Chapter 9, Problem 9.27P

A neutron in a nuclear reactor makes an elastic, head- on collision with the nucleus of a carbon atom initially at rest. (a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (b) The initial kinetic energy of the neutron is 1.60 X 10^-13J. Find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. (The mass of the carbon nucleus is nearly 12.0 times the mass of the neutron.)

To determine

The fraction of the kinetic energy of the neutron which is transferred to the carbon nucleus.

Answer to Problem 9.27P

The fraction of the kinetic energy of the neutron which is transferred to the carbon nucleus is $0.284$.

Explanation of Solution

Given info: The initial kinetic energy of the neutron is $1.60\times {10}^{-13}\text{J}$ and the mass of the carbon nucleus is $12.0$ times the mass of the neutron.

Write the equation of initial kinetic energy of the neutron.

$K_{\text{ni}}=\frac{1}{2}{m}_{\text{n}}{v}_{\text{ni}}^2$ (1)

Here,

$K_{\text{ni}}$ is the initial kinetic energy of the neutron.

$m_{\text{n}}$ is the mass of the neutron.

$v_{\text{ni}}$ is the initial speed of the neutron.

Write the equation of final kinetic energy of the carbon nucleus.

$K_{\text{cf}}=\frac{1}{2}{m}_{\text{c}}{v}_{\text{cf}}^2$ (2)

Here,

$K_{\text{cf}}$ is the final kinetic energy of the carbon nucleus.

$m_{\text{c}}$ is the mass of the carbon nucleus.

$v_{\text{cf}}$ is the final speed of the carbon nucleus.

Divide equation (2) by equation (1).

$\begin{array}{c}\frac{K_{\text{cf}}}{K_{\text{ni}}}=\frac{\left(\frac{1}{2}{m}_{\text{c}}{v}_{\text{cf}}^2\right)}{\left(\frac{1}{2}{m}_{\text{n}}{v}_{\text{ni}}^2\right)}\\ \frac{K_{\text{cf}}}{K_{\text{ni}}}=\frac{m_{\text{c}}{v}_{\text{cf}}^2}{m_{\text{n}}{v}_{\text{ni}}^2}\end{array}$ (3)

Write the equation of conservation of momentum.

$p_{\text{i}}=p_{\text{f}}$ (4)

Here,

$p_{\text{i}}$ is the initial momentum.

$p_{\text{f}}$ is the final momentum.

Write the equation of initial momentum.

$p_{\text{i}}=m_{\text{n}}{v}_{\text{ni}}+m_{\text{c}}{v}_{\text{ci}}$

Here,

$m_{\text{n}}$ is the mass of the neutron.

$v_{\text{ni}}$ is the initial velocity of the neutron.

$m_{\text{c}}$ is the mass of the carbon nucleus.

$v_{\text{ci}}$ is the initial velocity of the carbon nucleus.

Write the equation of final momentum.

$p_{\text{f}}=m_{\text{n}}{v}_{\text{nf}}+m_{\text{c}}{v}_{\text{cf}}$

Here,

$v_{\text{nf}}$ is the final velocity of the neutron.

$v_{\text{cf}}$ is the final velocity of the carbon nucleus.

Substitute $\left(m_{\text{n}}{v}_{\text{ni}}+m_{\text{c}}{v}_{\text{ci}}\right)$ for $p_{\text{i}}$ and $\left(m_{\text{n}}{v}_{\text{nf}}+m_{\text{c}}{v}_{\text{cf}}\right)$ for $p_{\text{f}}$ in equation (4) to find $v_{\text{cf}}$.

$\begin{array}{c}\left(m_{\text{n}}{v}_{\text{ni}}+m_{\text{c}}{v}_{\text{ci}}\right)=\left(m_{\text{n}}{v}_{\text{nf}}+m_{\text{c}}{v}_{\text{cf}}\right)\\ v_{\text{cf}}=\frac{\left(m_{\text{n}}{v}_{\text{ni}}+m_{\text{c}}{v}_{\text{ci}}\right)-m_{\text{n}}{v}_{\text{nf}}}{m_{\text{c}}}\end{array}$

Initially the carbon nucleus is at rest due to which the initial velocity of the carbon nucleus $v_{\text{ci}}$ is equal to $0$. The mass of the carbon nucleus $m_{\text{c}}$ is $12.0$ times the mass of the neutron $m_{\text{n}}$.

Substitute $12m_{\text{n}}$ for $m_{\text{c}}$ and $0$ for $v_{\text{ci}}$ in the above equation to find $v_{\text{cf}}$.

$\begin{array}{c}{v}_{\text{cf}}=\frac{\left\{m_{\text{n}}{v}_{\text{ni}}+\left(12m_{\text{n}}\right)\left(0\right)\right\}-m_{\text{n}}{v}_{\text{nf}}}{\left(12m_{\text{n}}\right)}\\ =\frac{m_{\text{n}}\left(v_{\text{ni}}-v_{\text{nf}}\right)}{12m_{\text{n}}}\\ v_{\text{cf}}=\frac{v_{\text{ni}}-v_{\text{nf}}}{12}\end{array}$ (5)

When an elastic head on collision occurs between two objects then velocity after the collision is same as the velocity before the collision but in negative directions.

