22.101 Applied Nuclear Physics (Fall 2006)

Lecture 16 (11/8/06)

Neutron Interactions: Q-Equation , Elastic Scattering

References :

R. D. Evans, Atom ic Nucleus (McGraw-Hill New York, 1955), Chap. 12.

W. E. Meyerhof, Elements of Nuclear Physics (McGraw-Hill, New York, 1967), Sec. 3.3.

Since a neutron has no charge it can easily enter into a nucleu s and cause a reaction. N e utrons interact prim arily with the nucleus of an atom , except in the special case of m a gnetic scattering where the interaction involves the neutron spin and the m a gnetic mom ent of the atom . Beca use m a gnetic scattering is of no interest in this class, we can neglect the interaction between neutrons and electrons and think of atom s and nuclei interchangeably. Neutron reactions can take place at any energy, so one has to pay particular attention to the energy variation of the interaction cross section. In a nuclear reactor neutrons can have energies ranging from 10 -3 ev (1 mev) to 10 7 ev (10 Mev). T his m eans our study of neutron interactions, in prin ciple, will have to cover an energy range of 10 ten orders of m a gni tude. In practice we will lim it ourselves to two energy ranges, the slowing down region (ev to Kev) and the therm al region (around 0.025 ev).

For a given energy region – therm al, epitherm al, resonance, fast – not all the possible reactions are equally im portant. W hich reaction is important depends on the target nucleus and the neutron energy. Generally speaking the im portant types of interactions, in the order of increasing com plexity from the standpoint of theoretical understanding, are:

(n,n) – elastic scattering. There are two processes, potential scattering which is neutron interaction at th e surface of the nucleus (no penetr ation) as in a b illiard ball-like collision, and resonance scattering which involves the for m ation and decay of a com pound nucleus.

(n, ) -- rad iative cap ture.

(n,n’) -- inelastic scattering. This re action involves the excitation of nuclear levels.

(n,p), (n, ), … -- charged particle em ission. (n,f) -- fission.

If we were interested in fission reactors , the reactions in the order of im portance would be fission, capture (in fuel and other reactor m a terials), scattering (elastic and inelastic), an d fission pro duct decay b y  -em ission. In this lecture we will mostly study elastic (or potential) scattering. The other reactions all involve com pound nucleus for m ation, a process we will discuss briefly arou nd the end o f the sem ester.

The Q-Equation

Consider the reaction, sketched in Fig. 15.1, where an incom i ng particle, labeled 1, collides w ith a target n u cleus (2 ), resulting in the em ission of an outgoing particle (3), with the residual nucleus (4) recoilin g. For sim plicity we assum e the target nucleus to be

Fig. 15.1. A two-body collision between incident particle 1 and targ et p a rticle 2, wh ich is at res t, leading to th e em ission of particle 3 at an angle  and a recoiling residual particle 4.

at rest, E 2 = 0. This is often a good approxim ation because fo r a target at room tem perature E 2 is 0.025 ev; except for incom ing neutrons in the therm al energy region, E 1 typically will be m u ch greater than E 2 . W e will derive an equ ation relatin g the outgo ing energy E 3 to the outgoing angle  using the conservation of total energy an d linear mom entum , and non-rela tivis tic kinem a tics,

( E 1  M 1 c 2 )  M 2 c 2 ( E 3  M 3 c 2 )  ( E 4  M 4 c 2 ) (15.1)

p p  p

1 3 4

(15.2)

Rewriting the m o mentum equation as

p 2 ( p  p ) 2

4 1 3

= p 2  p 2  2 p p cos  2 M E

(15.3)

1 3 1 3 4 4

and recalling

Q ( M 1  M 2  M 3  M 4 ) c 2

E 3  E 4  E 1 (15.4)

we obtain

 M 3   M 1  2

Q E 3  1  M

  E  1   

4   4 

4

cos  (15.5)

which is kno wn as the Q-equation. N o tice that th e energ ies E i and angle  are in the laboratory coordinate system (LCS), while Q is independent of coordinate system (since Q can be ex pressed in term s of m a sses which of course do not depend on coordinate system ). A typical situ ation is when the incident energy E 1 , the m a sses (and therefore Q- value) are all known, and one is interested in solving (15.5) f o r E 3 in term s of cos  , or vice versa.

