The Nuclear Reactor — Question 3

Section 43-3 · Physics pages 7196–7197

Question 3 An elastic collision shows why light nuclei can moderate neutrons A neutron of mass m and speed v collides head-on elastically with a stationary nucleus of mass Am. Use one-dimensional momentum and kinetic-energy conservation. m, v Am V Schematic; labels, not drawing lengths, specify the data. (a) Derive the fraction of neutron kinetic energy remaining after the collision. (b) Evaluate this limiting fraction for A=1 and A=12 and explain why it is not an average over all scattering angles.
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Question 3 - Solution An elastic collision shows why light nuclei can moderate neutrons m, v Am V (a) Method and calculation mv = mv′ + AmV, v2 = v′2 + AV 2 ⇒v′ = 1 −A 1 + Av. E′/E = (A −1 A + 1 )2 . (b) Interpretation and further calculation A = 1 : E′/E = 0; A = 12 : E′/E = 121/169 = .71598 . ∆E/E = 1 −E′/E is largest for the head-on transfer considered. Physical check For A greater than one the neutron reverses direction in this head-on case. General scattering leaves a distribution of outgoing energies.

Question 3

An elastic collision shows why light nuclei can moderate neutrons

A neutron of mass m and speed v collides head-on elastically with a stationary nucleus of mass Am. Use one-dimensional momentum and kinetic-energy conservation.

Diagram shown in the original SVG worksheet.

  1. Derive the fraction of neutron kinetic energy remaining after the collision.

  2. Evaluate this limiting fraction for A=1 and A=12 and explain why it is not an average over all scattering angles.

Question 3 - Solution

An elastic collision shows why light nuclei can moderate neutrons

Diagram shown in the original SVG worksheet.

(a) Method and calculation

mv=mv′+AmV,v2=v′2+AV2⇒v′=1−A1+Av.mv=mv\prime+AmV,\quad v^2=v\prime{}^2+AV^2\Rightarrow v\prime=\frac{1-A}{1+A}v.E′/E=(A−1A+1)2.\boxed{E\prime/E=\left(\frac{A-1}{A+1}\right)^2}.

(b) Interpretation and further calculation

A=1:E′/E=0;A=12:E′/E=121/169=.71598.\boxed{A=1:\ E\prime/E=0;\qquad A=12:\ E\prime/E=121/169=.71598}.ΔE/E=1−E′/Eis largest for the head-on transfer considered.\boxed{\Delta E/E=1-E\prime/E}\quad\text{is largest for the head-on transfer considered}.

Physical check

For A greater than one the neutron reverses direction in this head-on case. General scattering leaves a distribution of outgoing energies.