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In Introductory Nuclear Physics, Kenneth S. Krane describes neutron separation energy (page 65), and shows a list of measured values (Figure 5.2).

What experimental setup is used to measure these on stable isotopes?

For example, how does one measure the neutron separation of $\mathrm{^{56}Fe}$, which is calculated at 10.670 MeV? Perhaps it's done by shooting protons from a cyclotron at the $\mathrm{^{56}Fe}$ atom? Tuning proton energy until we observe $\mathrm{^{55}Fe}$ and a stray neutron, which should occur at the $10.670$ MeV mark.

But that's adding energy, rather than taking it away. And besides, the chance of hitting the nucleus is too low to observe anything. So how do we measure neutron separation energy?

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    $\begingroup$ If you know the binding energy of each nuclei, it comes out from math. The binding energy comes from an accurate measurement of the mass of each nuclei. Performing various scattering experiments tells you the cross-sections and energy levels within the nucleus. $\endgroup$ – Jon Custer Oct 16 at 13:38
  • $\begingroup$ Are you saying that the measurements of Figure 5.2 are actually calculated from binding energies measured in various scattering experiments? The text and figure are here and here. $\endgroup$ – MrMartin Oct 16 at 16:11
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    $\begingroup$ ”And besides, the chance of hitting the nucleus is too low to observe anything.” While hitting a single, selected nucleus with a single, selected beam particle would, indeed, be extremely difficult it is very easy to hit a macroscopic target with a beam generating many, many interactions. Often to the point that rate steps need to be taken to limit the rate. $\endgroup$ – dmckee Oct 16 at 18:03
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If you really, really wanted to do a direct measurement of the separation energy (instead of doing precision mass spectroscopy), you could use the

"shooting protons from a cyclotron at the $^{56}\mathrm{Fe}$ atom[s]"

approach.

But for practical reasons you would not look for the lowest energy at which you start getting neutrons out (which I would described as "direct threshold detection").

The practical difficulties with direct threshold detection are two:

  1. The cross-section rises from zero at threshold. So there is always a vanishingly small rate near threshold.

  2. At threshold the created neutron will have zero momentum in the CoM frame meaning that (a) in collider mode it leaves the interaction zone (if at all) with very little momentum making it extremely difficult to detect and (b) in fixed target mode it goes straight down-stream and lands in the beam-dump undetected (you can't use a 4-pi detector for this because your production rate is extremely small due to item (1), so you need a high-current beam).

What you do instead is use a variable beam energy above threhold but perform an exclusive measurement and reconstruct the neutron's CoM kinetic energy $T_n$ as a function of beam energy $E$. Then fit to find the threshold energy $$ E_\text{th} = \lim_{T_n \to 0} E \;.$$

This means that the rate is non-triviall and the ejected neutrons are energetic enough to deviate from the beam direction and arrive at a detector with enough enegy to be easily detected.

It is an entirely practical measurement, but it won't give you the kind of precision that comes from mass-spectroscopic methods.

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  • $\begingroup$ Thanks for the in-depth answer, and for the comment on macroscopic targets. Since we're not working with a single atom, wouldn't this approach lead to protons being absorbed, instead of releasing neutrons? Adding a proton to $\mathrm{^{56}Fe}$ produces $\mathrm{^{57}Co}$, with a proton separation energy of just 6.370 MeV $\endgroup$ – MrMartin Oct 17 at 9:00
  • $\begingroup$ You absolutely will get backgroud processes. You'd have that problem even if you could do a single nuclear target experiment. And it won't just be proton knock-out but also nuclear excitation with gamma relaxation, probably pion production, and (if we go very far above threshold) other stuff like cluster knock-out or the production of heavier mesons. Filtering the backgrounds is just part of the experimentor's art, and issues related to that filtering will contribute to the difficulty of obtaining high precision in this kind of measurement. $\endgroup$ – dmckee Oct 17 at 15:24

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