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## Are there fissionable isotopes of any element with short-lived radioactive isotopes?

 Quote by Morbius Andrew, No - I am saying that each nucleon contributes about 40 MeV to the nuclear potential well; that's NOT the binding energy. First, when we are talking about the "binding energy per nucleon"; we are talking about an average. Each nucleon has a binding energy dependent upon its energy, just as each electron in an atom has its own "ionization potential" - the amount of energy you have to add to that electron to kick it out of the atom. However, we are not considering the energy of the individual nucleons based on their position in the nuclear shell structure. We're just considering an average for convenience. We could just as well talk about the total binding energy - which is just the average multiplied by the number of nucleons. In order to get the true binding energy for a nucleus; you really have to do a full solution of the equations of quantum mechanics with a model for the nuclear potential. That's complicated. However, there is a rough formula that gives you the total binding energy for a nucleus with "Z" protons and "N" electrons. The first two terms of the formula are: Binding Energy = 15.8 MeV * (N + Z) - 0.174 MeV * Z * (Z-1) / (N+Z)^(1/3) .... There are more terms; but these two suffice for the present. Each nucleon, be it a proton or neutron makes an equal contribution to the first term. The protons make a contribution to the second term which is an "unbinding energy" because the sign is negative. This second term increases in magnitude roughly as Z squared. So for the heavier nuclei, the ones with higher Z; it is this negative "unbinding energy" that decreases the binding energy per nucleon as the atomic number Z increases. I didn't say that the Coulomb force didn't contribute. I'm saying MOST of the energy is from the nuclear force. The fact that the nuclear force "virtually disappears" as the fission fragments fly apart while the Coulomb force has a longer range doesn't mean the contribution to the energy due to the nuclear force is small. It just means that the nuclear force has less time to transfer its energy to the particles.
This is very useful to me. Thanks.

[I think you meant N "neutrons" not "electrons"]. So, putting in the values for U-235 (N = 143 and Z= 92) I get:

$$E_{binding} = 15.8*(235) - .174*(92*91)/235^{.33} = 3713 - 236 = 3477 \text{MeV}$$

So am I correct in concluding that the electromagnetic energy is about 7% (236/3713) of the total energy?
 ...So even though the nuclear force has a shorter distance to transfer the nuclear energy to the fission fragments - it still has more energy to transfer than the Coulomb force, and has to transfer that energy at a higher rate. It can do that because the nuclear force is over two orders of magnitude more powerful than the Coulomb force.
I understand your analogy. The fact that nuclear force is much greater offsets the fact that it operates over a shorter distance, so the energy well is still very deep.

What I am having difficulty understanding is how that relates to the energy that is released when the nucleus splits. The nuclear force is trying to keep the nucleus together so energy is needed to overcome it. The greater the binding energy of a nucleon, the more energy is needed to separate it from the nucleus so less energy is released in the form of kinetic energy of the fission parts.
 Radioactivity certain IS energy from the nuclear force. Consider the alpha decay of U-238. U-238 ---> Th-234 + He4 If you look up the nuclear masses of the above reactant and products; you will seen that you have "LOST" about 0.00458 amu of mass which is the equivalent of about 4.27 MeV of energy. The result of the alpha decay of U-238 is an alpha particle with 4.27 MeV of energy. [ The Th-232 gets a little energy to conserve momentum, but because it is so much more massive, the bulk of the energy goes to the alpha particle.] That's MUCH more than can be attributed to Coulomb repulsion. Most of the energy of the alpha is due to the difference in mass deficits.
Since E=mc^2, the sum of the rest masses of the alpha particle and Th-234 must be less than the U-238 nucleus since energy is released. But that would be the case whether it was electromagnetic energy or nuclear energy that was given off. The mass of an atom increases when it absorbs a photon and an electron moves to a higher energy level - just not very much. It is perceptible in the case of nuclear events becuase there is so much more energy involved in the release of energy from the nucleus.

