Chapter 6, Problem 69AP

A neutron in a reactor makes an elastic head-on collision with a carbon atom that is initially at rest. (The mass of the carbon nucleus is about 12 times that of the neutron.) (a) What fraction of the neutron’s kinetic energy is transferred to the carbon nucleus? (b) If the neutron’s initial kinetic energy is 1.6 × 10⁻¹³ J, find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.

To determine

The magnitude of fraction of kinetic energy transferred to the carbon nucleus.

Answer to Problem 69AP

Solution:

The magnitude of the fraction of kinetic energy is $0.28$.

Explanation of Solution

Given Info:

The mass of the carbon is twelve times as that of the neutron and the initial velocity of the carbon is zero.

Write the expression for the momentum conservation.

$m_n{v}_i+m_c{v}_i{}^{\prime }=m_n{v}_f+m_c{v}_f{}^{\prime }$

• $m_n$ is the mass of the neutron
• $m_c$ is the mass of the carbon
• $v_i$ is the initial velocity of the neutron
• $v_i{}^{\prime }$ is the initial velocity of the carbon
• $v_f$ is the final velocity of the neutron
• $v_f{}^{\prime }$ is the final velocity of the carbon

Use 0 for $v_i{}^{\prime }$ to rewrite the above expression.

$\begin{array}{c}{m}_n{v}_i+m_c\left(0\right)=m_n{v}_f+m_c{v}_f{}^{\prime }\\ m_n{v}_i=m_n{v}_f+m_c{v}_f{}^{\prime }\end{array}$ (I)

Write the relation between the mass of the carbon and neutron.

$m_c=12m_n$ (II)

Rewrite the (I) using (II).

$\begin{array}{c}{m}_n{v}_i=m_n{v}_f+12m_n{v}_f{}^{\prime }\\ v_i=v_f+12v_f{}^{\prime }\end{array}$ (III)

Write the expression for head on collision.

$v_i-v_i{}^{\prime }=-\left(v_f-v_f{}^{\prime }\right)$

Use 0 for $v_i{}^{\prime }$ to rewrite the above expression.

$\begin{array}{c}{v}_i-0=-\left(v_f-v_f{}^{\prime }\right)\\ v_i=-\left(v_f-v_f{}^{\prime }\right)\\ v_i=v_f{}^{\prime }-v_f\\ v_f{}^{\prime }=v_i+v_f\end{array}$ (IV)

Use (IV) in (III) to calculate $v_f$.

$\begin{array}{c}{v}_i=v_f+12\left(v_i+v_f\right)\\ v_i=v_f+12v_i+12v_f\\ -13v_f=11v_i\\ v_f=-\left(\frac{11}{13}\right)v_i\end{array}$ (V)

Substitute (V) in (IV) to calculate $v_f{}^{\prime }$.

$\begin{array}{c}{v}_f{}^{\prime }=v_i-\frac{11}{13}{v}_i\\ =\frac{2}{13}{v}_i\end{array}$

Conclusion:

The magnitude of the final velocity of the carbon is $\frac{2}{13}{v}_i$.

Write the formula to calculate the kinetic energy of neutron.

$KE=\frac{1}{2}{m}_n{v}_i{}^2$

• KE is the kinetic energy of the neutron

Write the formula to calculate the final kinetic energy of the carbon.

$K{E}^{\prime }=\frac{1}{2}{m}_c{\left(v_f{}^{\prime }\right)}^2$

• $K{E}^{\prime }$ is the kinetic energy of the carbon

Substitute $12m_n$ for $m_c$ and $\frac{2}{13}{v}_i$ for $v_f{}^{\prime }$ to rewrite the above expression.

$\begin{array}{c}K{E}^{\prime }=\frac{1}{2}\left(12m_n\right){\left(\frac{2}{13}{v}_i\right)}^2\\ =\frac{1}{2}{m}_n\left(\frac{48}{169}{v}_i{}^2\right)\\ =\frac{1}{2}{m}_n{v}_i{}^2\left(\frac{48}{169}\right)\end{array}$

Use KE for $\frac{1}{2}{m}_n{v}_i{}^2$ to calculate the ratio of kinetic energy of carbon to neutron in the above equation.

$\begin{array}{c}K{E}^{\prime }=KE\left(\frac{48}{169}\right)\\ \frac{K{E}^{\prime }}{KE}=\frac{48}{169}\\ =0.28\end{array}$

Therefore, the magnitude of the fraction of kinetic energy transferred to the carbon is $0.28$.

To determine

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

Answer to Problem 69AP

Solution:

The magnitude of the final kinetic energy of the neutron and kinetic energy of the carbon respectively $\text{1}\text{.2}\times {\text{10}}^{-13}\text{J}$ and $\text{4}\text{.5}\times {\text{10}}^{-14}\text{J}$.

Explanation of Solution

Given Info:

The initial kinetic energy of the neutron is $1.6\times {10}^{-13}\text{J}$.

Write the expression for the ratio of kinetic energy of the carbon to initial kinetic energy of the neutron.

$\frac{K{E}^{\prime }}{KE}=0.28$

Substitute $1.6\times {10}^{-13}\text{J}$ for KE in the above expression to calculate $K{E}^{\prime }$.

$\begin{array}{c}\frac{K{E}^{\prime }}{\left(1.6\times {10}^{-13}\text{J}\right)}=0.28\\ K{E}^{\prime }=0.28\left(1.6\times {10}^{-13}\text{J}\right)\\ =0.448\times {10}^{-13}\text{J}\end{array}$

The difference between the initial kinetic energy of the neutron and final kinetic energy of the carbon would give the final kinetic energy of the neutron.

Write the formula to calculate the final kinetic energy of the neutron.

$K{E}_f=KE-K{E}^{\prime }$

• $K{E}_f$ is the final kinetic energy of the neutron

Substitute $1.6\times {10}^{-13}\text{J}$ for KE and $0.448\times {10}^{-13}\text{J}$ for $K{E}^{\prime }$ to calculate $K{E}_f$.

$\begin{array}{c}K{E}_f=1.6\times {10}^{-13}\text{J}-0.448\times {10}^{-13}\text{J}\\ \text{=1}\text{.152}\times {\text{10}}^{-13}\text{J}\\ \approx \text{1}\text{.2}\times {\text{10}}^{-13}\text{J}\end{array}$

Conclusion:

Therefore, the magnitude of the final kinetic energy of the neutron and kinetic energy of the carbon respectively $\text{1}\text{.2}\times {\text{10}}^{-13}\text{J}$ and $\text{4}\text{.5}\times {\text{10}}^{-14}\text{J}$