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SLOWING DOWN

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					         Neutron Slowing-Down

                Vasily Arzhanov
               Reactor Physics, KTH




                  Overview
     •   Lethargy
     •   Continues Slowing-Down Model
     •   Discontinues Slowing-Down Model
     •   Slowing-Down with Absorptions
     •   Resonance Escape Probability
     •   Neutron Spectrum in Reactor



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                                               1
     Principles of a Nuclear Reactor
      E                                                                                              Leakage
                                                                  N2
     ?
                                                                  N1


                                                       N                              Fast fission




                                                                                                                 Slowing down
                                                     k≡ 2
                    ν n/fission
     Energy




                                                       N1                             Resonance abs.
                                     ν=?
                                                              Non-fissile abs.         Non-fuel abs.

     ?
                                        Fission

                                           ?
                                                                       Leakage
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                                    Lethargy Variable
                        E0
                                     E0
          E0           α
                        ∫ ln         E
                                        p( E)dE
                                                          α
ξ ≡ ln       =                                    = 1+         ln α
                         Eo
                                                                             Does not depend on energy!!!
          E                    E0
                                                         1−α
                               ∫
                              α Eo
                                     p( E)dE
                                                                                              Eref              dE
                                                                             u( E) ≡ u ≡ ln          ; du ≡ −
                                                                                               E                 E
     E0      ⎛ E Eref ⎞                                                                       Eref ∼ 10MeV
        = ln ⎜ 0        = u( E) − u( E0 ) ≡ Δu( E)
             ⎜ Eref E ⎟
ln
     E                ⎟
             ⎝        ⎠

         E0
                                                                        u ⎯⎯⎯ u + ξ ,
                                                                           1 coll
                                                                                  →                   on average
       ∫ Δu(E) p(E)dE                                α
 ξ=
      α  Eo
                                         = Δu = 1 +     ln α                   →
                                                                        u ⎯⎯⎯ u + Δumax , at most
                                                                             1 coll
               E0
                                                    1−α
               ∫
              α Eo
                     p( E)dE                                                       E
                                                                        Δumax = ln 0 = ln α −1
                                                                                  α E0
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                                                                                                                                2
                   Lethargy Scale
                                                                            1 average collision

                                                           α E0                                E0
                                                                                                                   Energy
                                                                                    ΔE                           1−α
                                                                                                       ΔE =          E0
                                                                                                                  2
         u                                                                           ξ
                                                                                                                   Lethargy
                              Δumax = ln α −1
         u                                                                           ξ
                                                                                                                   Lethargy
                                                                      Δu                       u0

It is tempting to assume: the number of                                            Δu                 Strictly speaking
                                                                      n1 =
                                                                       coll
                                                                                                      this is not true!!!
collisions per 1 neutron to traverse Δu is                                         ξ

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                 Warning Message
                                    Eref                                    dE
                          u ≡ ln                  ; du ≡ −
                                            E                                E

   Average number       du       dE   dE   dE                                                  Average number
     of collisions to        =      =    =                                                     of collisions to
         traverse du     ξ       ξ E ΔE 1 − α E                                                traverse dE
                                           2
                                   1

                                  0.9

                                  0.8



                                                      ξ
                                  0.7


   1−α                            0.6

ξ=     Erroneous result!!!        0.5

    2                             0.4

                                  0.3

                                  0.2
                                                (1 − α ) 2
                                                                                                                     A −1⎞
                                                                                                                             2

                                                                                                                α =⎛
                                  0.1

                                   0
                                                                                                                   ⎜     ⎟
                                                                                                                   ⎝ A +1⎠
                                        0       0.1       0.2   0.3   0.4    0.5   0.6   0.7    0.8   0.9   1




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                                                                                                                                     3
 Average Number of Collisions
A very important             The average number of collisions made by a neutron in
quantity:                    traversing a lethargy interval u1 to u2 where 0 ≤ u1 ≤ u2 < ∞.
                     Δu
           n1 =
            coll

                      ξ       This formula is true for H!!