Eq. (15.5) is actually not an equation for determ ining the Q-v alue; this is already known because all four particles in the reac tion and their rest m a sses are prescribed

beforehand. W h at then is the quantity that one can solve (15.5) to obtain? W e should think of the Q-equation as a relation connect ing the 12 degrees of freedom in any two- body collision problem , where two particles collide (as reactants) to give rise to two other particles (as products). The pr oblem is com ple tely specified when the velocities of the fours particles, or 12 deg rees of freed om (each velocity has 3 degrees of freedom ), are determ ined. Clearly not every single degree of freedom is an unknown in the situations of interest to us. Suppose we enum er ate all the degrees of freedom to see which is given (known) and which is a variable. First if the direction of travel and energy of the incom ing particle are given, usually the case, this specifies 3 d e grees of freedom .

Secondly it is custom ary to take the target nucleus to be stationary, so another 3 degrees of freedom are specified. Since conservations of energy and mom entum must hold in any collision (th ree cond itio ns since m o m entum and energy a r e related), th is leaves thre e degrees of freedom that are not specif ied in the problem . If we further assum e the em ission of the outgo ing particle (particle 3) is az im uthally sy mmetric (th at is, em ission is equally probably into a cone subtended by the angle  ), then only two degree of freedom are left. This w a y of counting shows that the ou tco m e of the collision is com pletely determ ined if we just specify another degree of freedom . Wha t variable should we take? Because we are often inte rested in knowing the energy or direction of travel of the outgoing particle, we can choose this last variable to be either E 3 or the scattering an gle  . In other words, if we know either E 3 or  , then everyth ing else (energy and direction) about the collision is determ ined. Kee p ing this in m i nd, it should com e as no surprise that what we will do with (1 5.5) is to turn it in to a relation between E 3 and  .

Thus far we have used non-relativistic expressions for the kinem atics. To turn (15.5) in to the relativ istic Q-equatio n we can simply replace the rest m a ss M i by an

effective m a ss, M eff M

 T i

/ 2 c 2 , and use the expression p 2 2 MT  T 2 / c 2 instead

of p 2 2 ME . For photons, we take M eff h  / 2 c 2 .

Inspection of (15.5) shows that it is a quadratic equation in th e variable x = .

An equation of the for m ax 2  bx  c 0 has two roots,

x    b  b 2 4 ac  / 2 a (15.6)

which m eans there are in general tw o possible solutions to the Q-equation, 

solution to b e physically acceptable, it m u st be real and positive. Thus th ere are

. For a

situations w h ere the Q-equation gives one, two, or no physical solutions [ c f. Evans, pp. 413-415, M e yerhof, p. 178]. For our purposes w e will focus on neutron collisions, in particular th e case of elastic (Q = 0) and in elastic (Q < 0) neu tron scattering. W e will exam ine these two processes briefly and then return to a m o re detailed discussion of elastic scattering in the laboratory and center-of-mass coord inate system s.

Elastic vs. Inelastic Sca ttering

Elastic scattering is the sim plest pro cess in neu tron interactio ns; it can be analyzed in com plete detail. This is also an im portan t process because it is the prim ary m echanis m by which neutrons lose energy in a reactor, from the instant th ey are em itted as fast neutrons in a fission event to when they appear as therm al neutrons. In this case, there is no excitation of the nucleus, Q = 0; whatever energy is lost by the neutron is gained by th e reco iling target nu cleu s. Let M 1 = M 3 = m (M n ), and M 2 = M 4 = M = A m . Then (15.5) becom es

E  1  1   E  1  1   2

cos  0 (15.7)

3  A  1  A  A

   

Suppose we ask under w h at condition is E 3 = E 1 ? W e see that this can occur only wh en

 = 0, corresponding to forward scattering (no interaction). For all finite  , E 3 has to be less th an E 1 , which is reasonable because som e energy has to be given to the energy of recoil, E 4 . One can show that the m a xim um ene rgy loss by the neutron occurs at   , which corresponds to backward scattering,

E 3 =  E 1 , 

 A  1  2

 A  1 

(15.8)

 

Eq.(15.7) is the starting point for the analysis of neutron m oderation (slowing down) in a scattering medium . W e will return to it later in this lecture.

Inelas tic sca tter ing is the process by which the incom ing neutron excites the target nucleus so it leav es the groun d state and g o es to an ex cited state at an energy E * above the ground state. Thus Q = -E* (E* > 0). We again let the neutron m a ss be m a n d the target nucleus m a ss be M (ground state) or M* (excited state), with M* = M + E*/c 2 . Since th is is a reaction w ith negative Q, it is an endotherm ic process requiring energy to be supplied before the reaction can take place. In the case of scattering th e only way energy can be supplied is through the kinetic energy of the incom ing particle (neutron). Suppose we ask what is m i ni m u m ener gy required for the reaction, the th reshold energy? To find this, we look at the situation where no energy is given to the outgoing particle, E 3

~0 and  ~ 0. Then (15.5) gives

 M 4  M 1 

 E *  E th 



 , o r E th ~ E * ( 1  1/ A ) (15.9)

4 

where we have denoted the m i ni m u m value of E 1 as E th . Thus we see the m i ni m u m kinetic energy required for react ion is always g r eater than th e excitation energy of th e nucleus. Where does the difference between E th and E* go? The answer is that it go es into th e center-of-m ass energy, the fraction of the kinetic energy of the incom ing neutron (in th e labo ratory coo rdinate) that is not available for reactio n.

