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Lesson 17 - Neutron response functions

In this and the next lesson, we will be using cross sections (for neutrons) and   linear interaction coefficients (for photons) to compute response functions.   We need to keep in mind for both lessons that we are talking about kerma response functions for material  that is present  in the model by which we compute the flux that the response function is multiplied by to get the kerma.   Another way of saying the same thing is: 

We are building response functions to be used with fluxes that ARE influenced by the detector material.

This lesson, concerning neutron response functions, is the easier (and shorter) of the two because of the fact that the common notation in the "neutron community" follows the exact form given in the previous lesson:

wpe1B7.gif (1845 bytes)

where  wpe1B6.gif (1006 bytes) is the energy transferred to the material medium by secondary charged particles due to neutron reaction type j of isotope i.

The secondary charged particle of most concern for above-thermal neutron reactions is the recoiling nucleus itself.  We assume is that the flux calculation itself takes care of the neutral particles -- the scattered neutrons and the gamma rays that are emitted from excited nuclei.  For the case of elastic scatter, the recoiling nucleus contains all of the initial energy of the particle minus the energy carried off by the scattering neutron. 

 

Isotropic elastic scattering

We learned earlier in the course that the average energy of the isotropically (in the COM system) elastic scattered neutron is halfway between the initial energy and the minimum energy of the scattered neutron, which means, of course, that the average energy of the recoil nucleus is one-half the energy lost, i.e.,

wpe1CF.gif (1272 bytes)

wpe1C8.gif (1546 bytes)

wpe1CA.gif (1491 bytes)

wpe1C9.gif (1765 bytes)

wpe1CB.gif (1676 bytes)

wpe1CC.gif (1881 bytes)

wpe1CD.gif (1428 bytes)

Therefore,

wpe1D9.gif (1386 bytes)

 


Example: Find the deposited energy for an elastic scattering reaction in U-235 for a 3 MeV neutron.  Assume the neutron scatters isotropically in the COM system.

Answer:  Using the above formula with A=235 gives us 0.0253 MeV.   (Boy, that number looks familiar...)


 

Anisotropic elastic scattering

If the collision is not isotropic, the deposited energy changes to:

wpe1DA.gif (1515 bytes)

where wpe1D2.gif (1003 bytes) is the average cosine of the angle of scatter in the COM system.  You must be sure to account for this anisotropy by using data (such as that found in Table 5.3).

( It makes sense that the deposited energy would decrease if the average scattering angle is forward, because -- as we learned in our study of kinematics of neutron collisions, the forward scatters are the ones in which the neutron loses the least energy.)

 


Example: Find the deposited energy for a 2 MeV neutron suffering an elastic scatter with Fe-55.  Do NOT assume that the neutron scatters isotropically in the COM, but use the iron data in Table 5.3

Answer:  Interpolation in Table 5.3 gives us a anisotropic correction term of:

wpe1DB.gif (2395 bytes)

Therefore, we have:

wpe1DD.gif (1689 bytes)


 

Inelastic scattering

For inelastic scatter (which we will assume is isotropic), the deposited energy can be found from:

wpe1D4.gif (1609 bytes)

where, as before, we have:

wpe1D8.gif (1466 bytes)


NOTE: Remember that Q -- and therefore wpe1D6.gif (875 bytes) -- are NEGATIVE numbers.


On average, then, inelastic scattering reactions result in LESS energy being deposited in the material, but no lower than half as much (since wpe1D6.gif (875 bytes) ranges from -1 to 0).

 

 


Example: Find the deposited energy for a 2 MeV neutron suffering an inelastic scatter with Fe-56 if it excites the excited state at 1.675 MeV.  Assume that the neutron scatters isotropically in the COM.

Answer:  For this situation, we have:

wpe1DE.gif (1433 bytes)

giving us

wpe1DF.gif (1866 bytes)


 


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