I am not sure how the chart is made, so I don't know if they know
that or not. It might also be that we aren't supposed to know
exactly how many prompt neutrons come out.
> I AM starting to understand that the fraction of the nucleons that are
> protons, in each of the nuclide products, is actually not very important
> in the nuclear properties (as opposed to the chemical or radioactive
> properties). So, once you specify the mass number A1 for one of the two
> products, the most important additional parameter is perhaps the number
> of neutrons emitted: once you specify that, then A2 is fixed, and what
> the two Z numbers of the products are is perhaps of secondary importance
> (except in issues of the stability and corrosiveness of the products, of
> course). So maybe what would be nice to see would be a chart giving the
> (say) three most likely A1 values (where A1 is the mass number of the
> heavier of the two nuclides produced), and for each of them, show the
> corresponding A2 values (together with their likelihoods, and total
> energies released), for zero, one, two, or three emitted neutrons.
I believe so. The distribution among those with a given mass
should be about what you get taking a random scoop out of a
bag that is 39% protons. That is, binomial distribution.
I didn't check the numbers, though, to see if that agrees.
> I think one BIG outage, in my understanding, is why the emitted neutrons
> are always high-energy neutrons. It's pretty understandable why the two
> nuclides would be flying apart, given the electrostatic repulsion of
> those two closely-spaced clumps of protons, but not so obvious for
> neutrons. I think my next step is to spend a lot more quality time with
> one of my nuclear physics books.
I think it isn't so obvious why they would not have (somewhat)
high energy. The two fragments fly apart at very high speed.
It seems most unlikely that some neutrons would just sit there.
Or, consider that they might stick to one fragment just a little
longer, in which case they have about the speed of that fragment.
The CM frame is special to us, but not so special to the neutrons.
-- glen
Tom Roberts
> I think one BIG outage, in my understanding, is why the emitted neutrons
> are always high-energy neutrons.
It's a combination of nuclear physics and kinematics. The strong force holding
the nucleus together is MUCH stronger than the electrostatic repulsion [#]. And
it is basic kinematics that when an object breaks up into a combination of
low-mass and high-mass secondaries, most of the energy goes into the low-mass
secondaries. This is rigorous for a 2-body decay, but remains statistically true
for multi-body decays (such as nuclear fissions) -- the low-mass secondaries
have much more phase space available to them than the large-mass ones, and most
of their phase space has higher energy.
[#] Think about it -- the nucleus does remain together, with a large
net positive charge. In view of quantum tunneling, the ratio must
be huge.
> It's pretty understandable why the two
> nuclides would be flying apart, given the electrostatic repulsion of
> those two closely-spaced clumps of protons, but not so obvious for
> neutrons.
The electrostatic repulsion between the large fission fragments is almost
negligible compared to the strong interaction that caused the fission in the
first place.
BTW the reason there is no obvious listing of "the most probable fission
reactions" is that there are so many of them that no single channel dominates.
Moreover, for most purposes it is the distribution of nuclear fragments that is
of interest (e.g. for determining the evolution of the nuclear fuel in a reactor
over periods of months or years).
Note that in reactors, U238 capture of neutrons is important, because a) it
reduces the neutron flux for fissions, and b) it decays U239->Np239->Pu239 which
is long-lived and fissionable, providing more fuel. Remember that a typical
reactor fuel rod has U235/U238 ~ 5%, so there is a lot of U238 present. Indeed
breeder reactors surround the operating core with blankets of U238 to exploit
this, using the neutrons that escape from the core and would be otherwise
unused. Th232 can also be bred into fissionable fuel, U233.
The operation of a nuclear reactor is always balanced right at criticality, in
such a way that the amount of cooling determines the power output of the core,
an equilibrium that finds the critical point. Control rods are far too slow to
maintain this equilibrium, but they do set the operating point in conjunction
with the amount of cooling. It is an engineering miracle that these complicated
systems can be operated at all, much less operated so safely over so many years
(the failures are notable for their rarity).
Tom Roberts
Mike Fontenot
> " Nucleonics, April, 1958, Fission-product Yields
> from U,Th,and Pu"
> Or go to a library for that issue. Find graph inside for U, Th, Pu
> fission yields.
Thanks for that reference.
As far as I could tell, that article isn't online anywhere. But I'll
see if it's available at my local university library.
Do I understand you correctly that the referenced article, and that
graph, give information about the PAIR of nuclei immediately produced by
one or more of the relatively likely reactions? I.e., does it associate
a given lighter fragment with a given heavier fragment? The yield
information I've seen so far never seems to correlate the lighter and
heavier fragments ... they just show the relative probabilities among
the heavier fragments, and likewise for the lighter fragments, without
relating the two.
--
Mike Fontenot
glen herrmannsfeldt
> Mike Fontenot wrote:
>> I think one BIG outage, in my understanding, is why the emitted neutrons
>> are always high-energy neutrons.
