How Does Carbon 14 Decay so well

Mar 2019
799
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cosmos
With respect to QM, my humble opinion always be "try to understand what QM talking from the simplest equation. It might be a good thing."
 
Apr 2017
528
130
Radioactive decay can be influenced, by bombarding with neutrons, e.g. in a nuclear bomb...
It appeared to me that statement was wrong ...This should be my subject , I took a degree course in Nuclear Engineering (QMC London) ...

My understanding was that when , say U 235 absorbs a thermal neutron it simply becomes U236 .. And U236 has a very short half life . pico seconds , and instantly splits emitting 2 or 3 neutrons ....

I'm a bit rusty so did a search and find U236 is stable !!! So what's going on???

I can't seem to find on line what precisely happens with this neutron absorption , all I find is this ...

" When 235U absorbs a thermal neutron, one of two processes can occur. About 82% of the time, it will fission; about 18% of the time, it will not fission" Wikipedia

which is very unsatisfactory ... these are thermal neutrons being absorbed , all with effectively no energy.
 
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Oct 2017
602
309
Glasgow
My understanding was that when , say U 235 absorbs a thermal neutron it simply becomes U236 .. And U236 has a very short half life . pico seconds , and instantly splits emitting 2 or 3 neutrons ....

I'm a bit rusty so did a search and find U236 is stable !!! So what's going on???

I can't seem to find on line what precisely happens with this neutron absorption , all I find is this ...

" When 235U absorbs a thermal neutron, one of two processes can occur. About 82% of the time, it will fission; about 18% of the time, it will not fission" Wikipedia
In nuclear reactions (or particle collisions), there is almost always an intermediate state which exists for a short time before breaking up into reaction products.

The \(\displaystyle ^{235}U\) nucleus absorbs a neutron. There will then be an intermediate state (which we can call excited \(\displaystyle ^{236}U\)). Then there will be reaction kinetics, where kinetic energy of the thermal neutron is exchanged between the \(\displaystyle ^{235}U\) nucleons via a range of interactions. Following this, the probability of yielding stable \(\displaystyle ^{236}U\) as a reaction product is not 1.

which is very unsatisfactory ... these are thermal neutrons being absorbed , all with effectively no energy.
For a given isotope, there are often "branches" where there will be a percentage chance of a decay proceeding down one branch compared to another. In the case of two branches, the branching ratio is defined as the ratio of amounts of reaction products (on terms of number). So, if you have an 82% chance to fission, the ratio of stable \(\displaystyle ^{236}U\) to fission products is approximately 1:4, or a branching ratio of 0.25. Even in stars, these branching ratios are only influenced by the presence of the particles required for a reaction. E.g. is an isotope requires neutrons or protons, then the branching ratio depends on the neutron density or proton density in the stellar interior. If the nuclear reaction requires photons (photodisintegration), then the photons need to exceed the threshold energy for capture, which effectively makes the reaction dependent on the star's interior temperature.
 
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Jun 2016
1,239
588
England
When I put my post in, I was initially just following the QM line,
But then I thought "what about neutron bombardment, does that not trigger decay?"
(and so added my final sentence to that effect).
Oz and Benit's posts clear that question for me.