Slowing Down the Neutrons Slowing Down the Neutrons

Slowing down the neutrons

Clearly, an obvious way to make a reactor work, and to make use of this characteristic of the 235U(n,f) cross-section, is to slow down the fast, ﬁssion neutrons. This can be accomplished, for example, when the neutrons collide with nuclei and scatter in some substance (a moderator).

It moderates the neutrons (slows them down).

If this could be achieved, neutron-induced ﬁssion would be very much more likely.

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Slowing down the neutrons

Say that the neutrons could be slowed down until they are in thermal equilibrium with the surrounding material in the reactor.

They would then have the same distribution of energies as the molecules in the material.

This is the same distribution of energies as for the atoms or molecules of a gas, say, in thermal equilibrium at the same temperature.

What would such a distribution look like?

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1 Thermal neutron distribution

Originally Boltzmann analysed the statistics of a number of particles at different energies in thermal equilibrium (actually for the molecules in a gas).

This produced the Maxwell-Boltzmann distribution. 1 2 E is the neutron # 4 & 1 # E & n(E) = n% ( E 2 exp% ) ( energy in Joules $ "kBT ' $ kBT '

T is the temperature of the moderator n(E)dE is the numbe!r o f and hence of the neutron distribution in neutrons with energies equilibrium with the moderator. between E and E + dE. Thus one talks of “300 K neutrons” for example. The total number of neutrons, n. kB is Boltzmann’s constant. -23 -1 kB = 1.38 x 10 J.K

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Thermal equilibrium

Once again, to express this as a ﬂux distribution we multiply by v.

Maxwell-Boltzmann . ﬂux distribution at The average energy at an 300K. absolute temperature T is 3/2kT

The most probable energy at an absolute temperature T is 1/2kT. (0.025eV for T = 300 K)

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2 Slowing neutrons down So, we have decided that we have to slow the neutrons down from the ﬁssion spectrum energy distribution to something like the thermal distribution. In order to see how this can be done let’s consider what happens in an elastic collision (elastic scattering).

Let’s work out the energy of a particle of mass m after it collides with a stationary particle of mass, M.

We will examine this process in some detail remembering that the momentum and the kinetic energy are both conserved in such an elastic collision.

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Elastic scattering

recoiling energy, Er nucleus mass, M neutron nucleus φ

mass, M θ mass, m energy, Ef energy, E0 scattered mass, m neutron

Before collision After collision

The energy of the scattered neutron depends on the scattering angle, θ, and the mass of the scattering nucleus.

We apply the equations for conservation of momentum and energy.

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3 This i s the p Elastic scattering hysics recoiling nucleus nucleus φ neutron θ mass, m mass, M scattered neutron Before collision After collision Conservation of momentum in the x direction mvm = mv "m cos# + Mv "M cos$ Conservation of momentum in 0 = mv "m sin# + Mv "M sin$ the y direction 1 1 1 ! mv 2 = mv " 2 + Mv " 2 or Conservation of energy 2 m 2 m 2 M

! E0 = E f + Er

Remember we want to ﬁnd the ﬁnal energy of the neutron, Ef in terms of the masses, the scattering angle and of course, the initial energy, E . ! θ , 0 ! 57 Lecture 32 © J. Watterson, 2007

Now , doin g the Elastic scattering maths Notice that we can get rid of the angle φ by squaring and adding the momentum equations, because cos 2 " + sin 2 " = 1 . 1 First, it probably makes it easier if we E = mv2 for non-relativistic speeds express everything in terms of energies using: 2 and energies. p = mv

! 2 2 2 1 2 so p = m v = 2m mv = 2mE 2 and! p = mv = 2mE ! Conservation of momentum in the 2mE0 = 2mE f cos" + 2MEr cos# x direction Using this we get for the two ! momentum equations Conservation of momentum in the 0 = 2!mE f sin" + 2MEr sin# y direction

We want to get r!id of the φ 2MEr cos" = 2mE0 # 2mE f cos$ terms, so lets put them on the left hand side. ! 2MEr sin" = # 2mE f sin$ 2 Squaring and adding: 2ME cos2 " + sin2 " = 2mE # 2mE cos$ + 2mE sin2 $ ! r ( ) ( 0 f ) f

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4 Doin g the m Elastic scattering aths

2 2 2 2 Our last equation was: 2MEr (cos " + sin ") = ( 2mE0 # 2mE f cos$) + 2mE f sin $ Now since: cos2 " + sin2 " =1 for any angle.

