Question 8.11: MODERATING FISSION NEUTRONS IN NUCLEAR REACTOR The fission o...

IDENTIFY and SET UP:

We ignore external forces, so momentum is conserved in the collision. The collision is elastic, so kinetic energy is also conserved. Figure 8.26 shows our sketch. We take the x-axis to be in the direction in which the neutron is moving initially. The collision is head-on, so both particles move along this same axis after the collision. The carbon nucleus is initially at rest, so we can use Eqs. (8.24) (v_{A 2 x}=\frac{m_{A}-m_{B}}{m_{A}+m_{B}} v_{A 1 x}) and (8.25) (v_{B 2 x}=\frac{2 m_{A}}{m_{A}+m_{B}} v_{A 1 x}); we replace A by n (for the neutron) and B by C (for the carbon nucleus). We have m_{\mathrm{n}}=1.0 \mathrm{u}, m_{\mathrm{C}}=12 \mathrm{u}, \text { and } v_{\mathrm{n} 1 x}=2.6 \times 10^{7} \mathrm{~m} / \mathrm{s}. The target variables are the final velocities v_{\mathrm{n} 2 x} \text { and } v_{\mathrm{C} 2 x}.

EXECUTE:

You can do the arithmetic. (Hint: There’s no reason to convert atomic mass units to kilograms.) The results are

EVALUATE: : The neutron ends up with \left|\left(m_{\mathrm{n}}-m_{\mathrm{C}}\right) /\left(m_{\mathrm{n}}+m_{\mathrm{C}}\right)\right|=\frac{11}{13} of its initial speed, and the speed of the recoiling carbon nucleus is \left|2 m_{\mathrm{n}} /\left(m_{\mathrm{n}}+m_{\mathrm{C}}\right)\right|=\frac{2}{13} of the neutron’s initial speed. Kinetic energy is proportional to speed squared, so the neutron’s final kinetic energy is \left(\frac{11}{13}\right)^{2} \approx 0.72 of its original value. After a second head-on collision, its kinetic energy is (0.72)^{2}, or about half its original value, and so on. After a few dozen collisions (few of which are head-on), the neutron speed will be low enough that it can efficiently cause a fission reaction in a uranium nucleus.