Chapter 43, Problem 58AP

(a)

To determine

Prove that $\frac{K_1}{K_{\text{tot}}}=\frac{m_2}{m_1+m_2}$ and $\frac{K_2}{K_{\text{tot}}}=\frac{m_1}{m_1+m_2}$.

Answer to Problem 58AP

It is proved that $\frac{K_1}{K_{\text{tot}}}=\frac{m_2}{m_1+m_2}$ and $\frac{K_2}{K_{\text{tot}}}=\frac{m_1}{m_1+m_2}$.

Explanation of Solution

Two fragments must move in opposite direction for the momentum conservation. Write the equation for conservation of momentum.

  $m_1{v}_1 =m_2{v}_2$

Here, $m_1,m_2$ are the masses of fragments and $v_1,v_2$ are velocities of fragments.

Rewrite the above relation in terms of $v_2$.

    $v_2=\left(\frac{m_1}{m_2}\right)v_1$

Write the equation for kinetic energy of mass $m_1$.

  $K_1=\frac{1}{2}{m}_1{v}_1^2$

Here, $K_1$ is the kinetic energy of mass $m_1$.

Write the equation for kinetic energy of mass $m_2$.

  $K_2=\frac{1}{2}{m}_2{v}_2^2$

Here, $K_2$ is the kinetic energy of mass $m_2$.

Rewrite the above expression by substituting $\left(\frac{m_1}{m_2}\right)v_1$ for $v_2$.

  $\begin{array}{c}{K}_2=\frac{1}{2}{m}_2{\left(\left(\frac{m_1}{m_2}\right)v_1\right)}^2\\ =\left(\frac{m_1}{m_2}\right)\frac{1}{2}{m}_1{v}_1^2\end{array}$

Rewrite the above equation by substituting $K_1$ for $\frac{1}{2}{m}_1{v}_1^2$.

  $K_2=\left(\frac{m_1}{m_2}\right)K_1$

Write the equation for total kinetic energy.

    $K_{total}=K_1+K_2$

Here, $K_{total}$ is the total kinetic energy.

Conclusion:

Write the equation for the fraction of total kinetic possessed by mass $m_1$.

  $\frac{K_1}{K_{total}}=\frac{K_1}{K_1+K_2}$

Rewrite the above expression by substituting $\left(\frac{m_1}{m_2}\right)K_1$ for $K_2$.

  $\begin{array}{c}\frac{K_1}{K_{total}}=\frac{K_1}{K_1+\left(\frac{m_1}{m_2}\right)K_1}\\ =\frac{m_2}{m_1+m_2}\end{array}$

Write the equation for the fraction of total kinetic possessed by mass $m_2$.

  $\frac{K_2}{K_{total}}=1-\frac{K_1}{K_1+K_2}$

Rewrite the above expression by substituting $\frac{m_2}{m_1+m_2}$ for $\frac{K_1}{K_{total}}$.

  $\begin{array}{c}\frac{K_2}{K_{total}}=1-\frac{m_2}{m_1+m_2}\\ =\frac{m_1}{m_1+m_2}\end{array}$

Therefore, the It is proved that $\frac{K_1}{K_{\text{tot}}}=\frac{m_2}{m_1+m_2}$ and $\frac{K_2}{K_{\text{tot}}}=\frac{m_1}{m_1+m_2}$.

(b)

To determine

The disintegration energy for nuclear fission.

Answer to Problem 58AP

The disintegration energy for nuclear fission is $177.4\text{ MeV}$.

Explanation of Solution

Write the equation for disintegration energy.

    $Q=\left(m_U-m_{Br}-m_{La}\right)c^2$

Here, $Q$ is the disintegration energy, $m_U$ is the mass of ${}_{92}^{236}U$, $m_{Br}$ is the mass of ${}_{35}^{87}Br$, $m_{La}$ is the mass of ${}_{57}^{149}La$, and $c$ is the speed of light in vacuum.

Conclusion:

Substitute $236.045562\text{u}$ for $m_U$, $86.920711\text{u}$ for $m_{Br}$, $148.934370\text{u}$ for $m_{La}$, and $\text{931}\text{.5 }\text{MeV}/\text{u}$ for $c^2$ in the above equation to find $Q$.

  $\begin{array}{c}Q=\left(236.045562\text{u}-86.920711\text{u}-148.934370\text{u}\right)\times \left(\text{931}\text{.5 }\text{MeV}/\text{u}\right)\\ =177.4\text{MeV}\end{array}$

Therefore, the disintegration energy for nuclear fission is $177.4\text{ MeV}$.

