91Ó°ÊÓ

Given that Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mass 2.0 u

Mass of neutron${\mathrm{m}}_{\mathrm{A}}=1\mathrm{unit}$${\mathrm{m}}_{\mathrm{A}}=1\mathrm{unit}$

Mass of deuteron unit${\mathrm{m}}_{\mathrm{B}}=2\mathrm{unit}$

The heart of a nuclear power plant is its nuclear reactor. They contain and manage nuclear chain reactions, or fission, which is a physical process that generates heat. The steam produced by the heat spins a turbine to produce energy.

Let u be the initial speed and v be the final speed of the neutron.

So $$\begin{gathered} \mathrm{v}=\left(\frac{{\mathrm{m}}_{\mathrm{A}}-{\mathrm{m}}_{\mathrm{B}}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}\right)\mathrm{u} \\ \mathrm{v}=\left(\frac{1-2}{1+3}\right)\mathrm{u} \\ \mathrm{v}=-\frac{\mathrm{u}}{3} \end{gathered}$$

Hence, the speed of neutron is one-third of its initial speed after the collision.

Let ${\mathrm{K}}_2$be kinetic energy after the collision and${\mathrm{K}}_1$ is kinetic energy before the collision.

$$\begin{gathered} {\mathrm{K}}_2=\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}^2 \\ {\mathrm{K}}_1=\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}{\mathrm{u}}^2 \end{gathered}$$

So

$$\begin{gathered} {\mathrm{K}}_2=\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}^2 \\ =\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}{\left(\frac{\mathrm{u}}{3}\right)}^2 \\ =\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}\frac{{\mathrm{u}}^2}{3} \\ =\frac{1}{9}\left(\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}{\mathrm{u}}^2\right) \\ =\frac{1}{9}{\mathrm{K}}_1 \end{gathered}$$

Kinetic energy is one-ninth of original kinetic energy.

Let n be the number of collisions that will reduce the speed of neutron to 1/59000 of its original value.

Now, $\mathrm{v}={\left(\frac{1}{3}\right)}^{\mathrm{n}}\mathrm{u}$

So

$$\begin{gathered} {\left(\frac{1}{3}\right)}^{\mathrm{n}}=\frac{1}{59000} \\ {3}^{\mathrm{n}}=59000 \\ \mathrm{nlog}3=\log 59000 \\ \mathrm{n}=10 \end{gathered}$$

After 10 collisions the speed of neutron will reduce to 1/59000 of its original value.