We have: 2ME = 2mE " 2 2mE 2mE cos# + 2mE cos2 # + 2mE sin2 # ! r ( 0 0 f f ) f

! = 2mE0 " 2 $ 2m E0 E f cos# + 2mE f using cos2 " + sin2 " =1 as well

Remember, we want Ef in terms of E0 and θ (as well as M and m) so we have to get rid of Er. ! We do this by going back to the original equations and E E E E E E using the one! fo r conservation of energy: 0 = f + r so r = 0 " f

Then substituting for Er and M E " E = mE " 2m E E cos# + mE cancelling out the 2, we have: ( 0 f ) 0 0 f f ! ! Now we have an equation that only contains Ef, E0 and cosθ (as well as M and m) but it is not in a convenient form – we would like to have E by itself on one side. ! f 59 Lecture 32 © J. Watterson, 2007

Doin g the m Elastic scattering aths 2 In order to do this, we We can ﬁnd x in "b ± b " 4ac ax 2 + bx + c = 0 x = remember that if we have a terms of a, b and c: 2a quadratic equation:

into this form with So let’s get M !E " E = mE " 2mE " 2m E E cos# + mE E playing the part ( 0 f ) 0 0 0 !f f f of x.

Move the terms to the left "(M + m)E f + 2m E0 cos# E f + (M " m)E0 = 0 a!nd group the factors ! or (M + m)E f " 2m E0 cos# E f " (M " m)E0 = 0

We can now apply the 2m E cos" ± 4m2 E cos2 " + 4 M 2 # m2 E ! 0 0 ( ) 0 above formula for a E f = quadratic to write: 2(M + m) ! 2 2 2 2 2m E0 cos" ± 2 m E0 cos " + (M # m )E0 = 2(M + m)

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5 Doin g the Elastic scattering maths

Continuing to simplify 2 2 2 2 m E0 cos" ± m E0 cos " +(M # m )E0 this equation: E = f (M + m) Take out of the 2 2 2 2 E0 m E0 cos" ± E0 m cos " +(M # m ) = square root. step-by-step. (M + m)

1 E0 $ 2 2 2 2 2 ' = m cos" ± m cos " + M # m ! M + m %& ( ) () 1 E $ 2 2 2 2 2 ' = 0 m cos" ± m {1# sin "}+ M # m M + m %& ( ) ()

1 E0 $ 2 2 2 2 2 2 ' = m cos" ± m # m sin " + M # m M + m %& ( ) ()

1 E0 $ 2 2 2 2 ' = m cos" ± M # m sin " M + m %& ( ) () 1 $ 2 2 ' m E 0 * M - 3 = 0 &cos " ±2 + . # sin 2 " 5 ) M + m & , m / ) % 1 4 (

Final ly, the Elastic scattering result 1 0 2 2 3 m E * - Thus we obtain: 0 2 # M & 2 5 E f = cos" ±, $ ' ) sin "/ M + m2 % m ( 5 1 + . 4

! 1 2 0 2 2 3 m 2 E * # M & - 0 2 2 5 or E f = 2 cos" + , $ ' ) sin " / M + m 2 % m ( 5 es and ( ) 1 + . 4 th sid ing bo s sign quar minu l S g the ingfu oppin mean dr s not se it i becau For a !ne utron scattering off m =1 Because the masses always a nucleus with mass number occur in ratios. A we can take: M = A

6 Elastic scattering

Initial energy of the neutron.

1 2 E0 2 2 2 E f = 2 cos" + (A # sin ") (A +1) [ ]

Final energy of the neutron, after sca!tt ering. Mass number of the Scattering angle – scattering nucleus angle through which the neutron is scattered

Elastic scattering

1 2 for a 1 MeV neutron and E0 $ 2 2 2 ' If we plot: E = cos" + A # sin " f 2 %& ( ) () different scattering (A +1) substances we obtain:

As you can see all the neutrons that would be scattered through 90º or more are basically stopped (and thermalised) in one collision with a hydrogen nucleus (proton). In the case of carbon the maximum energy loss is only about 300 keV even for a scattering angle of 180º (π) so several collisions would be necessary to thermalise the neutrons. 64 Lecture 32 © J. Watterson, 2007

7 Elastic scattering

Exercise 36

A neutron with an energy of 1 MeV is scattered by the moderator in a light water reactor. In other words in a reactor where the moderator is

ordinary H2O.

Suppose that it is scattered through an angle of 45º. Calculate: 1. Its energy after scattering if it were scattered by the oxygen. 2. Its energy after scattering if it were scattered by the hydrogen.

In the second case, what is the energy of the proton after the scattering?