(By the way, I didn't mean to include beta particles, which is obviously not due to proton-proton repulsion. I was really thinking just of alpha decay).

AM

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 Quote by Andrew Mason This is very useful to me. Thanks. [I think you meant N "neutrons" not "electrons"].
Yes - I meant neutrons - corrected above.

 So, putting in the values for U-235 (N = 143 and Z= 92) $$E_{binding} = 15.8*(235) - .174*(92*91)/235^{.33} = 3713 - 236 = 3477 \text{MeV}$$ So am I correct in concluding that the electromagnetic energy is about 7% (236/3713) of the total energy?
I gave you only two terms of a much more complex expression. There are other terms
not included. These terms such as the asymmetry term are also negative and tend to
"unbind" the nucleus.

But if you want to compare the "unbinding energy" due to electrostatic repulsion to the
attractive part of the nuclear force; you can use those terms.

The complete expression is:

$$E_B = 15.8 A - 18.3 A^{(2/3)} - 0.714 {Z (Z-1) \over A^{(1/3)}} - 23.2 { (N - Z)^2 \over A } + \delta$$

where $$\delta = 0$$ for an odd A nucleus like U-235

 I understand your analogy. The fact that nuclear force is much greater offsets the fact that it operates over a shorter distance, so the energy well is still very deep. What I am having difficulty understanding is how that relates to the energy that is released when the nucleus splits. The nuclear force is trying to keep the nucleus together so energy is needed to overcome it. The greater the binding energy of a nucleon, the more energy is needed to separate it from the nucleus so less energy is released in the form of kinetic energy of the fission parts.
The nuclide with the greater binding energy is going to be a product nuclide.
The reactant nuclide has less binding energy.

If we call our zero energy level; the total amount of energy of the constituent particles;
then the bound state nucleus has a negative energy on that scale, and the magnitude
is the binding energy.

So when a nuclide goes from a state with lesser binding energy to a state with greater
binding energy; it means the product nuclide has an energy level that is even lower
than that of the reactant nuclide. The difference in the energy level of the reactant and
product - which is also equal to the difference in the binding energies is the "Q" of the
reaction - the energy available to be distributed as kinetic energy.

If a nucleus has greater binding energy; it means that the energy of that nucleus is
LOWER on a absolute energy scale than a nucleus that has lower binding energy.

Don't think of binding energy as an amount of energy in the nucleus - think of it
as an energy deficit - in fact it's equal to a quantity called the "mass deficit"
converted to energy units as per Einstein's famous equation.

A more stable nucleus - which is one with higher binding energy - will be
a product of an exothermic reaction. A less stable nucleus; one with lesser
binding energy - will be the reactant.

 Since E=mc^2, the sum of the rest masses of the alpha particle and Th-234 must be less than the U-238 nucleus since energy is released. But that would be the case whether it was electromagnetic energy or nuclear energy that was given off.
Yes - what one really needs to calculate is the Coulomb potential energy for the
separated charges. [That's becaus I saved a post before I was finished because
I didn't want to lose the work so far like I did when Mozilla "evaporated" on me.]

Dr. Gregory Greenman
Physicist

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 Quote by Morbius If we call our zero energy level; the total amount of energy of the constituent particles; then the bound state nucleus has a negative energy on that scale, and the magnitude is the binding energy. So when a nuclide goes from a state with lesser binding energy to a state with greater binding energy; it means the product nuclide has an energy level that is even lower than that of the reactant nuclide. The difference in the energy level of the reactant and product - which is also equal to the difference in the binding energies is the "Q" of the reaction - the energy available to be distributed as kinetic energy. If a nucleus has greater binding energy; it means that the energy of that nucleus is LOWER on a absolute energy scale than a nucleus that has lower binding energy.
Right. Similar to gravitational potential energy being negative (0 at infinity) because the force is attractive so matter gains kinetic energy as it is 'caught' in the force field.