S.M. Dancoff has shown: the average gain in lethargy,               Δu   , by a neutron after
n collisions with nuclei of mass A is given by

                                      Δu = n ⋅ ξ
For the general case, A > 1, the corresponding calculations are very complicated,
no simple approximating formulas have been found yet. However, it was shown
that this formiula is reasonably accurate, particularly when
                                 Δu   u −u
                                     = 2 −11                   1
                                Δumax ln α
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Frequency Function in Lethargy
              We have already found
            the frequency function in
                                                  1
                                     angle: p(θ ) = sin θ
                                                  2
                                                  1
                                  cosine: p(η ) =
                                                  2
                                                  ( A + 1)2         1
                                  energy: p( E) =           =
                                                    4 AE0     ( 1 − α ) E0

         p(u)du = − p( E)dE
                   Eref              Eref              e −(u−u0 )
         u0 = ln          ; u = ln          ⇒ p(u) =
                   E0                 E                 1−α

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                                                                                                4
         Slowing-Down Features
           of Some Moderators
             N=
                     (
                  ln E f Eth     ) = ln ( 2 ⋅ 10   6
                                                       0.025 )
                                                                 =
                                                                     18.2
                         ξ                     ξ                      ξ
         Moderator           ξ          N               ξΣs          ξΣs/Σa
           H 2O          0.927          19.6           1.36                 62
           D 2O          0.510          35.7           0.18           5860
           9Be           0.207          88.0           0.15               138
            12C          0.158         115.3           0.06               166
           238U          0.008       2171.6            .0040          0.011

         N - number of collision to thermal energy
         ξΣs - slowing down power
         ξΣs/Σa - moderation ratio (quality factor)
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         Infinite Medium Model
• Purpose: to describe neutron motion in
  energy space, not in physical space.
• Medium is
    – homogeneous
    – isotropic
    – infinite in extent
• Neutron sources
    – are uniformly distributed in space
    – emit neutrons isotropically

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                                                                                      5
                       Basic Quantities
                                          Eref
            mv 2 2 = E;        u ≡ log
                                              E


              φ ( E) ≡ v ⋅ n( E) ⎡# ( cm 2 ⋅ s ⋅ J ) ⎤
                                 ⎣                   ⎦

                                             unit volume per unit time
           φ ( E)dE = Total track length per energies between E and Eof dE.
                      neutrons with kinetic                            +

                                 number
             Σ s ( E) = Expectedof energyof elastic-scattering collisions between
                        neutrons          E and nuclei per unit distance of travel.


             Σ a ( E) = Σγ ( E) + Σ f ( E);       φ ( E)dE = φ ( v)dv = −φ (u)du


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          Continues Slowing Down
                       E

                                                      The actual (discontinues)
            Slowing
                                                      slowing down neutron-energy
          down track
                                                      track is very difficult to handle.




                                                                Distance
                       E

                                                      Much attention has been devoted
Slowing down track                                    to find a continues function which
    in a medium of                                    will serve as a good average over
heavy nuclear mass                                    all the possible step functions.


                                                                Distance
 HT2008                                Slowing down                                        12




                                                                                                6
         Continues Slowing Down
                 Model
• The energy loss is continues in time process.
• Neutron energy changes are due to collisions
  occurring in arbitrary small (differential)
  decrements.
• To describe this process the concept of slowing
  down density, q(E), is defined.
• The model is fairly accurate for heavy nuclei.
• On the contrary, the model is not that useful for
  light nuclei; a more accurate description is
  needed.

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         Slowing Down Density
                                                       Energy          It does
                                                                       not
                                                  q0       E0          count


 q(E) = Number of neutrons per unit volume per
    unit time whose energies are changed from          E        q(E)
   some value above E to some value below E.

             Simple steady state model:
   • Neutrons are introduced at energy E0                  0       It counts
   uniformly in space and continuously in time.
   • Neutrons are removed at zero energy.
   • Medium is a pure scatterer, Σa = 0.