(c)

To determine

The splitting of disintegration energy between the two primary fragments.

Answer to Problem 58AP

The splitting of disintegration energy for Br is $112.0\text{ MeV}$ and for La is $65.4\text{ MeV}$.

Explanation of Solution

Write the equation to find the disintegration energy for Br.

  $K_{\text{Br}}=Q\frac{m_{\text{La}}}{m_{\text{Br}}+m_{\text{La}}}$

Here, $K_{\text{Br}}$ is the disintegration energy for Br.

Write the equation to find the disintegration energy for La.

  $K_{\text{La}}=Q-K_{\text{Br}}$

Conclusion:

Substitute $177.4\text{MeV}$ for $Q$, $149\text{u}$ for $m_{\text{La}}$, and $87\text{u}$ for $m_{\text{Br}}$ in the equation for $K_{\text{Br}}$.

    $\begin{array}{c}{K}_{\text{Br}}=\left(177.4\text{MeV}\right)\frac{149\text{u}}{87\text{u}+149\text{u}}\\ =112.0\text{MeV}\end{array}$

Substitute $177.4\text{MeV}$ for $Q$ and $112.0\text{MeV}$ for $K_{\text{Br}}$ in the equation for $K_{\text{La}}$.

  $\begin{array}{c}{K}_{\text{La}}=177.4\text{MeV}-112.0\text{MeV}\\ \text{=}65.4\text{MeV}\end{array}$

Therefore, the splitting of disintegration energy for Br is $112.0\text{ MeV}$ and for La is $65.4\text{ MeV}$.

(d)

To determine

The speed of fragments just after the fission.

Answer to Problem 58AP

The speed of Br nuclie and La nuclei are $1.58\times {10}^7\text{m}/\text{s}$ and $9.20\times {10}^6\text{m}/\text{s}$ respectively.

Explanation of Solution

Write the equation to find the speed of Br nuclie.

  $v_{\text{Br}}=\sqrt{\frac{2K_{\text{Br}}}{m_{\text{Br}}}}$

Here, $v_{\text{Br}}$ is the speed of Br nuclie.

Write the equation to find the speed of La nuclie.

  $v_{\text{Br}}=\sqrt{\frac{2K_{\text{Br}}}{m_{\text{Br}}}}$

Here, $v_{\text{Br}}$ is the speed of Br nuclie.

  $v_{\text{La}}=\sqrt{\frac{2K_{\text{La}}}{m_{\text{La}}}}$

Here, $v_{\text{La}}$ is the speed of Br nuclie.

Conclusion:

Substitute $112.0\text{ MeV}$ for $K_{\text{Br}}$ and $87\text{u}$ for $m_{\text{Br}}$ in the equation to find $v_{\text{Br}}$.

    $\begin{array}{c}{v}_{\text{Br}}=\sqrt{\frac{2\left(112.0\text{ MeV}\left(\frac{{10}^6\text{eV}}{\text{1}\text{MeV}}\right)\left(\frac{1\text{J}}{1.60\times {10}^{-19}\text{eV}}\right)\right)}{87\text{u}\left(\frac{1\text{kg}}{1.66\times {10}^{-27}\text{u}}\right)}}\\ =\sqrt{2.50\times {10}^{14}{\text{m}}^2/{\text{s}}^2}\\ =1.58\times {10}^7\text{m}/\text{s}\end{array}$

Substitute $65.4\text{MeV}$ for $K_{\text{La}}$ and $149\text{u}$ for $m_{\text{La}}$ in the equation to find $v_{\text{Br}}$.

    $\begin{array}{c}{v}_{\text{Br}}=\sqrt{\frac{2\left(112.0\text{ MeV}\left(\frac{{10}^6\text{eV}}{\text{1}\text{MeV}}\right)\left(\frac{1\text{J}}{1.60\times {10}^{-19}\text{eV}}\right)\right)}{87\text{u}\left(\frac{1\text{kg}}{1.66\times {10}^{-27}\text{u}}\right)}}\\ =\sqrt{84.64\times {10}^{12}{\text{m}}^2/{\text{s}}^2}\\ =9.20\times {10}^6\text{m}/\text{s}\end{array}$

Therefore, the speed of Br nuclie and La nuclei are $1.58\times {10}^7\text{m}/\text{s}$ and $9.20\times {10}^6\text{m}/\text{s}$ respectively.