Question 9.2.Q2: Elastic scattering of neutrons interacting with atomic nucle......
Elastic scattering of neutrons interacting with atomic nuclei of absorber is the most probable interaction of neutrons with absorbing medium for neutrons with kinetic energy E_K ≤ 2 MeV. The other possible interactions are inelastic scattering, neutron capture, spallation, and fission. In the energy range below 10 MeV the elastic scattering of neutrons can be treated classically, as shown schematically in Fig. 9.2 for two-particle elastic collision between projectile m_1 (neutron) moving with velocity υ_1 and a stationary target m_2 (nucleus of absorber), with θ the scattering angle of the projectile, 𝜙 the recoil angle of the target, and b the so-called impact parameter. After collision projectile m_1 continues with velocity u_1 and the target recoils with velocity u_2.
(a) Use the classical principles of energy and momentum conservation to derive an expression for energy transfer ΔE_K from the projectile m_1 to the stationary target m_2 in laboratory coordinate system.
(b) Using the general expression for ΔE_K derived in (a), express \Delta E_K for the specific example of neutron elastic scattering on a nucleus of the absorbing medium. Also state the energy transfer fraction f_{tr}, maximum energy transfer fraction \left(f_{\mathrm{tr}}\right)_{\max }, and mean energy transfer fraction \bar{f}_{tr} for elastic scattering of neutrons on target nuclei of atomic mass A.
(c) Using the expression derived in (b) for kinetic energy transfer ΔE_K from neutron to target nucleus in elastic scattering, determine mean energy \overline{\Delta E_{\mathrm{K}}} transferred from neutron to target nucleus in elastic scattering.
(d) Determine an expression for mean energy \overline{E_K^{\mathrm{n}}} that a neutron carries from an elastic scattering event with target nucleus of atomic weight A. Calculate post-elastic scattering neutron kinetic energy as a fraction of incident neutron energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 for the following target nuclei: hydrogen (A = 1), cadmium (A = 112), and uranium (A = 238).
(a) The kinetic energy transfer ΔE_K from projectile m_1 to the target m_2 is determined classically in laboratory coordinate system using the conservation of energy and momentum laws as follows (see Fig. 9.2):
(1) Conservation of kinetic energy:
\left(E_{\mathrm{K}}\right)_0+0=\frac{1}{2} m_1 v_1+0=\frac{1}{2} m_1 u_1+\frac{1}{2} m_2 u_2 (9.16)
where \left(E_K\right)_0 is the kinetic energy of the projectile (incident particle) m_1.
(2) Conservation of momentum along abscissa (x) axis (note: the incident particle is moving along the abscissa axis in the positive direction):
m_1 v_1=m_1 u_1 \cos \theta+m_2 u_2 \cos \phi (9.17)
(3) Conservation of momentum along ordinate (y) axis:
0=m_1 u_1 \sin \theta-m_2 u_2 \sin \phi (9.18)
Equations (9.17) and (9.18) can, respectively, be written as follows
\left(m_1 v_1-m_2 u_2 \cos \phi\right)^2=\left(m_1 u_1 \cos \theta\right)^2 (9.19)
and
\left(m_1 u_1 \sin \theta\right)^2 \equiv m_1^2 u_1^2-m_1^2 u_1^2 \cos ^2 \theta=\left(m_2 u_2 \sin \phi\right)^2 (9.20)
Inserting (9.19) into (9.20) gives the following expression
m_2^2 u_2^2=m_1^2 u_1^2-m_1^2 v_1^2+2 m_1 v_1 m_2 u_2 \cos \phi, (9.21)
which, after inserting (9.16) multiplied by 2m_1, reads
m_2^2 u_2^2=2 m_1 v_1 m_2 u_2 \cos \phi-m_1 m_2 u_2^2 (9.22)
or
2 m_1 v_1 \cos \phi=\left(m_1+m_2\right) u_1 (9.23)
Since \Delta E_{\mathrm{K}}=\frac{1}{2} m_2 u_2^2 [i.e., energy transfer \Delta E_K from the projectile (incident particle) m_1 with kinetic energy \left(E_K\right)_0 to the target of mass m_2 is equal to recoil energy of the target \frac{1}{2} m_2 u_2^2], we get the following general expression for energy transfer \Delta E_K.
