Question:(a) A neutron of mass ${\mathbf{m}}_{\mathbf{n}}$and kinetic energy K makes a head-on elastic collision with a stationary atom of mass . Show that the fractional kinetic energy loss of the neutron is given by $\frac{\mathbf{∆}\mathbf{K}}{\mathbf{K}}\mathbf{=}\frac{\mathbf{4}{\mathbf{m}}_{\mathbf{n}}\mathbf{m}}{{\left(m+m_n\right)}^{\mathbf{2}}}$.

Find role="math" localid="1661942719139" $\frac{\mathbf{∆}\mathbf{K}}{\mathbf{K}}$for each of the following acting as the stationary atom:

(b) hydrogen,

(c) deuterium,

(d) carbon, and

(e) lead.

(f) If K=1.00MeV initially, how many such head-on collisions would it take to reduce the neutron’s kinetic energy to a thermal value (0.25 eV) if the stationary atoms it collides with are deuterium, a commonly used moderator? (In actual moderators, most collisions are not head-on.)

The kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

Let the initial velocity of the neutron is $V_{ni}$.

Let the final velocity of the neutron is $V_{nf}$.

The initial velocity of the target nucleus, $V_i=0$

Let the final velocity of the target nucleus is $V_f$.

According to the law of conservation of linear momentum,

role="math" localid="1661943465507" $$\begin{gathered} m{v}_i+m_n{v}_{ni}=m{v}_f+m_n{v}_{nf} \\ {m}_n{v}_{ni}=m{v}_f+m_n{v}_{nf} \\ {v}_{nf}=\frac{m_n{v}_{nf}-m{v}_f}{m_n} \end{gathered}$$

................... (1)

According to the law of conservation of energy,

$\begin{array}{r}\frac{1}{2}m{v}_i^2+\frac{1}{2}{m}_n{v}_{ni}^2=\frac{1}{2}m{v}_f^2+\frac{1}{2}{m}_n{v}_{nf}^2\\ \frac{1}{2}{m}_n{v}_{ni}^2=\frac{1}{2}m{v}_f^2+\frac{1}{2}{m}_n{v}_{nf}^2\\ m_n{v}_{ni}^2=m{v}_f^2+m_n{v}_{nf}^2\end{array}$ ................... (2)

Substitute the equation in the equation, and we get,

$\begin{array}{r}{m}_n{v}_{ni}^2=m{v}_f^2+m_n{\left(\frac{m_n{v}_{ni}-m{v}_f}{m_n}\right)}^2\\ m_n{v}_{ni}^2=m{v}_f^2+m_n\left[\frac{m_n{v}_{ni}^2+m{v}_f^2-2\left(m_n{v}_{ni}\right)m{v}_f}{m_n^2}\right]\\ m_n{v}_{ni}^2=m{m}_n{v}_f^2+m_n^2{v}_{ni}^2-2\left(m_n{v}_{ni}\right)\left(m{v}_f\right)\end{array}$

Solving further as:

$\begin{array}{r}{m}^2{v}_f^2+m{m}_n{v}_f^2-2\left(m_n{v}_{ni}\right)\left(m{v}_f\right)=0\\ m^2{v}_f^2+m{m}_n{v}_f^2=2\left(m_n{v}_{ni}\right)\left(m{v}_f\right)\\ m{v}_f^2+m_n{v}_f^2=2\left(m_n{v}_{ni}{v}_f\right)\\ \left(m+m_n\right)v_f^2=2\left(m_n{v}_{ni}\right)v_f\\ v_f=\frac{2m_n{v}_{ni}}{\left(m+m_n\right)}\end{array}$

The initial kinetic energy of the neutron,

$K_i=\frac{1}{2}{m}_n{v^2}_{ni}$

The final kinetic energy is$K=\frac{1}{2}m{v}_f^2$.

