Question 8.10: Moderator in a nuclear reactor Here we will look at an examp...
Moderator in a nuclear reactor
Here we will look at an example of subatomic elastic collisions. High-speed neutrons are produced in a nuclear reactor during nuclear fission processes. Before a neutron can trigger additional fissions, it has to be slowed down by collisions with nuclei of a material called the moderator. In some reactors the moderator consists of carbon in the form of graphite. The masses of nuclei and subatomic particles are measured in units called atomic mass units, abbreviated u, where 1 u = 1.66 \times 10^{-27} kg. Suppose a neutron (mass 1.0 u) traveling at 2.6 \times 10^7 m/s makes an elastic head-on collision with a carbon nucleus (mass 12 u) that is initially at rest. What are the velocities after the collision?
SET UP Figure 8.18 shows our sketches. We use the subscripts n and C to denote the neutron and carbon nucleus, respectively.
SOLVE Because we have a head-on elastic collision, all velocity and momentum vectors lie along the x axis, and we can use Equations 8.10 and 8.13, with m_n = 1.0 u, m_C = 12 u, and \upsilon_{n,i} = 2.6 \times 10^7 m/s.
We’ll get two simultaneous equations that we’ll need to solve for \upsilon_{n, f} and \upsilon_{C, f}.
First, conservation of the x component of momentum gives
m_A\upsilon _{A,i}=m_A\upsilon _{A,}+m_B\upsilon _{B,f}. (8.10)
\upsilon _{B,f}-\upsilon _{A,f}=\upsilon _{A,i} (relative velocity before and after collision). (8.13)
m_n(\upsilon _{n,i})+m_C(\upsilon _{C,i})=m_n(\upsilon _{n,f})+m_C(\upsilon _{C,f}),
(1.0 u)(2.6 \times 10^7 m/s)+(12.0 u)(0)
= (1.0 u)\upsilon _{n,f}+(12.0 u)\upsilon _{C,f}.
Second, the relative-velocity relationship yields
\upsilon _{n,i}-\upsilon _{C,i}=-(\upsilon _{n,f}-\upsilon _{C,f}),
2.6 \times 10^7 m/s=\upsilon _{C,f}-\upsilon _{n,f}.
These are two simultaneous equations for \upsilon _{n,f} and \upsilon _{C,f} . We’ll let you do the algebra; the results are
\upsilon _{n,f}=-2.2 \times 10^7 m/s
\upsilon _{C,f}=0.40 \times 10^7 m/s (final x velocities).
REFLECT The neutron ends up with \frac{11}{13} of its original speed, and the speed of the recoiling carbon nucleus is \frac{2}{13} of the neutron’s original speed. Kinetic energy is proportional to speed squared, so the neutron’s final kinetic energy is \left(\frac{11}{13} \right)^2 , or about 0.72 of its original value. If the neutron makes a second such collision, its kinetic energy is (0.72)², or about half of its original value, and so on. After many such collisions, the neutron’s kinetic energy is reduced to a small fraction ( \frac{1}{100} or less) of its initial value.
Practice Problem: If the neutron’s kinetic energy is reduced to \frac{9}{16} of its initial value in a single collision, what is the mass of the moderator nucleus? Answer: 7.0 u.
