Neutron Slowing Down Kinematics

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					      Neutron Slowing Down: Kinematics


                     B. Rouben
                McMaster University
      Course EP 6D03 – Nuclear Reactor Analysis
                  (Reactor Physics)
                   2009 Jan.-Apr.




2009 January                                      1
                        Contents
   We start the discussion of the energy dependence
    of the neutron flux by deriving the kinematics of
    neutron-nucleus collisions in the neutron
    slowing-down process.
   Reference: Duderstadt & Hamilton, Section 2.I-D
    pp. 34-45




    2009 January                                  2
                   Kinematics of Collisions
   Kinematics  relationship between pre- and
    post-collision momentum and energy in
    neutron-nucleus scattering.
   We label the neutron n and the moderator
    nucleus N.
   Using the scale of atomic masses, it is a very
    good approximation to treat the neutron as
    having mass 1 and the nucleus as having mass
    A (the moderator’s atomic mass):
         mn = 1
         mN = A
    2009 January                                 3
          Stationary Nuclei & Centre-of-Mass System
   We also make the assumption that the moderator nuclei
    are at rest. This is a good approximation for very fast-
    moving neutrons (recall that a 1-MeV neutron travels at
    13,800 km/s).
   The moderator nuclei are stationary in the reactor’s
    frame of reference, which is by convention called the
    “lab system” – which we shall denote by a subscript L.
   We need also to consider the “center-of-mass system”
    of the neutron and the nucleus – which we shall denote
    by a subscript CM.


    2009 January                                        4
                   Collision in L and CM Systems
   Consider the collision in both the L and the CM
    systems:




    2009 January                                      5
                      Velocities Before Collision
   In the L and CM systems, let the velocities of the neutron and
                                                   
    nucleus respectively be vnL , vnCM , vNL  0, and vNCM     (1)

   With respect to the L system, the center-of-mass system is itself
    moving, with velocity
                                  
              vnL  AvNL vnL  0   vnL
          vC                                            ( 2)
                  1 A      A 1    A 1
   The velocities in L and CM are then related by:
                                     A 
                   vnCM  vnL  vC        vnL
                                      1 A
                                            1 
                   vNCM    vNL  vC  vC        vnL   (3)
                                               A 1
    2009 January                                                     6
                     Velocities After Collision
   Let the velocities and energies after the collision be denoted with a
    superscript apostrophe:
                        ' '        '         '
                        vnL , vnCM , v NL , and v NCM      ( 4)
   By conservation of momentum and energy, it is easy to show that
    after the collision the speeds of the neutron and of the nucleus in
    CM are not changed, only their direction of motion (the notation
    for speeds is as for velocities, but without the arrow):
                    '               A                            1
             vnCM  vnCM  vnCM 
              '
                                         vnL , and vNCM  vNCM 
                                                    '
                                                                      vnL (5)
                                    A 1                         A 1
    (use was made of Eq. (3))
   i.e., in the collision, the neutron may be scattered in CM through
    the angle CM: the nucleus would then be moving after the
    collision in the direction given by angle CM -. In L the angle of
    scattering of the neutron would be different, say L. Note that the
    angle of motion of the nucleus in L after the collision is not L -!
      2009 January                                                              7
             Angles of Scattering in L and CM
   The velocity of the center of mass is not affected by the collision:
                                        '    '
                         vC  vnL  vnCM  vnL  vnCM   ( 6)

   The diagram below relates the L and CM angles of scattering.




   From the diagram
                  vnL sin  L  vnCM sin CM
                   '             '


                    and vnL cos L  vC  vnCM cosCM
                         '                 '
                                                               ( 7)

     2009 January                                                     8
         Relationship Between Scattering Angles
   By dividing the parts of Eq. (7) we get the relation
                     vnCM sin CM
                      '
                                         sin CM          sin CM
        tan  L                                                     ( 7)
                  vC  vnCM cosCM
                         '            vC
                                            cosCM
                                                       vC
                                                             cosCM
                                      '
                                     vnCM             vnCM
   But from the two parts of Eq. (3) we have
                    AvnL              v   1
          vnCM           AvC , i.e. C           (8)
                    A 1             vnCM A

   So that Eq. (7) becomes
                         sin CM
            tan  L                  ( 9)
                        1
                           cosCM
                        A

    2009 January                                                              9
         Relationship Between Scattering Angles
   From Eq. (9), we get
                         1           1
            cos L           
                       sec L   1  tan 2  L
                                   1
                                               1/ 2
                                        
                                        
                     1     sin CM
                                2
                                         
                      1              
                                       2

                        cosCM  
                                        
                         A           
                           1
                              cosCM
                          A
                                          1/ 2
                                                       10
                      1 2               
                      2  cosCM  1
                     A     A            

    2009 January                                              10
         Neutron Speed & Energy After Collision
   From the diagram on slide 11 we can relate the neutron speeds
    before and after the collision, using the cosine law:
                   vnL  vnCM  2vnCM vC cos  CM   vC
                    '2    '2      '                       2
                                                                 (11)
   This, together with Eqs. (2) and (5), gives:
                         2                                      2
             A  2            A        1                 1  2
      vnL  
       '2
                     vnL  2      vnL      vnL cosCM         vnL
             A  1          A 1     A 1               A  1
             A2  2 A cosCM  1 2
                                vnL                                 (12)
                    A  12



   The neutron energies in L, before and after, are in the same ratio:
                       A2  2 A cosCM  1
                   E 
                    '
                                           EnL        (13)
                    nL
                              A  12


