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Uranium 236

by avito009
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avito009
#1
Jun28-14, 11:07 AM
P: 6
If uranium 238 is more stable than uranium 235 because 3 extra neutrons add to strong force then uranium 236 having 1 extra neutron should have more strong force than uranium 235 so why does it decay so fast and why is it more unstable than uranium 235?
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PeterDonis
#2
Jun28-14, 12:37 PM
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Adding extra neutrons to a nucleus doesn't just add more attractive force because of the strong interaction; it also adds repulsive force because of the Pauli exclusion principle--no two neutrons can be in the same state. So neutrons (and protons) in a nucleus have to occupy "energy shells", in much the same way as electrons in an atom do:

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

So adding a neutron to a nucleus could make it more stable, if, for example, the added neutron fills an energy shell; but it could also make it less stable, if, for example, the added neutron has to go into a new shell that wasn't occupied in the old nucleus. And there are lots of possible intermediate cases as the numbers of nucleons in each shell get larger. There are also effects due to the interaction between the spins of the protons and neutrons, which have to be taken into account.
bcrowell
#3
Jun28-14, 09:08 PM
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Interesting question!

Uranium isotopes can decay by alpha decay, beta decay, or spontaneous fission. 235U and 236U both decay almost entirely by alpha decay. What matters is not just the stability of the parent nucleus but the stability of the parent *relative* to the daughter, which is measured by the alpha-decay energy E. It's generally observed that the log of the alpha-decay half-life is approximately a linear function of E^-1/2. For these two parents we have E=4.7 MeV and 4.6 MeV respectively. That would suggest that they would have nearly the same alpha-decay half-lives

However, it is always found that odd nuclei have slower alpha-decay rates than their even-even neighbors. The ratio of these quantities is called the hindrance factor. The hindrance factor is apparently mainly due to angular momentum. In 236U->232Th, we have a 0+ even-even nucleus going to another even-even nucleus with spin-parity 0+. But in 235U->231Th the ground states are 7/2- and 5/2+. The change in parity actually violates a parity selection rule, so I guess the 235U must decay to some excited state of 231Th, not the ground state. But anyway there is probably a difference in angular momentum, which has to be accounted for by the alpha particle. As an alpha particle with angular momentum L tunnels out through the Coulomb barrier, it also experiences a centrifugal barrier proportional to L(L+1). I think this centrifugal barrier is what causes the much longer half-life compared to 236U.

Note that the short half-lives of 237 and 239U are because they are unstable with respect to beta decay.

bcrowell
#4
Jun29-14, 06:04 PM
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Uranium 236

Some more info on the alpha decay of 235U to 231Th: http://www.nndc.bnl.gov/useroutput/A...A52FF2B_1.html

4.7 MeV was the ground state->ground state Q value of the decay, but that decay has a hindrance factor of 2.5x10^3 because of the mismatch in spin and parity between the parent (7/2-) and daughter (5.2+); this decay accounts for only 5% of the alpha decay.

The strongest alpha decay branch (58%) is to a state in 231Th with an excitation energy of 205 keV, which has tentatively been assigned spin-parity 7/2-. It has a hindrance factor of only 5.9, which is presumably the reason for the 7/2- assignment. The alpha decay energy for this branch is 4.40 MeV.

Now let's compare with 236U: http://www.nndc.bnl.gov/useroutput/A...F29366A_1.html . Its alpha decay goes mostly by ground-state to ground-state decay, and has an energy of 4.57 MeV. This is considerably more than the 4.40 MeV for the predominant alpha from 235U. Its hindrance factor is 1.0.

So basically 235U has no good choices for decay. It can decay to the ground state of 231Th, but that has a high hindrance factor due to spin and parity. It can decay to an excited state with the same spin and parity, but that has a lower alpha energy and therefore a much lower decay rate.
avito009
#5
Jul5-14, 07:19 AM
P: 6
I found a simpler explanation

When a neutron is absorbed by U 235, the compound nucleus U 236 is formed. The new compound nucleus will be unstable due to its excited state as it absorbs the binding energy of the neutron. If the Uranium 236 atom is allowed to settle to its ground state it will be very stable and has a half of approximately 25 million years. The stability of an atomic nucleus is determined by its binding energy - the amount of energy required to disrupt it. Any time a neutron or proton is captured by an atomic nucleus, the nucleus rearranges its structure. If energy is released by the rearrangement, the binding energy decreases. If energy is absorbed, the binding energy increases.

The isotopes important for the large scale release of energy through fission are uranium-235 (U-235), plutonium-239 (Pu-239), and uranium- 233 (U-233). The binding energy of these three isotopes is so low that when a neutron is captured, the energy released by rearrangement exceeds it. The nucleus is then no longer stable and must either shed the excess energy, or split into two pieces. Since fission occurs regardless of the neutron's kinetic energy (i.e. no extra energy from its motion is needed to disrupt the nucleus), this is called "slow fission".
mfb
#6
Jul5-14, 08:11 AM
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P: 11,925
Fission rates and options have nothing to do with alpha decay rates.


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