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                                                                                 7
                         Steady State
                                             Energy
                                                               No leakage
                                                               No absorption
  q( E) = 0;    E > E0
                                                               No accumulation

 q( E0 ) = q0             Source:  q0             E0

                                                       dqin = dqout
                                        E + dE

                                            E
                                                       dqout


                           Removal                0



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           Slowing Down Density
                 Solution
                     Energy
                                                  q( E + dE) = dqin + q
                                                         q( E) = q + dqout
                        dqin
                                                  q( E + dE) = q( E)
            E + dE                                      dq( E)
                                                               =0
                                  q                      dE
                 E

                         dqout                           q( E) = q0



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                                                                                      8
           Flux Solution in Lethargy
                                             Total number of scatterings per
                     Σ s (u)φ (u)du =        unit volume, per unit time due to
                                             neutrons with lethargy u about du

  Number of neutrons per                     Average number of
unit volume, per unit time                   scatterings made by a
 that slow past lethargy u                   neutron in traversing du

                                    du       Total number of scatterings per
                             q(u)        =   unit volume, per unit time due to
                                    ξ        neutrons with lethargy u about du


                             du                             q0
 Σ s (u)φ (u)du = q(u)                        φ (u) =              const
                             ξ                          Σ s (u)ξ
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           Flux Solution in Energy
                                                     dE
                                             du = −
                                                      E
                                             φ ( E)dE = −φ (u)du

             The continues slowing                          q0      q     1
             down model predicts:            φ ( E) =             = 0 ∝
                                                        ξΣ s ( E)E ξΣ s E E
           φ (u)                               φ ( E)




                                         u                                 E

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                                                                                      9
    Discontinues Slowing Down
              Model
•   It accounts for discontinuity of neutron trajectory.
•   Infinite medium model.
•   Target nuclei are at rest (E > 1 eV).
•   Scattering is isotropic in CoM (E < 1 MeV).
•   System is at steady state.
•   Neutron source is uniform and monoenergetic:
    q0 neutrons of E0 appear per unit volume and
    time from uniformly distributed sources
    throughout the medium.
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            Collision Intervals
               Source
                                    E0

                                    E              1st collision interval
                  Neutron energy




                                    αE0
                                                   2nd collision interval
                                   α2E   0
                                                   3d       collision interval
                                   α3E0


                                                                              A −1⎞
                                                                                      2

                                                                       α =⎛
                                      0
                                                                          ⎜       ⎟
               Removal                                                      ⎝ A +1⎠

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                                                                                               10
           Solution in 1st Interval
                                                    Σ ( E′ ) φ ( E′ )
                                                   E0
                                                                           q
         Equation         Σs ( E) φ ( E) =         ∫ (1 − α ) E′ dE′ + (1 − α ) E
                                                            s                             0

                                                    E                                           0



Collision density          F1 ( E) ≡ Σ s ( E ) φ ( E )

                                                 F1 ( E′)
                                         E0
                                                                         q0
         Equation           F1 ( E) =    ∫ (1 − α ) E′ dE′ + (1 − α ) E
                                         E                                        0



                                                                         1 ( 1−α )
                                              q0         ⎛ E0 ⎞
                            φ1 ( E) =                                                         α E0 ≤ E ≤ E0
                                      (1 − α ) E0Σs ( E) ⎜ E ⎟
         Solution                                                                     ;
                                                         ⎝ ⎠


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     Solution in Other Intervals
                                                 α n−1 E0                 Eα
                                                            Fn ( E′)dE′        F ( E′)dE′
    Recurrence relation            Fn ( E) =        ∫E
                                                                        + ∫ n −1
                                                            (1 − α ) E′ α n−1E (1 − α ) E′
                                                                              0



                                             α ( 1−α )
                                 q0 ( E0 )               ⎧ α 1 (1−α ) ⎛ E ⎞
                                                         ⎪                             1 ( 1−α ) ⎪
                                                                                                  ⎫
   F2 ( E) ≡ Σ s ( E)φ ( E) =                            ⎨           ln ⎜   ⎟ + ⎡1 − α
                                                                                ⎣
                                                                                                ⎤⎬
                                                                                                ⎦
                                ( 1 − α ) E1 (1−α )      ⎪ ( 1 − α ) ⎝ α E0 ⎠
                                                         ⎩                                        ⎪
                                                                                                  ⎭