> It's a combination of nuclear physics and kinematics.
> The strong force holding the nucleus together is MUCH stronger
> than the electrostatic repulsion [#]. And it is basic kinematics
> that when an object breaks up into a combination of low-mass and
> high-mass secondaries, most of the energy goes into the low-mass
> secondaries. This is rigorous for a 2-body decay, but remains
> statistically true for multi-body decays (such as nuclear fissions)
> -- the low-mass secondaries have much more phase space available
> to them than the large-mass ones, and most of their phase space
> has higher energy.
I believe, though, that it isn't quite that high. In the case of
alpha decay, the energy is inverse to mass, as the two have equal
(opposite) momentum and so more energy to the smaller mass.
For fission, it might be 200MeV, but most goes to the charged
fragments. The neutrons are in the low MeV range, which is still
much higher than the thermal (eV) range, but much lower than the
fission fragment energy.
(snip)
> The electrostatic repulsion between the large fission fragments is
> almost negligible compared to the strong interaction that caused
> the fission in the first place.
I thought Meitner's calculation was that the electrostatic repulsion
was just about enough to overcome the binding energy. Well, in
her liquid drop explanation, to overcome the surface tension.
But even so, it isn't that the neutrons are just sitting still
in the nucleus. They are fermions, and so have a fermi energy,
speeding, bouncing, around inside the nucleus.
> Control rods are far too slow to maintain this equilibrium,
> but they do set the operating point in conjunction with the
> amount of cooling. It is an engineering miracle that these
> complicated systems can be operated at all, much less operated
> so safely over so many years (the failures are notable for
> their rarity).
Well, for reactors there are delayed neutrons to keep the reaction
going at just below the prompt criticality level. But there
is also poisoning, such as by Xe135 (which I just learned about
in detail while following this thread), which complicates
the equilibrium. Not understanding Xe135 seems to be a
primary cause of the Chernobyl explosion.
-- glen
brad
> As far as I could tell, that article isn't online anywhere.
Elsevier?
> Do I understand you correctly that the referenced article, and that
> graph, give information about the PAIR of nuclei immediately produced by
> one or more of the relatively likely reactions?
Yes.
> a given lighter fragment with a given heavier fragment? The yield
> information I've seen so far never seems to correlate the lighter and
> heavier fragments ... they just show the relative probabilities among
> the heavier fragments, and likewise for the lighter fragments, without
> relating the two.
The graph shows mass nos. from 70 - 160 with dual % curves that
correlate
light and heavy yields by their appearances at the same height.
I assume this isn't more of what you've found already.
Brad
Tom Roberts
> Mike Fontenot wrote:
>> I think one BIG outage, in my understanding, is why the emitted
>> neutrons are always high-energy neutrons.
> It's a combination of nuclear physics and kinematics. The strong force
> holding the nucleus together is MUCH stronger than the electrostatic
> repulsion [#]. And it is basic kinematics that when an object breaks up
> into a combination of low-mass and high-mass secondaries, most of the
> energy goes into the low-mass secondaries. This is rigorous for a 2-body
> decay, but remains statistically true for multi-body decays (such as
> nuclear fissions) -- the low-mass secondaries have much more phase space
> available to them than the large-mass ones, and most of their phase
> space has higher energy.
I just realized another reason: what are called "slow" neutrons have kinetic
energies in the thermal range, a fraction of an eV. But the nuclear fission
releases on the order of 10 MeV in energy. So it's highly unlikely that any
secondary will have less than an eV of kinetic energy, because there is many
millions of times more energy available than that, to be shared among a handful
of secondaries.
Tom Roberts
Mike Fontenot
> BTW the reason there is no obvious listing of "the most probable fission
> reactions" is that there are so many of them that no single channel
> dominates.
I still think it's a well-defined question to ask "What is the most
probable reaction when Pu-239 fissions?". The distribution of possible
reactions IS apparently fairly broad, but it still has some maximum,
even though that probability may be fairly small.
But my original phrasing of the question WAS naive. I asked for the
most likely chain reaction. I've come to understand that it's not
possible to exert much (if any) control over which reaction you will get
in the fissioning of any particular Pu-239 nucleus, and it makes no
sense to imagine that there will be an entire chain reaction where every
Pu-239 breaks into the same two product nuclei. My original wording
really was motivated by the need to specify that I was only interested
in the most probable reaction, among all those reactions that produce at
least two emitted neutrons (so that they are capable of being the first
link in a chain reaction).
--
Mike Fontenot
Mike Fontenot
> On May 28, 3:16 pm, Mike Fontenot<mlf...@comcast.net> wrote:
>
>> As far as I could tell, that article isn't online anywhere.
>
> Elsevier?
>
I'm not familiar with that. From a quick look, I think I've got to be a
subscriber to get anything out of it. I think my local university
library may be my best bet.