 Don't think of binding energy as an amount of energy in the nucleus - think of it as an energy deficit - in fact it's equal to a quantity called the "mass deficit" converted to energy units as per Einstein's famous equation.
Yes, I realize that the binding energy represents the energy required by a nucleon to get over the lip of the binding energy well. So if it is given that amount of kinetic energy it is home free, especially if it is a proton, in which case it gets an electromagnetic propulsive kick as well.

That is my whole problem here. If I understand what you are saying (and I do not doubt that you are right, it may be just me) that the neutron's capture by the nucleus gives the neutron such enormous kinetic energy that it is enough not only to get about 90 nucleons over the top of their binding energy well, but has enough left over to give the two fission parts kinetic energy that greatly exceeds the energy from proton-proton coulomb repulsion.
 [That's because I saved a post before I was finished becauseI didn't want to lose the work so far like I did when Mozilla "evaporated" on me.]
Yeah. That can be a problem. I copy a long reply to Notepad and save it, then paste it back in later.

AM

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 Quote by Andrew Mason Right. Similar to gravitational potential energy being negative (0 at infinity) because the force is attractive so matter gains kinetic energy as it is 'caught' in the force field.
Andrew,

Exactly.

 Yes, I realize that the binding energy represents the energy required by a nucleon to get over the lip of the binding energy well. So if it is given that amount of kinetic energy it is home free, especially if it is a proton, in which case it gets an electromagnetic propulsive kick as well.
Yes - if the proton can escape - more likely an alpha particle - it gets additional energy
due to the electrostatic repulsion of the alpha and the remaining nucleus.

 That is my whole problem here. If I understand what you are saying (and I do not doubt that you are right, it may be just me) that the neutron's capture by the nucleus gives the neutron such enormous kinetic energy that it is enough not only to get about 90 nucleons over the top of their binding energy well, but has enough left over to give the two fission parts kinetic energy that greatly exceeds the energy from proton-proton coulomb repulsion.
There's more to it than just the energy. For example, let's take a target nucleus that
has Z protons and N neutrons. A = Z + N. Let's assume Z is odd and N is even.
Then A will be odd, and the $$\delta$$ in the expression above is zero. The nucleus will
have a binding energy given by the above.

Now the nucleus captures a neutron. We now have to compute a new binding energy
for the compound nucleus. In this case, the value of A --> A + 1; so the new A is even.
The number of neutrons N --> N+1; which is now odd. When A is even; but both Z and
N are odd; the value of $$\delta$$ is negative. [The $$\delta$$ term is called the "pairing term"; and is
due to a quantum mechanical effect.]

So in effect, the compound nucleus "looses" some binding energy. [ This doesn't mean
that any real energy is being destroyed; it just means that the ground state energy for the
compound nucleus is higher than it would have been otherwise.]

So you now have a nucleus that not only has a bunch of energy that the new neutron
brought in - but also its ground state energy has changed. These two effects coupled
together may mean that the new compound nucleus is unstable. The nucleus has to
now find a stable state.

Most likely, a nucleus that absorbs a neutron will be unstable with respect to $$\beta-$$ decay.
That extra neutron will turn into a proton, an electron, and an anti-neutrino; and the latter
two will be ejected and the nuclide transmutes to one with the next higher atomic number
Z+1 due to the new proton.

Note that the daughter nuclide with atomic number Z+1 will have more electrostatic repulsion
than the parent which had atomic number Z. There will be MORE Coulomb "unbinding energy".
However, the nuclear effects are more important than the Coulomb effect. Since the original
value of Z was odd; the new value Z+1 will be even. The neutron number N will go back to
its original value which is even. So the daughter nucleus will have an even number of protons,
and an even number of neutrons. This is more important for stability of the nucleus than is
the extra Coulomb repulsion. The nucleus is more stable even with the additional Coulomb
"unbinding energy" because the nuclear effects "trump" the Coulomb effects.

So the extra energy the neutron brings in doesn't have to be enough to get out of the
"old" potential well of the target nucleus - it has to be enough to get out of the well of
the "new" nucleus.