                                   Eα
                                        F∞ ( E′)dE′                               q
   Fn ( E) ⎯⎯⎯ F∞ ( E) =
            n→∞
                →                  ∫E   ( 1 − α ) E′
                                                     ⇒ F∞ ( E) ≡ Σ s ( E)φ∞ ( E) = 0
                                                                                  ξE
                                                                                                   q0
                                                                                  φ∞ ( E) =
                                                                                                ξ EΣ s ( E)

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                                                                                                                   11
    Collision Density in Energy
               80


               70
                                                                                       q0
                                          Asymptotic form F∞ ( E) =
               60
                                                                                       ξE
               50


               40


               30                               F3 ( E)
                                                              F2 ( E)
               20                                                               F1 ( E)

               10


                0
                    0     0.1       0.2         0.3     0.4     0.5       0.6    0.7      0.8   0.9       1
                                                                                                                E E0
                            4th           3rd         2nd interval                  1st interval


                          α4 α3                  α2                   α                                   1
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              Solution in Lethargy
          Collision density in lethargy may be obtained by
             (1) Repeating the derivation in lethargy scale

             (2) Using the relationship:                       F( E)dE = − F(u)du;                    du = − dE E

          q0 = 1 ⇒ Normalized density or Placzek functions

                         1 α (1−α )u
          F1 (u) =          e
                        1−α
                                                                          u
                                α (1−α )( u − n ln1 α )         1                                      α (1−α )( u −u′)
         Fn+1 (u) = Cn+1e                                 −
                                                              1 −α        ∫ α F (u′ − ln 1 α )e
                                                                      n ln1
                                                                                n                                         du′




HT2008                                                Slowing down                                                          24




                                                                                                                                 12
 Collision Density in Lethargy
                    2.6


                    2.4                                                            A= 4                             1
                                                                                                         F∞ (u) =       = 2.35
                                                                                                                    ξ
                    2.2


                    2.0
                                                                                       F3 (u)
                    1.8                                    F2 (u)

                    1.6              F1 (u)
                                                                                   A= 2                             1
                    1.4                                                                                  F∞ (u) =       = 1.38
                                                                                                                    ξ
                    1.2

                                                                                   A= 1                             1
                    1.0                                                                                  F∞ (u) =       =1
                          0
                                              ln
                                                   1
                                                                        2 ln
                                                                               1
                                                                                          3 ln
                                                                                                 1                  ξ
                                                   α                           α                 α
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     Slowing Down in Mixtures
         Mixture of nuclides:                 Ai , Ni ,σ s ,i
                                                                                   2
                                                ⎛ A −1⎞                                                   αi
         Σ s ,i = Niσ s ,i                 αi = ⎜ i      ⎟                                   ξi = 1 +          ln α i
                                                ⎝ Ai + 1 ⎠                                              1 − αi

         qi (u) = ξi Σ s ,i (u)φ (u)
         q(u) = ∑ qi (u) = ∑ ξ i Σ s ,i (u)φ (u) = ξ ⋅ Σ s (u)φ (u)
                      i                        i

               ∑ξ Σ       i   s ,i   (u)     ∑ ξ Σ (u)       i     s ,i
         ξ =    i
                                           =       i

                     Σ s (u)                 ∑ Σ (u)   i
                                                                 s ,i


HT2008                                                 Slowing down                                                          26




                                                                                                                                  13
Slowing-Down with Absorption
                                                                                            Σa
                                          Absorption                   Pr {Absorption} =
                                                                                            Σt
         Each collision
                                                                                           Σs
                                          Scattering                   Pr {Scattering} =
                                                                                           Σt

           Neutrons may be removed from the slowing-down stream.
           Further complication: neutrons may appear due to fission.
           Consequence: q(E) ≠ const
           First step: Slowing-down in absorbing medium of pure hydrogen.



