Are there fissionable isotopes of any element with shortlived radioactive isotopes?by bananan Tags: element, fissionable, isotopes, radioactive, shortlived 

#37
Dec606, 04:49 PM

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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 protonproton coulomb repulsion. AM 



#38
Dec706, 08:33 AM

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Exactly. due to the electrostatic repulsion of the alpha and the remaining nucleus. 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 [tex]\delta[/tex] 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 [tex]\delta[/tex] is negative. [The [tex]\delta[/tex] 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 [tex]\beta[/tex] decay. That extra neutron will turn into a proton, an electron, and an antineutrino; 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. the system doesn't reevaluate 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 



#39
Dec706, 09:07 AM

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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 U235. AM 



#40
Dec706, 09:56 AM

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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. [tex]\beta[/tex] 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 [tex]\beta[/tex] 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 U235 is fissile, it will fission with low energy "thermal" neutrons; U238 is fissionable, it will fission only with neutrons with kinetic energy above a certain threshold. However, for low energy neutrons; U238 will not fission. But U238 has the same 92 protons that U235 has. If it was only about the Coulomb energetics, the U239 compound nucleus formed when U238 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 U236 formed by the absorption of a neutron by U235; will decay by fission. But U239 created by low energy neutrons doesn't fission!! Instead U239 forms first Np239, which then forms Pu239. So we go from Z=92 ultimately to Z=94. The electrostatic repulsion "unbinding energy" is proporttional to Z*(Z1). 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 U238 should fission with low energy neutrons like U235 does!! But U238 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. under Unix / Linux works differently than Explorer under Windows. Dr. Gregory Greenman Physicist 



#41
Dec906, 11:46 AM

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I knew that I had read Feynman somewhere saying that fission energy was electrical energy. After a bit of digging, I found the passage: AM 



#42
Dec906, 03:04 PM

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The formula above accounts for that because of the cube root of "A" in the denominator. [tex]{Z(Z  1) \over {A^{1/3}}[/tex] Lets look at the individual terms in the expression for binding energy: [tex]E_B = 15.8 A  18.3 A^{(2/3)}  0.714 {Z (Z1) \over A^{(1/3)}}  23.2 { (N  Z)^2 \over A } + \delta[/tex] ________________U235______U238_______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 U238 more stable!!! 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, U235 is really not ready to fall apart, which is why it has a halflife of 705 MILLION years. [U238 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 



#43
Dec1106, 08:22 AM

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AM 



#44
Dec1106, 09:00 AM

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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 



#45
Dec1106, 12:58 PM

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AM 



#46
Dec1106, 01:30 PM

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A "stationary state" is one that can exist "indefinitely"  i.e. a "stable" state. If there is no "stationary state"  then the wavefunction is timedependent; that is whatever state the quantum system is in will decay. http://en.wikipedia.org/wiki/Ground_state There are multiple ways for U236 nucleus to deexcite; which gives rise to the MULTIPLE reactions that a U235 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) U235 doesn't always fission when hit by a neutron; fission is one possible reaction. Dr. Gregory Greenman Physicist 



#47
Dec1106, 07:54 PM

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#48
Dec1106, 09:46 PM

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Here's a plot of the total crosssection, along with elastic scatter, fission, and (n,gamma). As can be seen; for low energy neutrons, fission is the dominant decay mode. However, for high energy neutrons, above the resonace region; fission loses out to elastic scatter by a wide marging. 


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