You really have to due a quantum mechanical description of the nucleus with all the
shell structure; just like one has to do with electrons in an atom.

 Yeah. That can be a problem. I copy a long reply to Notepad and save it, then paste it back in later.
Yes - also discovered that if you make a mistake in a "tex" equation, and you edit it;
the system doesn't re-evaluate the tex expression after you edit it. You have to copy
the entire post to an editor [ I use Emacs instead of Notepad because I'm working in
Unix], delete the faulty post, and post a new reply using the saved text from the editor.

Dr. Gregory Greenman
Physicis

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Thanks very much for all of this. It has been very helpful and a very enjoyable discussion..

I have to run but let me just put this thought out:

In order to release energy, the fission parts have to have more binding energy than the original nucleus (as in fusion: the binding energy of the product - fused nucleus - has to be greater than the original parts if the fusion releases energy). This is possible for U because the fission parts can both be heavier than iron so the total binding energy of the fission parts is higher (which I think is the short answer to the original question).

But isn't the reason the binding energies of the fission parts increase due to electrostatic repulsion? That is, there is a smaller negative electrostatic repulsion term in the binding energy equation for the fission parts compared to original U-235.

 Quote by Morbius Yes - also discovered that if you make a mistake in a "tex" equation, and you edit it; the system doesn't re-evaluate the tex expression after you edit it. You have to copy the entire post to an editor [ I use Emacs instead of Notepad because I'm working in Unix], delete the faulty post, and post a new reply using the saved text from the editor.
Now I understand why your posts have these extra carriage returns!! The system DOES re-evaluate the tex expression, it just doesn't reload your screen. If you use the advanced editor it should reload it for you. Otherwise, just reload the screen in your browser and the revaluated tex graphic will show up..

AM

Recognitions:
 Quote by Andrew Mason In order to release energy, the fission parts have to have more binding energy than the original nucleus
Andrew,

The fission products have to have more binding energy than the COMPOUND nucleus -
not the orignal nucleus. Just as in the case with the neutron capture; it's the binding
energy of the compound nucleus that is important.

What happens after the nucleus absorbs a neutron; and which decay channel it follows
is dependent on the compound nucleus; not the original nucleus.

One option for decay is for the nucleus to eject the new neutron. That's a reaction
called "compound elastic scatter" - scatter because the neutron may not come out
going the same direction as when it went in. [When it comes out not going the
same direction, the target nucleus has to recoil to conserve momentum, hence
the ejected neutron doesn't have all the energy it came in with - some goes to
the kinetic energy of the recoiling target. The target nucleus is left in the same
internal state that it had before the collision.]

The nucleus could also keep some of the energy of the neutron, and be left in an
excited state after ejecting the neutron. In this case, the neutron will have lost
even more energy than it does in an elastic collision. In this case, the reaction is
"compound inelastic scatter". All inelastic scatter is compound inelastic scatter.
However, not all elastic scatter is compound elastic scatter; because there is
also the possiblility of "potential scatter" [sometimes called "shape scatter"]; in
which the neutron scatters off the nuclear potential of the target nucleus without
forming a compound nucleus.

 But isn't the reason the binding energies of the fission parts increase due to electrostatic repulsion? That is, there is a smaller negative electrostatic repulsion term in the binding energy equation for the fission parts compared to original U-235.
Yes - that's PART of it. However, I again refer you to the previous post where I discussed
$$\beta-$$ decay. There the nucleus decayed in a manner which INCREASES the
electostatic repulsion in favor of nuclear effects.

I'm just saying that you can't say that the path to fission depends on just the
electrostatic effects. That's part - but not the whole story. If the nucleus does fission;
then some of the energy of the fission products is due to their repulsion.

However, the nuclear effects are more important than the Coulomb effect, and it is often
the case, as we see in $$\beta-$$ decay that the nuclear effects will trump the Coulomb effect.
So you can't say that fission is energetically selected based on just the Coulomb repulsion.