>
> The graph shows mass nos. from 70 - 160 with dual % curves that
> correlate
> light and heavy yields by their appearances at the same height.
> I assume this isn't more of what you've found already.
>
Sounds different from anything I've seen so far ... I'll try to get a
copy. Thanks.
--
Mike Fontenot
Mike Fontenot
> Moreover, for most purposes it is the distribution of nuclear
> fragments that is of interest (e.g. for determining the evolution of the
> nuclear fuel in a reactor over periods of months or years).
>
Thanks for your response, Tom.
In my original question, I was interested strictly in the Pu-239 fission
in the pit of a bomb. So my interest was only in the immediate fission
products.
--
Mike Fontenot
glen herrmannsfeldt
(snip)
> I just realized another reason: what are called "slow" neutrons
> have kinetic energies in the thermal range, a fraction of an eV.
> But the nuclear fission releases on the order of 10 MeV in energy.
Closer to 200MeV.
> So it's highly unlikely that any secondary will have less than
> an eV of kinetic energy, because there is many millions of times
> more energy available than that, to be shared among a handful of
> secondaries.
Also, the energy of the neutron inside the nucleus is in the 10MeV
range. From Heisenburg uncertianty, its momentum is on the order
of hbar/d where d is the size of the nucleus. Then its energy
is p*p/(2m).
Neutrons pair up in the nucleus similar to the way electrons
pair up in atoms. Even Z and even N nuclides are more stable.
When U238 absorbs a neutron, it goes into a higher nuclear
energy level, but when U235 absorbs one it fills an existing
level. (Pairs with one already there.)
That, and the attraction of the nuclear force, supplies
much more energy to the nucleus of U235 than U238.
The pairing also affects the likely fission fragments, as
again even Z and even N are favored.
-- glen
Tom Roberts
> In my original question, I was interested strictly in the Pu-239 fission
> in the pit of a bomb. So my interest was only in the immediate fission
> products.
The criticality of a nuclear bomb or reactor is characterized by k_eff, the
effective ratio of neutrons available to generate new fissions per incident
neutron. Note k_eff depends on the geometry, the density, the composition of the
nuclear material, the neutron energy spectrum, etc.
In a bomb, the maximum possible value of k_eff is just over 2, the mean number
of neutrons emitted by a single fission, but it is always reduced by leakage and
other neutron losses. The chain reaction will not sustain itself unless k_eff >
1. If you only consider one reaction, with a probability of a few percent, you
have no hope of computing k_eff anywhere near that large (of course a real
system will ignore your restriction to just one reaction). You must consider the
entire system, including all of the different fission reactions. For bombs, you
need not worry about the beta decays and their effects on the fuel composition,
but for a reactor that is essential, as the system is engineered so their
delayed neutrons maintain the system right at k_eff = 1.
As I have said before, it is an engineering miracle that such systems can be
made to work. Because they are so complicated -- the Manhattan project only
scratched the surface, and yet were able to achieve three detonations without
any fizzles.
Tom Roberts
Ken S. Tucker
I needed a license to transport in NA radioactive isotopes to
calibrate
particle detectors. The instructor was a Nevada Test Site monitor when
A-bombs were atmospheric. What they used is plain vacuum cleaners on
peoples rooves to suck up the dust into filters.
The material in the filters was then analysed.
Sr 90 was was the main worry.
Decay modes (and times) in particle physics is still statistical.
In the 60's was a mess of Mesons and a hiatus of Hyperons,
with many probabilities of decay modes and time, measured in
clould and bubble chambers.
The exact specs of a Pu239 bomb are still secret.
Regards
Ken S. Tucker
t-sqrd
For fission yields see: http://ie.lbl.gov/fission.html
or for discussion on fission see IAEA http://www-nds.ipen.br/
(use link to fission yields)
The emitted neutrons are in a ~750Kev maxwell-boltzman distribution.
Higher energy projectile neutrons usually add the the number of
neutrons emitted rather than their temperature. The total fission
yield of ~ 180 Mev is mostly from coulomb replusion of the fragments.
tom
Ken S. Tucker
Yes that makes sense to me Tom.
Let a positive nucleus be designated as (+N), and a neutron as n,
then
(+N) + n -> (+A) + (+B) , the A B being the fragment results from
the n impacting.
The A and B, after braking the Strong Nucler Force Bond, will exhibit
Coulomb repulsion by
(+A) <= => (+B).
That's my basic understanding.
Regards
Ken S. Tucker
brad
> Anyone on this newsgroup know what the highest-probability fission chain
> reaction of plutonium-239 is?
>
> --
> Mike Fontenot
Just an aside I thought you might find interesting.
Washington University In St. Louis (2004, November 4). Like Old
Faithful: Researchers Describe How Natural Nuclear Reactor Worked.
ScienceDaily. Retrieved June 4, 2011, from http://www.sciencedaily.com
/releases/2004/10/041030182729.htm
Brad