Relations between Outgoing Energy and Scattering Angle

We return to the Q-equation for elastic scattering to obtain a relation between the energy of the outgoing neutron, E 3 , and the angle of scattering,  . Again regarding

(15.5) as a q u adratic equ ation for th e variab le , we have

1

A  1

 cos    A 2  sin 2   1 / 2  (15.11)

This is a perfectly good relation betw een E 3 and  (with E 1 fixed), although it is not a sim ple one. Nonetheless, it shows a one-to-one correspondence between these two variab les. T h is is what we m eant when we said that the prob lem is reduced to only degree of freedom . Whe never we are given either E 3 or  we can imm ediately determ ine the other variable. The reason we said that ( 15.11 ) is not a simple relatio n is that we can obtain another relation between energy and scattering ang le, except in this case th e scattering an gle is the an gle in the center-of-m ass coordinate system (CMCS), whereas  in (15.11) is the scatterin g angle in th e labor atory coordinate system (LCS). To find th is sim pler relation we first review th e c onnection the two coordinate system s.

Relation between LCS a n d CMCS

Suppose we start with the velocities of the incoming neutron and target nucleus, and those of the outgoing neutron and recoi ling nucleus as shown in the Fig. 15.2.

M

m

v 1

C

X v 0

v 4

M



v 3

(a) LCS

m

V 4

m

V 1 = v 1 - v 0

C

X

V 2 = v 0

M

X  c

| V 4 | = | V 2 | =  0

| V 3 | = | V 1 |

(b) CMCS

V 3

V 4

v 0

v

4

 c



v 0

v 3

(c)

V 3

After Collision

Befor e Collision

Fig. 15.2. Elastic scattering in LCS (a) and CMCS (b), and the geom etric relation

between LCS and CMCS post-coll ision velocity vectors (c). Figure by MIT OCW.

In this diagram we denote the LCS and CMCS velocities by lower and upper cases respectively, so V i = v i – v o , where v o = [1/(A+1)] v 1 is th e velocity of the center-of-mass. Notice th at the scattering angle in C MCS is labeled as  c . W e see that in L C S the center-

of- m ass m o ves in the direction of the incom ing neutron (with th e target nucleus at rest), whereas in CMCS the target nu cleu s m oves toward the center-of-m ass which is statio nary by definitio n. One can show (in a p r oblem set) that in CMC S the post-collision velo cities have the sam e magnitude as the pre-collision velocities, the only effect of the collisio n being a rotation, from V 1 to V 3 , and V 2 to V 4 .

Part (c) of Fig. 15.1 is particularly useful for deriving the relations between LCS and CMCS velocities an d angles. Perhaps the m o st im portan t relation is that between the outgoing speed v 3 and the scattering angle in C MCS ,  c . We can write

1 mv 2 1 m  V  v  2

2 3 2 3 o

= 1 m  V 2  v 2  2 V v cos 

 (15.12)

2 3 o 3 o c

or

E 1 E   1      1    cos 

 (15.13)

3 2 1 c

where   ( A  1) / ( A  1)  2 . Compared to (15.11), (15.13) is clearly sim pler to m a nipulate. These two relations m u st be equivalent since no approxim ations have been m a de in either derivation. Taking the square of (15.11) gives

E 1 E  cos 2   A 2  sin 2   2 cos   A 2  sin 2   1/ 2  (15.14)

3 ( A  1) 2 1

To dem onstrate the equivalence of (15.13) and (15.14) one needs a relatio n between th e two scattering angles,  and  c . This can be obtained from Fig. 15.1(c) by w r iting

cos   v o  V 3 cos  c  / v 3

= 1  A cos  c (15.15)

The relations (15.13), (15.14), and (15.15) all demonstrate a one-to-one correspondence between energy and angle or angle and angle. T h ey can be used to transform distributions from one variab le to ano ther, as we will dem onstrate in the d i scuss ion of energy and angular distribution of elastically scattered neutrons in the following lecture.