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For example, U-235 fission cross section looks like this:


(source: science20.com)

As I understand it, the resonances peaks correspond to discrete quantum states of the excited compound nucleus. As you go higher, the density of states is too high to be resolved and you get that continuum.

But at thermal energies (left part of the graphic), I don't really understand what's going on, since the available states should be low. Consequently. I expect the cross section to be low too.

Is it a tail of a resonance peak corresponding to low energy states? Is the 1/v behavior dominating the decline of that resonance peak?

I'm expanding the question a little, since I'm not satisfied with the answers. Here is what I believe should be happening (the example is done with the absorption of a neutron by Indium-115):

enter image description here

Left is before absorption, right is after. The orange level is not a level in the compound nucleus, so absorption would be diminished.

This also happens with Uranium-238, so the question is not about fission only.

enter image description here

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  • $\begingroup$ I'm not a nuclear physicist so I'm reluctant to post this as an answer. My intuitive guess is that for faster neutrons the overlap of the wave function with the nucleus happens for a shorter period of time making the capturing of a proton less likely. A really slow neutron doesn't need to overcome the coulomb barrier and can just "sit" in the nucleus for a really long time (effectively becoming captured). $\endgroup$
    – Spencer
    Mar 8, 2015 at 3:56
  • $\begingroup$ @Spencer Thanks. I think I understand that general behavior, but I wanted an explanation from the point of view of discrete quantum states. Even if the neutron was slow and "sat" there for a long time, if it doesn't have the "correct" energy I don't see how it could be captured. $\endgroup$
    – 4nt
    Mar 8, 2015 at 4:03
  • $\begingroup$ Hmm... You gotta be careful there. Just by being there the neutron has already perturbed the energy levels of the nucleus; remember it also contributes to the Hamiltonian. So they won't necessarily be the same as they were before it arrived. Also I'm not sure there is such a thing as too little energy to be bound; just by being free in the first place it should have more energy than the bound nucleons; and if it doesn't there is always tunneling. Furthermore those resonance peaks correspond to differences in energy levels not the magnitude of the energy levels themselves. $\endgroup$
    – Spencer
    Mar 8, 2015 at 4:28
  • $\begingroup$ I believe Ali's answer below is correct. Note "fission" means that the nucleus is at a metastable state.In a sense the continuum in energy is similar as for a low energy electron to be caught in the continuum of the bands in a metal (that is how we discharge our fingers when the pad on the lap top stops working :) ) .kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_7/4_7_2.html $\endgroup$
    – anna v
    Mar 8, 2015 at 4:38
  • $\begingroup$ It seems pretty clear that out of the 2 possibilities the OP laid out, it is the 1/v behavior dominating. We just have a multitude of justifications for that behavior operating at different levels of physics. I read this question as asking for a genuine quantum explanation of that 1/v behavior, which is a perfectly valid question. $\endgroup$ Mar 8, 2015 at 4:42

2 Answers 2

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This is because U-235 is fissile, that is you only have to deliver the neutron to the nucleus for the magic to happen. Unlike U-238 where just delivering it doesn't do the job, there you also have to impart the nucleus with the neutrons kinetic energy.

Once we know this, it becomes clear that for low energy neutrons, their de Broglie wavelength is very big. So the cross-section is effectively determined by the quantum size of the neutron, rather than any other dynamics, so roughly $\sigma \approx \pi \lambda_{dB}^2\approx \frac{1}{E}$

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    $\begingroup$ all capture crossections have this behavior, kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_7/4_7_2.html $\endgroup$
    – anna v
    Mar 8, 2015 at 4:44
  • $\begingroup$ Look at Fig F in the link you provided, for neutron cross section with U-238 $\endgroup$
    – Ali Moh
    Mar 8, 2015 at 4:48
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    $\begingroup$ Yes, there is a threshold for non fissile ones. It is the "fissile" that makes the difference as you state in your answer. $\endgroup$
    – anna v
    Mar 8, 2015 at 4:56
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The energy Eigenstates of the final nucleus (after the neutron has been captured) form a complete set. That means that any wave function can be written as a superposition of these states; in particular we can express the incoming neutrons wave function in terms of these states,

$$ \psi_{in}(x,t) = \sum_{n=1}^\infty a_n(t) \psi_n(x) e^{i E_n t}.$$

Lets say the neutron is captured at time $T$. The coefficients $a_n(T)$ correspond to the probability amplitudes that, when measured, the neutron will be in the bound state with energy $E_n$.

The incoming kinetic energy of the neutron is going to be an average of the $E_n$,

$$ E_{Kinetic} = \sum_n \vert a_n(T) \vert^2 E_n,$$

if this incoming energy isn't exactly equal to one of the energy levels there is no problem; this just means that there is a finite probability of landing in higher and lower energy levels. This phenomenon is sometimes described as the uncertainty in the energy.


There is a slight problem here because the odds are that the neutron is prepared in some state of definite momentum like $e^{ipz}$ which doesn't actually live in our hilbert space. A way around this is to approximate the incoming wave function as,

$$ \psi_{in} = A e^{ipz} e^{-r^2/L^2}, $$

and to take the limit as $L\rightarrow \infty$ at the end (this is motivated by the theory of generalized functions ) .

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  • $\begingroup$ @Ant, I think I understand what your asking now. Does this help? $\endgroup$
    – Spencer
    Mar 8, 2015 at 6:00
  • $\begingroup$ Yes, this is much clearer, thanks. I will leave the question open until tomorrow though. Just in case someone wants to add something. $\endgroup$
    – 4nt
    Mar 8, 2015 at 6:04
  • $\begingroup$ Cool, I'm sure an expert in the field can provide a much more in depth explanation. $\endgroup$
    – Spencer
    Mar 8, 2015 at 6:05
  • $\begingroup$ One more thing, when you say: "if this incoming energy isn't exactly equal to one of the energy levels there is no problem; this just means that there is a finite probability of landing in higher and lower energy levels.", do you mean that the coefficients a_n(t) will change with time so that the average kinetic energy E_Kinetic could deviate from the incoming one? $\endgroup$
    – 4nt
    Mar 9, 2015 at 15:51
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    $\begingroup$ This doesn't answer the question. The OP wants to know the reason for the high cross-section at low energies. $\endgroup$
    – user4552
    Sep 7, 2019 at 14:17

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