Hence, the fractional kinetic energy loss of the neutron is given by,

$$\begin{gathered} \frac{∆K}{K}=\frac{\left(\frac{1}{2}m{v}_f^2\right)}{\left(\frac{1}{2}{m}_n{v}_{ni}^2\right)} \\ =\frac{\mathrm{m}\left({\displaystyle \frac{2{\mathrm{m}}_{\mathrm{n}}{\mathrm{v}}_{\mathrm{ni}}^2}{\mathrm{m}+{\mathrm{m}}_{\mathrm{n}}}}\right)}{m_n{v}_{ni}^2} \\ =\frac{4{\mathrm{m}}_{\mathrm{n}}\mathrm{m}}{{\left(\mathrm{m}+{\mathrm{m}}_{\mathrm{n}}\right)}^2} \end{gathered}$$

..................... (5)

Hence, the fractional kinetic energy loss of the neutron is $\frac{4{\mathrm{m}}_{\mathrm{n}}\mathrm{m}}{{\left(\mathrm{m}+{\mathrm{m}}_{\mathrm{n}}\right)}^2}$.

The mass of a neutron,$m_n=1.0\mathrm{u}$.

The mass of a hydrogen atom,$m=1.0\mathrm{u}$.

Hence, the fractional kinetic energy loss of the neutron is given by,

$$\begin{gathered} \frac{∆\mathrm{K}}{\mathrm{K}}=\frac{4\left(1.0\mathrm{u}\right)\left(1.0\mathrm{u}\right)}{{\left(1.0\mathrm{u}+1.0\mathrm{u}\right)}^2} \\ =\boxed{1.0} \end{gathered}$$

Hence, the fractional kinetic energy loss of the neutron is 1.0.

The mass of a deuterium atom is 2.0u.

Hence, the fractional kinetic energy loss of the neutron is given by,

$$\begin{gathered} \frac{∆\mathrm{K}}{\mathrm{K}}=\frac{4\left(2.0\mathrm{u}\right)\left(2.0\mathrm{u}\right)}{{\left(2.0\mathrm{u}+2.0\mathrm{u}\right)}^2} \\ =\boxed{0.89} \end{gathered}$$

Hence, the fractional kinetic energy loss of the neutron is 0.89.

The mass of a carbon atom is 12.0u.

Hence, the fractional kinetic energy loss of the neutron is given by,

$$\begin{gathered} \frac{∆\mathrm{K}}{\mathrm{K}}=\frac{4\left(12.0\mathrm{u}\right)\left(1.0\mathrm{u}\right)}{{\left(12.0\mathrm{u}+1.0\mathrm{u}\right)}^2} \\ =\boxed{0.28} \end{gathered}$$

Hence, the fractional kinetic energy loss of the neutron is 0.28.

The mass of the lead atom is 207.0u.

Hence, the fractional kinetic energy loss of the neutron is given by

$$\begin{gathered} \frac{∆\mathrm{K}}{\mathrm{K}}=\frac{4\left(207.0\mathrm{u}\right)\left(1.0\mathrm{u}\right)}{{\left(207.0\mathrm{u}+1.0\mathrm{u}\right)}^2} \\ =\boxed{0.019} \end{gathered}$$

Hence, the fractional kinetic energy loss of the neutron is 0.019.

The initial energy is,

$$\begin{gathered} K=1.00\mathrm{MeV} \\ =\left(1.00\mathrm{MeV}\right)\left(\frac{{10}^6\mathrm{eV}}{1\mathrm{MeV}}\right) \\ =1.00\times {10}^6\mathrm{eV} \end{gathered}$$

The energy after n collisions is,

$K_n=0.025eV$

The energy of the neutron is reduced during the collision by the factor $1-0.89=0.11$.

So the energy after n collision is given by,

$K_n={\left(0.11\right)}^{n}K$

Applying the natural logarithm on both sides and solving for as:

role="math" localid="1661944630272" $\begin{array}{r}n=\frac{\ln \left(\frac{K_n}{K}\right)}{\ln \left(0.11\right)}\\ =\frac{\ln \left(\frac{0.025\mathrm{eV}}{1.00\mathrm{MeV}}\right)}{\ln \left(0.11\right)}\\ =7.93\\ \approx \boxed{8}\end{array}$

Hence, the energy first falls below 0.025 eV on the eighth collision.