HT2008                                        Slowing down                                       27




Slowing-Down with Absorption
        in Hydrogen
                   Scattered out                        Scattered in



                                              Σ s ( E′) φ ( E′)
                                         E0
                                                                           q0
Previous result:    Σs ( E)φ ( E) =      ∫                      dE′ +
                                         E      ( 1 − α ) E′          ( 1 − α ) E0

                                         E0
                                              Σ s ( E′ ) φ ( E′ )           q0
            Now:    Σt ( E ) φ ( E ) =   ∫                          dE′ +
                                         E
                                                      E′                    E0
                          F ( E)




HT2008                                        Slowing down                                       28




                                                                                                      14
            Solution for Hydrogen
          d           Σ ( E)                     q      ⎛ E0 Σ ( E′ ) dE′ ⎞
            F( E) = − s         F( E) ⇒ FH ( E) = 0 exp ⎜ − ∫ a           ⎟
         dE          Σt ( E ) E                   E     ⎝ E Σt ( E′ ) E′ ⎠
                                                        ⎜                 ⎟


                                                    q0          ⎛ E0 Σ a ( E′ ) dE′ ⎞
                                        φH ( E) =           exp ⎜ − ∫               ⎟
                                                  EΣt ( E )     ⎜ E Σt ( E′ ) E′ ⎟
                                                                ⎝                   ⎠


                                                         ⎛ E0 Σ a ( E′) dE′ ⎞
          q( E) = EF( E)                qH ( E) = q0 exp ⎜ − ∫              ⎟
                                                         ⎝ E Σt ( E′) E′ ⎠

HT2008                              Slowing down                                29




     Non-Absorption Probability
                                                      qH ( E) qH ( E)
q0        E0      qH (E0 ) = q0      pH ( E; E0 ) =             =
                                                      qH ( E0 )   q0


                                                        ⎛ E0 Σ a ( E′ ) dE′ ⎞
                                     pH ( E; E0 ) = exp ⎜ − ∫               ⎟
                                                        ⎜ E Σt ( E′ ) E′ ⎟
                                                        ⎝                   ⎠
           E         qH ( E)
                                                         ⎛ u Σ a ( u′ ) du′ ⎞
                                       pH ( u; 0 ) = exp ⎜ − ∫              ⎟
                                                         ⎜ 0 Σ ( u′ ) ⎟
                                                         ⎝        t         ⎠
                 0


HT2008                              Slowing down                                30




                                                                                        15
 Slowing-Down in General Case
              Neutrons                Neutrons
              removed                 added

                            ⎧ E0 Σ s ( E′ ) φ ( E′ )           q0
                            ⎪∫                       dE′ +             ;       α E0 < E < E0
                            ⎪ E ( 1 − α ) E′               (1 − α ) E0
          Σs ( E) φ ( E ) = ⎨E α
Previous
                            ⎪ Σ s ( E′ ) φ ( E′ )
   result
                            ⎪ ∫ ( 1 − α ) E′ dE′;                                      E < α E0
                            ⎩E

                             ⎧ E0 Σ s ( E′ ) φ ( E′ )           q0
                             ⎪∫                       dE′ +             ;      α E0 < E < E0
                             ⎪ E ( 1 − α ) E′               (1 − α ) E0
 Present
          Σt ( E ) φ ( E ) = ⎨ E α
                             ⎪ Σ s ( E′ ) φ ( E′ )
situation
                             ⎪ ∫ ( 1 − α ) E′ dE′;                                     E < α E0
                             ⎩E
                                There is no analytic solution!
  HT2008                                    Slowing down                                       31




                      Adopted Solution
                                                      ⎡ 1u         Σ a (u′)         ⎤
                                      q(u) = q(0) exp ⎢ − ∫                      du′⎥
    Approximations:                                   ⎣ ξ 0 Σ s (u′) + Σ a (u′) ⎦
 1) Asymptotic range
 2) Small absorption                                   ⎡ 1 E0       Σ a ( E′)     dE′ ⎤
                                      q( E) = q(0) exp ⎢ − ∫                          ⎥
                                                       ⎣ ξ E Σ s ( E′) + Σ a ( E′) E′ ⎦


                                           q0               ⎡ 1u        Σ a (u′)        ⎤
                      φ (u) =                           exp ⎢ − ∫                    du′⎥
                                 ξ [ Σ s (u) + Σ a (u)]     ⎣ ξ 0 Σ s (u′) + Σ a (u′) ⎦