Although U-235 is fissile, it will fission with low energy "thermal" neutrons; U-238 is
fissionable, it will fission only with neutrons with kinetic energy above a certain
threshold.

However, for low energy neutrons; U-238 will not fission. But U-238 has the same
92 protons that U-235 has. If it was only about the Coulomb energetics, the U-239
compound nucleus formed when U-238 absorbs a low energy neutron, could split into
a couple of fission fragments of lower Z; and thus lower the stored electrostatic
repulsion energy in a manner identical to the way the U-236 formed by the absorption
of a neutron by U-235; will decay by fission.

But U-239 created by low energy neutrons doesn't fission!! Instead U-239 forms first
Np-239, which then forms Pu-239. So we go from Z=92 ultimately to Z=94. The
electrostatic repulsion "unbinding energy" is proporttional to Z*(Z-1). In going from
Z=92 to Z=94; the Coulomb repulsion "unbinding energy" INCREASES by 4.4%

The nucleus could minimize the Coulomb repulsion energy by fissioning; and
instead it INCREASES the Coulomb "unbinding energy" by 4.4%

If it was only about minimizing the Coulomb energy; then U-238 should fission with
low energy neutrons like U-235 does!! But U-238 doesn't fission with low energy neutrons;
it takes an alternate route which INCREASES Coulomb "unbinding energy". So whether
a nucleus fissions or not is not due solely to the Coulomb energetics.

 Now I understand why your posts have these extra carriage returns!! The system DOES re-evaluate the tex expression, it just doesn't reload your screen. If you use the advanced editor it should reload it for you. Otherwise, just reload the screen in your browser and the revaluated tex graphic will show up..
I've heard the carriage return comment on other forums too. Evidently Mozilla Firefox
under Unix / Linux works differently than Explorer under Windows.

Dr. Gregory Greenman
Physicist

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 Quote by Morbius However, the nuclear effects are more important than the Coulomb effect, and it is often the case, as we see in $$\beta-$$ decay that the nuclear effects will trump the Coulomb effect. So you can't say that fission is energetically selected based on just the Coulomb repulsion. Although U-235 is fissile, it will fission with low energy "thermal" neutrons; U-238 is fissionable, it will fission only with neutrons with kinetic energy above a certain threshold. However, for low energy neutrons; U-238 will not fission. But U-238 has the same 92 protons that U-235 has. If it was only about the Coulomb energetics, the U-239 compound nucleus formed when U-238 absorbs a low energy neutron, could split into a couple of fission fragments of lower Z; and thus lower the stored electrostatic repulsion energy in a manner identical to the way the U-236 formed by the absorption of a neutron by U-235; will decay by fission.... If it was only about minimizing the Coulomb energy; then U-238 should fission with low energy neutrons like U-235 does!! But U-238 doesn't fission with low energy neutrons; it takes an alternate route which INCREASES Coulomb "unbinding energy". So whether a nucleus fissions or not is not due solely to the Coulomb energetics.
But.... the addition of 3 more neutrons increases the size of the nucleus and average distance between protons, so it reduces the Coulomb potential within the nucleus (effectively increasing the nuclear binding energy). That is (partly) why it does not fission with low energy neutrons. It needs more energy to overcome the higher nuclear binding energy. But I appreciate that there are some quantum effects with that $\delta$ term which is >0 for U-238.