\Delta E_{\mathrm{K}}=\frac{4 m_1 m_2}{\left(m_1+m_2\right)^2}\left(E_{\mathrm{K}}\right)_0 \cos ^2 \phi (9.24)
showing that the energy \Delta E_K transferred from the projectile to the recoiling target is governed by the recoil angle 𝜙. With regard to recoil angle 𝜙 there are two special angles to consider:
(1) In an elastic, head-on collision, the recoil angle 𝜙 of the target is zero and \Delta E_{\mathrm{K}} \text { attains its maximum value }\left(\Delta E_{\mathrm{K}}\right)_{\max } for given projectile mass m_1, target mass m_2, and kinetic energy \left(E_K\right)_0 of the projectile expressed as a consequence of \cos 0^{\circ}=1 as
\left(\Delta E_{\mathrm{K}}\right)_{\max }=\frac{4 m_1 m_2}{\left(m_1+m_2\right)^2}\left(E_{\mathrm{K}}\right)_0 \cos ^2 0^{\circ}=\frac{4 m_1 m_2}{\left(m_1+m_2\right)^2}\left(E_{\mathrm{K}}\right)_0 . (9.25)
(2) In grazing angle interaction between projectile and target, the recoil angle 𝜙 of the target is 90°, resulting in no energy transfer \left(\Delta E_K = 0\right), as a consequence of cos 90° = 0.
(3) All other recoil angles are between 0° and 90° and the energy transfer from incident particle (neutron) to recoiling target (nucleus) is between \left(\Delta E_K\right)_{\max } and 0.
(b) For the specific case of neutron scattering on an absorber nucleus we simplify the general classical expression (9.24) making the following changes:
(1) Projectile m_1 is the incident neutron with mass m_n and incident kinetic energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 .
(2) Target is the nucleus of the attenuating medium with mass consisting of A nucleons (atomic mass number), i.e., m_2 ≈ Am_n where we assume that proton mass m_p and neutron mass m_n are approximately equal (m_p = 938.3 MeV/c² ≈ m_n = 939.6 MeV/c²).
With A the atomic weight of the target nucleus we now get the following simple expression for (9.24)
\Delta E_{\mathrm{K}}=\frac{4 m_{\mathrm{n}}\left[Z m_{\mathrm{p}}+(A-Z) m_{\mathrm{n}}\right]}{\left\{m_{\mathrm{n}}+\left[Z m_{\mathrm{p}}+(A-Z) m_{\mathrm{n}}\right]\right\}^2}\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 \cos ^2 \phi \approx \frac{4 A}{(1+A)^2}\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 \cos ^2 \phi, (9.26)
or we can state that the energy transfer fraction f_{\mathrm{tr}}=\Delta E_{\mathrm{K}} /\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 that is, the fraction of kinetic energy of the incident neutron transferred to the stationary target, is
f_{\mathrm{tr}}=\frac{\Delta E_{\mathrm{K}}}{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0}=\frac{4 A}{(1+A)^2} \cos ^2 \phi (9.27)
Like in the general case described by (9.24), we again have two special angles: 𝜙 = 0 and 𝜙 = 90°. For direct, head-on collision where the nucleus recoils with 𝜙 = 0 the energy transfer \Delta E_K and energy transfer fraction ftr attain its maximum possible value, respectively, since \cos 0^{\circ}=1 in (9.26) and (9.27)
\left(\Delta E_{\mathrm{K}}\right)_{\max }=\left.\Delta E_{\mathrm{K}}\right|_{\phi=0}=\frac{4 A}{(1+A)^2}\left(E_{\mathrm{K}}\right)_0 (9.28)
and
\left(f_{\mathrm{tr}}\right)_{\max }=\left.f_{\mathrm{tr}}\right|_{\phi=0}=\frac{\left(\Delta E_{\mathrm{K}}\right)_{\max }}{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0}=\frac{4 A}{(1+A)^2} . (9.29)
Note that for a direct hit (𝜙 = 0), the maximum energy transfer fraction \left(f_{\mathrm{tr}}\right)_{\max } that is transferred from the incident neutron to recoil nucleus has the following properties:
(1) Depends only on the atomic mass number A (atomic weight) of the target
nucleus.
(2) Is equal to 1 when the target nucleus is the hydrogen nucleus (proton) with
(A = 1).
(3) Decreases from \left(f_{tr}\right)_{max} = 1 as the target mass increases from A = 1.