                                             q0                 ⎡ 1 E0       Σ a ( E′)     dE′ ⎤
                      φ ( E) =                              exp ⎢ − ∫                          ⎥
                                 ξ [ Σ s ( E) + Σ a ( E)] E     ⎣ ξ E Σ s ( E′) + Σ a ( E′) E′ ⎦


  HT2008                                    Slowing down                                       32




                                                                                                    16
   Non-Absorption Probability
                            ⎡ 1u         Σ a (u′)         ⎤
                 p(u) ≡ exp ⎢ − ∫                      du′⎥
                            ⎣ ξ 0 Σ s (u′) + Σ a (u′) ⎦
                             ⎡ 1 E0       Σ a ( E′)     dE′ ⎤
                 p( E) ≡ exp ⎢ − ∫                          ⎥
                             ⎣ ξ E Σ s ( E′) + Σ a ( E′) E′ ⎦
                                                                                         Also called resonance
More generally                                                                           escape probability
                                       ⎡ 1 u2 Σ (u′) ⎤
                     p(u1 → u2 ) ≡ exp ⎢ − ∫ a        du′⎥                               Σa
                                       ⎢ ξ u1 Σt (u′)
                                       ⎣                 ⎥
                                                         ⎦
                                        ⎡ 1 E2 Σ ( E′) dE′ ⎤
                     p( E1 ← E2 ) ≡ exp ⎢ − ∫ a            ⎥
                                        ⎢ ξ E1 Σt ( E′) E′ ⎥
                                        ⎣                  ⎦                                                     E

HT2008                                                Slowing down                                          33




                            Resonance Escape
                                              ⎡ 1 E0      Σ a ( E)     dE ⎤
                            p( E ← E0 ) = exp ⎢ − ∫                       ⎥
                                              ⎣ ξ E Σ a ( E) + Σ s ( E) E ⎦
                     10 4
                                                              235
                                                                     U
                     10 3


                     10 2
         σ (barns)




                                                                                  fission
                     10 1


                     10 0
                                                                     capture
                 10 -1


                 10 -2 -3
                     10       10 -2   10 -1   10 0   10 1   10 2    10 3   10 4   10 5        10 6   10 7

                                                      Energy (eV)
HT2008                                                Slowing down                                          34




                                                                                                                     17
              Simplified Derivation
No absorption: q(u) = const                                    Energy         Lethargy
                                                                 E/α           u–lnα-1


                                                                    E           u
Relative change          Probability of                                         u+du
                                                               E ‒ dE
        in q due         absorption
    absorptions          upon a
                         collision


             dq du   Σa                                   ⎡ 1u        Σ a (u′)        ⎤
            − =    ⋅                        q(u) = q0 exp ⎢ − ∫                    du′⎥
              q ξ Σa + Σs                                 ⎣ ξ 0 Σ s (u′) + Σ a (u′) ⎦

                   Number of
                   collisions per
                   one neutron in
                   traversing du
 HT2008                              Slowing down                                      35




          Energy Dependent Flux
          φ ( E)                           q0                           Infinite medium,
                              φ ( E) =                                  no absorption,
                                          ξΣ s E                        constant Σs

                                                   q0 p( E)             Infinite medium
                              φ ( E) =
                                         ξ [ Σ s ( E) + Σ a ( E)] E     with absorption




                              Eres                      E0




 HT2008                              Slowing down                                      36




                                                                                            18
          Lethargy Dependent Flux

                                                                           Σ a (u)


φ (u)



                                                                                 q0 p(u)
                                                          φ (u) =
                                                                           ξ [ Σ s (u) + Σ a (u)]


    0
                                                                                u

 HT2008                          Slowing down                                                        37




   Neutron Velocity Distribution
          kB = 1.381×10-23 J/K = 8.617×10-5 eV/K

                                 v+dv                     Probability of being at
                   v                                      energy level E=mv2/2 :
  Velocity                                                                                 E
                                                                                     −
    space:                       4πv2dv                   p( E)dE = e                    k BT
                                                                                                dE