I knew that I had read Feynman somewhere saying that fission energy was electrical energy. After a bit of digging, I found the passage:
 Quote by Richard Feynman, Lectures on Physics, Vol. II, page 1-2 " There is another question: "What holds the nucleus together?" In a nucleus there are several protons, all of which are positive. Why don't they push themselves apart? It turns out that nuclei there are, in addition to electrical forces, nonelectrical forces, called nuclear forces, which are greater than the electrical forces and which are able to hold the protons together in spite of the electrical repulsion. The nuclear forces, however, have a short range - their force falls off much more rapidly than 1/r2. And this has an important consequence. If a nucleus has too many protons in it, it gets too big, and will not stay together. An example is uranium, with 92 protons. The nuclear forces act mainly between each proton (or neutron) and its nearest neighbor, while the electrical forces act over larger distances, giving a repulsion between each proton and all of the others in the nucleus. The more protons in a nucleus, the stronger is the electrical repulsion, until, as in the case of uranium, the balance is so delicate that the nucleus is almost ready to fly apart from the repulsive electrical force. If such a nucleus is just "tapped" lightly (as can be done by sending in a slow neutron), it breaks into two pieces, each with positive charge, and these pieces fly apart by electrical reuplsion. The energy which is liberated is the energy of the atomic bomb. The energy is usually called "nuclear" energy, but it is really "electrical" energy released when electrical forces have overcome the attractive nuclear forces.
Now I agree that he oversimplifies things a little here. I think he is ignoring the fact that the kinetic energy of the free neutrons that are produced in fission obviously cannot be the direct result of coulomb repulsion since there is none with a neutron. I am also aware that this was written 43 years ago and that Feynman was not a nuclear physicist - but he was a pretty smart guy. So I am thinking he can't be that far off here.

AM

Recognitions:
 Quote by Andrew Mason But.... the addition of 3 more neutrons increases the size of the nucleus and average distance between protons, so it reduces the Coulomb potential within the nucleus (effectively increasing the nuclear binding energy).
Andrew,

The formula above accounts for that because of the cube root of "A"
in the denominator.

$${Z(Z - 1) \over {A^{1/3}}$$

Lets look at the individual terms in the expression for binding energy:

$$E_B = 15.8 A - 18.3 A^{(2/3)} - 0.714 {Z (Z-1) \over A^{(1/3)}} - 23.2 { (N - Z)^2 \over A } + \delta$$

________________U-235______U-238_______Difference

"Volume"________3713.0______3760.4_________+47.4

"Surface"________-696.7______-703.4__________-6.8

"Coulomb"________-968.8______-964.3_________+4.7

"Asymmetry"______-256.8______-284.2________-27.5

"Pairing"_____________0.0_________0.2_________+0.2

Total_____________1790.7_____1808.7________+18.0

Except for the "pairing term" which is usually pretty small for large nuclei;
the Coulomb term is the LEAST important of any of the major term.

The nuclear "Volume" term is an order of magnitude larger than the
Coulomb term. It's only because some of the nuclear terms have
opposite signs, that the Coulomb term is as important as it is; which
it is still a minority player.

It's the extra NUCLEAR binding energy that makes the U-238 more stable!!!

 Now I agree that he oversimplifies things a little here. I think he is ignoring the fact that the kinetic energy of the free neutrons that are produced in fission obviously cannot be the direct result of coulomb repulsion since there is none with a neutron. I am also aware that this was written 43 years ago and that Feynman was not a nuclear physicist - but he was a pretty smart guy. So I am thinking he can't be that far off here.
Feynman is correct - but he's simplifying things a bit. The addition of a
free neutron to a nucleus is more than "tapping" slightly - because the
free neutron is falling into a deep potential well.

I again refer to my analogy with the stone and the well. Yes - you tap the
stone slightly to get it to fall into the well - but the loud sound you hear
didn't come from the energy imparted by your tap.

It came from the potential energy of the stone.

Likewise, U-235 is really not ready to fall apart, which is why it has a
half-life of 705 MILLION years. [U-238 is 4.5 BILLION]

It's the energy of a free neutron falling into a nuclear potential that
blows the nucleus apart.

Dr. Gregory Greenman
Physicist

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 Quote by Morbius It's the energy of a free neutron falling into a nuclear potential that blows the nucleus apart.
I can accept that since the binding energy of the additional neutron is released inside the nucleus that the nucleus will become excited. What I can't quite see is how this energy (which I calculate using the formula you have provided to be about 6.4 MeV) would be sufficient to break apart the nucleus, let alone propel the fission parts with such energy.