    2009 January                                                            11
                  Neutron Energy After Collision
   Since the numerator in Eq. (13) cannot be larger than the
    denominator, we see that the neutron cannot gain energy in L on
    account of the collision – as we suspected, i.e.
           EnL  EnL , i.e.
            '


           EnL  EnL
            ' max
                                  (14)
   It is convenient to rewrite Eq. (13) using the new variable
                    A 1
                              2

                                     (15)
                    A  1
   In terms of , Eq. (13) becomes
              (1   )  (1   ) cosCM
          E '
             nL                          EnL     (16)
                           2

    2009 January                                                  12
                     Neutron’s Final Energy Range
   The minimum final energy of the neutron in L is obtained in a
    backscattering collision (L= CM = ):

                     E'min
                             
                               1     1   (1) E     EnL   (17)
                      nL                               nL
                                         2
   Note that the neutron can lose all its energy in L only in a
    backscattering collision with a nucleus characterized by  = 0
    (A = 1), i.e., only hydrogen (1H1) can stop a neutron in a single
    (head-on) collision.
   In general, summarizing Eqs. (14) and (17), the neutron’s energy in
    L goes from EnL before the collision:
                         EnL  EnL  EnL
                                 '
                                                       (18 )

   i.e., the range of the final neutron’s energy has a width of (1-)EnL.

      2009 January                                                          13
                     Probability of Specific Energy
   What is the probability of the neutron ending up with any specific
    energy value in that range?
   To answer this question, note first that for typical moderators, i.e.,
    those with light nuclei (e.g., A  12), scattering is isotropic in the
    center-of-mass system for starting neutron energies EnL 10 MeV,
    therefore the probability of scattering is independent of the
    (differential) solid angle d = d cos(CM) dCM.
   And note also from Eq. (16) that
    EnL (the neutron' s energy after the collision) is linear in cos(CM ).
     '



   We can then conclude that the probability of attaining any value of
    the final neutron energy in L is also uniform in the allowed range,
    EnL EnL.
      2009 January                                                    14
               Average Value of Neutron’s Final Energy
   That is, the probability of scattering to any value of the final
    neutron energy has the constant value (a very important conclusion)
             P EnL  EnL  
                                         1
                       '
                                                    , EnL  EnL  EnL (19 )
                                                              '

                                 1   EnL
   This also means that the scattering cross section can be written
                                     s E 
             s EnL  EnL  
                        '
                                             , EnL  EnL  EnL ( 20 )
                                                       '

                                 1   EnL
   From Eq. (19), the average value of the final neutron energy in L is
                                     P EnL  EnL EnLdEnL
                               EnL
                     EnL  
                      'avge                    '    '   '
                             EnL

                                                                               ' 2 EnL
                                     1       EnL                    1         E
                             1   EnL E
                                                   EnLdEnL 
                                                     '   '                     nL
                                               nL               1   EnL    2   EnL

                          EnL   2 EnL 1  
                           2         2
                                            EnL                                         ( 21)
                          21   EnL    2
      2009 January                                                                                15
                     Average Energy Loss
   The average loss in energy of the neutron is therefore
                                1         1
    E      EnL  E
        avge
                      nL EnL 
                      'avge
                                      EnL        EnL (22)
                                  2           2
    Thus we can write that, on each collision, the average
    fractional loss of energy of the neutron is :
    E avge 1  
                             ( 23)
      E       2
   Examples:
   In a collision with a hydrogen nucleus (1H1,  = 0), a neutron
    loses on average half its energy.
   In a collision with a deuterium nucleus (2H1,  = 1/9), a neutron
    loses on average 4/9 of its energy.
     2009 January                                                  16
                               Neutron Lethargy
   The quantity u  “neutron lethargy” is a function of the neutron’s
    energy as it slows down, and is defined as
            E 
     u  ln  0      (24),
             E
     where E0 is defined as an upper lim it to the neutron energy ,
     and is often taken as E0  10 MeV       (25)
   With that definition, the neutron lethargy increases as the neutron
    slows down, as is appropriate for the name “letahrgy”.
   The gain in lethargy after a collision, denoted , is
                     E0        E        E 
      u  ln 
                           ln  0   ln  nL   log arithmic energy loss (26)
                     EnL 
                       '
                          
                                 E 
                                  nL 
                                             E' 
                                             nL 



     2009 January                                                               17
              Interactive Discussion/Exercises
                                                  '
   1): U sin g the probability distribution for EnL [ Eq.(18)], and the
        definition of the average gain in leth arg y after a collision,
        show that the latter is equal to

          1
                
                   ln   1 
                               A  12 ln  A  1 
                                                     (27)
               1               2A         A 1
  2) This formula cannot be used as is for hydrogen
  ( = 0). Show, by writing  = e-x and letting x,
  that it reduces in that case to  = 1.
 3) Note that the average gain in lethargy is constant
   from collision to collision. This can be used to
   determine the average number of collisions to
   thermalize a neutron (say, from 2 MeV to 1 eV).
    2009 January                                                   18
              Interactive Discussion/Exercises
   On average, how many collisions does it take for a 2-
    MeV neutron to be thermalized by a hydrogen
    moderator to an energy of 1 eV?
   Same question, but for a deuterium moderator
   Same question, but for a carbon (graphite) moderator
   Are the answers above sufficient to get a sense of the
    different total distances travelled by a neutron as it is
    thermalized by the different moderators, or do you need
    additional information, and, if so, which?



    2009 January                                         19
               END


2009 January         20