# Neutron microscopic cross section data

PBurke1985
Hi guys,

I am trying to calculate the macroscopic cross section (scattering and absorption) for high energy neutrons between 1 MeV and 10^7 MeV in Silicon. I am using the macroscopic cross section to determine the mean free path. To calculate the macroscopic cross section, i need to know the microscopic cross section for the given material, as a function of neutron energy, as given in the following link:

http://www.tpub.com/content/doe/h1019v1/css/h1019v1_113.htm"

Is there an analytical expression to generate this curve for a given material, or can i get this curve from some type of database online? I need the curve for both scattering and absorption.

Thanks, any help is appreciated,

Pete.

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Bob S
PBurke1985
Thanks for the reply Bob, this is exactly what i was looking for. The only issue is, i need the cross sections up to 1E13 eV. The maximum energy seems to be fixed at ~ 1E8 eV on the plot. I didn't see any options, but is there any way of increasing the limits? If not, do you have any idea how the cross section decays after 100 MeV ? Thanks very much.

Pete.

Bob S
Thanks for the reply Bob, this is exactly what i was looking for. The only issue is, i need the cross sections up to 1E13 eV. The maximum energy seems to be fixed at ~ 1E8 eV on the plot. I didn't see any options, but is there any way of increasing the limits? If not, do you have any idea how the cross section decays after 100 MeV ? Thanks very much.
Look at the cross section plots in:

http://pdg.lbl.gov/2002/contents_plots.html

First, look at np (same as pn). This is essentially barns per nucleon (proton). Now look at pp, which goes up to ~ 109 GeV (~1018 eV). At these energies, neutron and proton cross sections are about the same (minimum ~ 40 mb per nucleon at 100 GeV). For silicon, I would multiply per nucleon cross sections by ~282/3 = 9.3 (geometric cross section).

Bob S

PBurke1985
Thanks again for the info Bob. This is exactly what i need. My last question (honest), what does the 'initial state' np, pn and pp refer to? Sorry if this is a really basic question, but i'm very new to particle physics. I have only dealt with photon interactions with matter before and charged particle interactions are a lot more difficult to understand. Thanks.

Pete.

Bob S
My last question (honest), what does the 'initial state' np, pn and pp refer to? Sorry if this is a really basic question, but i'm very new to particle physics. I have only dealt with photon interactions with matter before and charged particle interactions are a lot more difficult to understand.
np refers to a neutron of energy E colliding with a proton (like in a liquid hydrogen target), while pn refers to a proton of energy e colliding with a neutron at rest (like a neutron in deuterium target). In the latter, the experimental measurement, the pd (proton on deuterium) cross section is measured, and then the contribution from pp (proton hitting a proton) has to be subtracted. The pp cross section is easily measured using a proton beam on a liquid hydrogen target.

A note on cross sections. Assume the radius of a proton is R = ~1.2 x 10-13 cm. The frontal geometric surface (target) area is thus πR2 = [STRIKE]31[/STRIKE] 45 x 10-27 cm2 = [STRIKE]31[/STRIKE] 45 millibarns.

Suppose you have a block of silicon with A=28 that is 1-cm-by-1-cm wide by z cm deep (volume z cm3). What is the probability of an incident proton (or neutron) colliding with a silicon nucleus in the block? The "volume" of the nucleus scales as A which is proportional to R3, but the frontal geometric "area" scales as R2, so the geometric frontal area scales as A2/3 = 9.3. So if the geometric area is [STRIKE]31[/STRIKE] 45 millibarns for 1 nucleon, then the geometric area of a silicon nucleus is ~9.3 x [STRIKE]31[/STRIKE] 45 millibarns = [STRIKE]290[/STRIKE] 400 millibarns per silicon nucleus. For the above silicon block, the density is 2.33 grams per cm3, so the mass of the block is 2.33 z grams. Because the gram molecular weight of silicon is 28 grams per 6 x 1023 atoms, the block contains 6 x 1023 x 2.33 z/28 silicon nuclei.

The probability of an incident proton colliding with one nucleus in the block is just the ratio of areas, which is (using [STRIKE]31[/STRIKE] 45 millibarns per nucleon or [STRIKE]290[/STRIKE] 400 millibarns per silicon nucleus)

probability ratio = 6 x 1023 x (2.33 z/28) x [STRIKE]290[/STRIKE] 400 millibarns/atom = [STRIKE]0.014[/STRIKE] 0.02 z per cm2.

So the "collision length" is about 1/[STRIKE]0.014[/STRIKE] 0.02 = [STRIKE]72[/STRIKE] 50 cm, or [STRIKE]31[/STRIKE] 115 grams per cm2. This is the thickness that reduces the incident proton flux by the factor 1/e.

The actual estimated collision length for silicon is ~ 70 grams per cm2 and interaction length ~108 grams per cm2, as shown in

http://pdg.lbl.gov/2009/reviews/rpp2009-rev-atomic-nuclear-prop.pdf

Bob S

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PBurke1985
Thanks Bob, those calculations make sense.

Cheers,

Pete.