AM

Recognitions:
 Quote by Andrew Mason I can accept that since the binding energy of the additional neutron is released inside the nucleus that the nucleus will become excited. What I can't quite see is how this energy (which I calculate using the formula you have provided to be about 6.4 MeV) would be sufficient to break apart the nucleus, let alone propel the fission parts with such energy.
Andrew,

Then you have to do a Quantum Mechancal treatment.

It's more complicated than just comparing the energy to the binding energy. After all,
a neutron brings in a few MeV; and the total binding energy is on the order of 2 GeV.
[The binding energy formulas are only approximate anyway.] You might not think that
a few MeV is enough to disturb a nucleus that has a couple GeV in binding energy;
but it does - which is why certain nuclides will fission.

As I stated in one of my previous posts - if you want to know how the nucleus is
going to react - then you need to do an analysis of the compound nucleus via
quantum mechanics.

Quantum Mechanics puts limits on how the nucleus can decay, and more importantly;
what states are stable. There are more variables to be considered than just the energy.
You need to conserve angular momentum, nuclear spin...

A given reaction "channel" as it is called may be favored energetically - that is, the
resultant energy state of the products will be minimized. However, if you can't conserve
angular momentum, or spin.... then that channel will be forbidden.

When a nucleus absorbs a neutron, you might be tempted to say that the small increase
in energy relative to the binding energy of the nucleus should result in the nucleus doing
nothing - we would just have a stable compound nucleus. However, that new configuration
of protons and neutrons may not be stable from the quantum mechanical point of view -
there's no "stationary state" for that configuration - so it has to decay; regardless of what
you might think would happen if you consider the energetics alone.

Dr. Gregory Greenman
Physicist

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 Quote by Morbius When a nucleus absorbs a neutron, you might be tempted to say that the small increase in energy relative to the binding energy of the nucleus should result in the nucleus doing nothing - we would just have a stable compound nucleus. However, that new configuration of protons and neutrons may not be stable from the quantum mechanical point of view - there's no "stationary state" for that configuration - so it has to decay; regardless of what you might think would happen if you consider the energetics alone.
If there is no stationary state for the compound nucleus, of $^{235}U + n$ how can $^{236}U$ exist at all? Is there path that quantum mechanics permits for the excited $^{236}U$ nucleus to get rid of energy without fissioning?

AM

Recognitions:
 Quote by Andrew Mason If there is no stationary state for the compound nucleus, of $^{235}U + n$ how can $^{236}U$ exist at all? Is there path that quantum mechanics permits for the excited $^{236}U$ nucleus to get rid of energy without fissioning?
Andrew,

A "stationary state" is one that can exist "indefinitely" - i.e. a "stable" state.

If there is no "stationary state" - then the wave-function is time-dependent; that is whatever
state the quantum system is in will decay.

http://en.wikipedia.org/wiki/Ground_state

There are multiple ways for U-236 nucleus to de-excite; which gives rise to the MULTIPLE
reactions that a U-235 nucleus can have when struck by a neutron:

(n,n) elastic scatter
(n,n') inelastic scatter
(n,2n)
(n,3n)
(n,4n)
(n,fission)
(n,gamma)

U-235 doesn't always fission when hit by a neutron; fission is one possible reaction.

Dr. Gregory Greenman
Physicist

 Quote by Andrew Mason Is there path that quantum mechanics permits for the excited $^{236}U$ nucleus to get rid of energy without fissioning?
Neutron absorption in most nuclides is followed by emission of a gamma-ray, and the reaction is usually referred to as an (n,$\gamma$) reaction. IIRC, about 16-17% of neutron absorptions by U-235 result in U-236* which decays promptly by gamma emission, otherwise fission occurs.

Recognitions:
 Quote by Astronuc Neutron absorption in most nuclides is followed by emission of a gamma-ray, and the reaction is usually referred to as an (n,$\gamma$) reaction. IIRC, about 16-17% of neutron absorptions by U-235 result in U-236* which decays promptly by gamma emission, otherwise fission occurs.