                                                                  mv 2
  Neutrons are at                         4π v 2              −
  thermal equilibrium   n( v ) = n0                   3
                                                          e       2 k BT

  with medium
                                      ( 2π kBT   m)   2




 HT2008                          Slowing down                                                        38




                                                                                                          19
         Maxwell Distribution for
            Neutron Density
                                                                                    2
                                                                            ⎛ v ⎞
                                                           4 v 2 − ⎜ v0 ⎟
         Thermal spectrum                       n( v) = n0        e⎝ ⎠                         Hard spectrum
                                                            π v03




                                                     1

                                                    0.9
 The most                                           0.8

 probable
                       2 k BT                       0.7

 velocity: v MP = v0 =                              0.6
                                            m       0.5

                                                    0.4

and                        2                        0.3
                       mv
energy:         E0 =      = k BT
                           0                        0.2
                        2                           0.1
                                                                       v0
                                                     0
                                                          0   1000   2000       3000    4000   5000       6000   7000   8000

HT2008                                           Slowing down                                                           39




               Maxwell Distribution
                for Neutron Flux
                       ⎛ v ⎞
                               2
                                                                                        2kBT        m
             4 v 3 −⎜ v0 ⎟
 φ ( v) = n0 3     e⎝ ⎠ =
                                                                vMP = v0 =                   = 2200   ( 20 °C )
                                                                                         m          s
            v0 π                                                            ∞

                                                                            ∫ vn(v)dv
                                    2
                           ⎛ v ⎞
                 4 v3     −⎜ ⎟
                                                                                                   2
         = n0 v0 4    e    ⎝ v0 ⎠
                                        =                            v=     0
                                                                                               =          v0 = 1.128v0
                v0 π                                                         ∞
                                                                                                      π
                       ⎛ v ⎞
                               2                                            ∫ n(v)dv
              4 v 3 −⎜ v0 ⎟                                                 0
         = φ0 4     e⎝ ⎠                                                                           3 2
             v0 π                                                                        v2 =        v0
                                                                                                   2
                                                                                        mv 2 3
                                                                                            = kBT
                              mv 2                                                       2   2
 Don’t forget :            E=                             dE = mvdv
                               2
HT2008                                           Slowing down                                                           40




                                                                                                                               20
    Under the Neutron Life-Time
             E

 10 MeV
                                                               E
                                              E      −
                              φ ( E) = φ0           e k BT ;                      kBT ≈ 2.2 MeV;    T ≈ 2.5 ⋅ 1010 K
                                          ( kBT )
                                                  2



0.1 MeV
                                                       q0 p( E)
                                  φ ( E) =
                                             ξ [ Σ s ( E) + Σ a ( E)] E
   1 eV
                                                                         E
                                                                   −
                                                     E
                                   φ ( E) = φ0                 e       k BT
                                                                              ;    kBT ≈ 0.025 eV;     T ≈ 300K
                                                  (k T )
                                                           2
                                                     B
         0


 HT2008                                            Slowing down                                                41




                                         Life Time
                                                    How long time does the neutron exist under
Slowing-down time = ts                              slowing-down phase respectively as thermal?

         Number of                                   Number of
    collisions in du                                 collisions in dt


                         du       dE 2dv vdt       2λ dv
                              =     =   =    ⇒ dt = s 2
                         ξ        ξE ξv   λs        ξ v

            2λs ( v) dv 2λ ⎛ 1 1 ⎞
         v0
                                     2 1                                            v(1 eV) = 1.39 · 106 cm/s
  ts =   ∫ ξ v2 ≈ ξ s ⎜ v1 − v0 ⎟ ≈ ξΣs v1
         v1                ⎝     ⎠
                                                                                     v(0.1 MeV ) = 4.4 · 108 cm/s



              λa        1
  tth =            =               Thermal life-length
              v        Σav
 HT2008                                            Slowing down                                                42




                                                                                                                       21
    Neutrons Slowing-Down Time
      and Thermal Life-Time
         Material   ts [μs]          tth [μs]
          H2O         1              2×102
          D2O         8              1.5×105
           Be         10             4.3×103
            C        25              1.2×104



HT2008                Slowing down              43




                    The END




HT2008                Slowing down              44




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