For grazing incidence where the nucleus recoils with 𝜙 = 90° the energy transfer fraction f_{tr} attains its minimum value of f_{tr} = 0, as a result of cos 0° = 0 in (9.27)
\left(f_{\mathrm{tr}}\right)_{\min }=\frac{\left.\Delta E_{\mathrm{K}}\right|_{\phi=0}}{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0}=0 \quad \text { or }\left.\quad \Delta E_{\mathrm{K}}\right|_{\phi=90^{\circ}}=0 (9.30)
(c) Mean kinetic energy transfer \overline{\Delta E_K} from neutron to target nucleus is calculated by averaging (9.26) over the recoil angle 𝜙 of the target nucleus as follows
\overline{\Delta E_{\mathrm{K}}}=\frac{4 A}{(1+A)^2}\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 \overline{\cos ^2 \phi}, (9.31)
with \overline{\cos ^2 \phi} determined as
\overline{\cos ^2 \phi}=\frac{\int_0^{\pi / 2} \cos ^2 \phi \mathrm{d} \phi}{\int_0^{\pi / 2} \mathrm{~d} \phi}=\frac{1}{\pi} \int_0^{\pi / 2} \frac{1+\cos (2 \phi)}{2} \mathrm{~d}(2 \phi)=\frac{1}{\pi}\left[\phi+\frac{\sin (2 \phi)}{2}\right]_0^{\frac{\pi}{2}}=\frac{1}{2} . (9.32)
Mean energy transfer \overline{\Delta E_{\mathrm{K}}} from neutron to target nucleus is thus given as follows
\overline{\Delta E_{\mathrm{K}}}=\frac{4 A}{(1+A)^2}\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 \overline{\cos ^2 \phi}=\frac{2 A}{(1+A)^2}\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 (9.33)
and similarly for the mean energy transfer fraction \bar{f}_{tr}.
\bar{f}_{\mathrm{tr}}=\frac{\overline{\Delta E_{\mathrm{K}}}}{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0}=\frac{2 A}{(1+A)^2} (9.34)
Equations (9.29) and (9.32) show that to moderate neutrons (that is to slow them down) with the fewest number of elastic collisions, target nuclei with as low as possible atomic weight A should be used as absorber medium. Thus, the most efficient nucleus for this purpose is hydrogen with A = 1 for which \left(\Delta E_{\mathrm{K}}\right)_{\max }=\left(E_{\mathrm{K}}\right)_0, \overline{\Delta E_{\mathrm{K}}}=\frac{1}{2}\left(E_{\mathrm{K}}\right)_0, \text { and } \bar{f}_{\mathrm{tr}}=\frac{1}{2} as seen by inserting A = 1 into (9.28), (9.29), and (9.32), respectively.
In Fig. 9.3 we plot, against atomic mass number A, the maximum energy transfer fraction \left(f_{\mathrm{tr}}\right)_{\max } given in (9.29) and the mean energy transfer fraction \bar{f}_{\mathrm{tr}} given in (9.32) and show that the two energy transfer fractions decline rapidly roughly as 1/A with increasing A from their high values of 1 and 0.5, respectively, for hydrogen and amount to only 0.017 and 0.0085, respectively, for uranium-238, as also indicated in Fig. 9.3. Several other nuclides are also identified in the graph.
(d) Mean kinetic energy \overline{E_{\mathrm{K}}^{\mathrm{n}}} of post-elastic collision neutron is calculated as mean energy transfer \overline{\Delta E_{\mathrm{K}}} from neutron to target nucleus subtracted from incident neutron energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0.
\overline{E_{\mathrm{K}}^{\mathrm{n}}}=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0-\overline{\Delta E_{\mathrm{K}}}=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0\left[1-\frac{2 A}{(1+A)^2}\right]=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 \frac{1+A^2}{(1+A)^2} (9.35)
Fractions of incident energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 that a neutron retains after elastic collision with a given target nucleus are calculated from (9.35) as follows: \overline{E_{\mathrm{K}}^{\mathrm{n}}} /\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0= \left(1+A^2\right) /(1+A)^2 \overline{E_{\mathrm{K}}^{\mathrm{n}}} /\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0=0.5 for H (A = 1); 0.858 for C (A = 12); 0.982 for Cd (A = 112); and 0.992 for U (A = 238).