Write the equation for velocity for head on elastic collision.

$v_{\text{ni}}-v_{\text{ci}}=-\left(v_{\text{nf}}-v_{\text{cf}}\right)$

Substitute $0$ for $v_{\text{ci}}$ in the above equation.

$\begin{array}{c}{v}_{\text{ni}}-\left(0\right)=-\left(v_{\text{nf}}-v_{\text{cf}}\right)\\ v_{\text{ni}}=-\left(v_{\text{nf}}-v_{\text{cf}}\right)\\ v_{\text{nf}}=v_{\text{cf}}-v_{\text{ni}}\end{array}$

Substitute $\left(v_{\text{cf}}-v_{\text{ni}}\right)$ for $v_{\text{nf}}$ in equation (5) to find $v_{\text{cf}}$.

$\begin{array}{c}{v}_{\text{cf}}=\frac{v_{\text{ni}}-\left(v_{\text{cf}}-v_{\text{ni}}\right)}{12}\\ 12v_{\text{cf}}=2v_{\text{ni}}-v_{\text{cf}}\\ 13v_{\text{cf}}=2v_{\text{ni}}\\ v_{\text{cf}}=\frac{2}{13}{v}_{\text{ni}}\end{array}$

Substitute $12m_{\text{n}}$ for $m_{\text{c}}$ and $\frac{2}{13}{v}_{\text{ni}}$ for $v_{\text{cf}}$ in equation (4).

$\begin{array}{c}\frac{K_{\text{cf}}}{K_{\text{ni}}}=\frac{\left(12m_{\text{n}}\right){\left(\frac{2}{13}{v}_{\text{ni}}\right)}^2}{m_{\text{n}}{v}_{\text{ni}}^2}\\ \frac{K_{\text{cf}}}{K_{\text{ni}}}=\frac{48}{169}\\ \frac{K_{\text{cf}}}{K_{\text{ni}}}=0.284\end{array}$

Conclusion:

Therefore, the fraction of the kinetic energy of the neutron which is transferred to the carbon nucleus is $0.284$.

To determine

The final kinetic energy of the neutron and the final kinetic energy of the carbon nucleus.

Answer to Problem 9.27P

The final kinetic energy of the neutron is $1.15\times {10}^{-13}\text{J}$ and the final kinetic energy of the carbon nucleus is $4.54\times {10}^{-14}\text{J}$.

Explanation of Solution

Given info: The initial kinetic energy of the neutron is $1.60\times {10}^{-13}\text{J}$ and the mass of the carbon nucleus is $12.0$ times the mass of the neutron.

The fraction of the kinetic energy of the neutron which is transferred to the carbon nucleus is $0.284$.

The fraction of the kinetic energy which stays with the neutron after it transfers its energy to the carbon nucleus is,

$x_{\text{f}}=1-\frac{K_{\text{cf}}}{K_{\text{ni}}}$

Here,

$x_{\text{f}}$ is the fraction of the kinetic energy which stays with the neutron after the energy transfer.

Substitute $0.284$ for $\frac{K_{\text{cf}}}{K_{\text{ni}}}$ in the above equation to find $x_{\text{f}}$.

$\begin{array}{c}{x}_{\text{f}}=1-0.284\\ x_{\text{f}}=0.716\end{array}$

Thus, the fraction of the kinetic energy which stays with the neutron is $0.716$.

Write the equation of final kinetic energy of the neutron.

$K_{\text{nf}}=x_{\text{f}}{K}_{\text{ni}}$

Here,

$K_{\text{nf}}$ is the final kinetic energy of the neutron.

Substitute $0.716$ for $x_{\text{f}}$ and $1.60\times {10}^{-13}\text{J}$ for $K_{\text{ni}}$ in above equation to find $K_{\text{nf}}$.

$\begin{array}{c}{K}_{\text{nf}}=\left(0.716\right)\left(1.60\times {10}^{-13}\text{J}\right)\\ =1.1456\times {10}^{-13}\text{J}\\ \approx \text{1}\text{.15}\times {\text{10}}^{-13}\text{J}\end{array}$

Thus, the final kinetic energy of the neutron is $1.15\times {10}^{-13}\text{J}$.

The neutron transferred a fraction of its energy to the carbon nucleus.

Write the final kinetic energy of the carbon nucleus when the energy is transferred from neutron.

$K_{\text{cf}}=\frac{K_{\text{cf}}}{K_{\text{ni}}}\cdot K_{\text{ni}}$

Substitute $0.284$ for $\frac{K_{\text{cf}}}{K_{\text{ni}}}$ and $1.60\times {10}^{-13}\text{J}$ for $K_{\text{ni}}$ in above equation to find $K_{\text{cf}}$.

$\begin{array}{c}{K}_{\text{cf}}=\left(0.284\right)\left(1.60\times {10}^{-13}\text{J}\right)\\ =4.54\times {10}^{-14}\text{J}\end{array}$

Thus, the final kinetic energy of the carbon nucleus is $4.544\times {10}^{-13}\text{J}$.

Conclusion:

Therefore, the final kinetic energy of the neutron is $1.15\times {10}^{-13}\text{J}$ and the final kinetic energy of the carbon nucleus is $4.54\times {10}^{-14}\text{J}$.