Chapter 46, Problem 69CP

(a)

To determine

Find the energy released in the reaction.

Answer to Problem 69CP

The energy released in the reaction is $\text{0}\text{.782 MeV}$.

Explanation of Solution

Write the equation for energy released.

    $\Delta E=\left(m_{\text{n}}-m_{\text{p}}-m_{\text{e}}\right)c^2$                                                                  (I)

Here, $\Delta E$ is the energy released, $m_p$ is the mass of proton, $m_n$ is the mass of neutron, $m_e$ is the mass of electron and $c$ is the speed of light.

Conclusion:

Substitute $\text{1}\text{.008 665 u}$ for $m_n$, $\text{1}\text{.007 825 u}$ for $m_p$ and $931.5\text{ MeV/u}$ for $c^2$ in equation I.

    $\begin{array}{c}\Delta E=\left(\text{1}\text{.008 665 u}-\text{1}\text{.007 825 u}\right)\left(931.5\text{ MeV/u}\right)\\ =0.782\text{ MeV}\end{array}$

Therefore, the energy released in the reaction is $\text{0}\text{.782 MeV}$.

(b)

To determine

Find the speed of proton and the electron after the reaction.

Answer to Problem 69CP

The speed of electron and the proton after the reaction are $0.919c$ and $382\text{ km/s}$ respectively. 

Explanation of Solution

Consider that the neutron at rest then the momentum conservation for the decay process implies that, $p_p=p_e$ .

Then, write the relativistic energy equation.

    $\sqrt{{\left(m_{\text{p}}{c}^2\right)}^2+p_{\text{p}}^2{c}^2}+\sqrt{{\left(m_{\text{e}}{c}^2\right)}^2+p_{\text{e}}^2{c}^2}=m_{\text{n}}{c}^2$                                                         (II)

Here, $p_p$ is the momentum of proton, $p_e$ is the momentum of electron.

Since, $p_p=p_e=p$, then the above equation becomes,

    $\begin{array}{l}\sqrt{{\left(m_{\text{p}}{c}^2\right)}^2+p^2{c}^2}=m_{\text{n}}{c}^2-\sqrt{{\left(m_e{c}^2\right)}^2+p^2{c}^2}\\ p^2{c}^2={\left[\frac{{\left(m_{\text{n}}{c}^2\right)}^2-{\left(m_{\text{p}}{c}^2\right)}^2+{\left(m_{\text{e}}{c}^2\right)}^2}{2m_{\text{n}}{c}^2}\right]}^2-{\left(m_{\text{e}}{c}^2\right)}^2\end{array}$                                               (III)

Write the equation for speed of electron.

    $\begin{array}{l}{p}_{\text{e}}c=\gamma m_{\text{e}}{v}_{\text{e}}c\\ \frac{\gamma v_{\text{e}}}{c}=\frac{p_{\text{e}}c}{m_{\text{e}}{c}^2}=\frac{1}{\sqrt{1-{\left(v_{\text{e}}/c\right)}^2}}\frac{v_{\text{e}}}{c}\\ 1-{\left(\frac{v_e}{c}\right)}^2={\left(\frac{v_e}{c}\right)}^2{\left(\frac{m_e{c}^2}{p_{e}c}\right)}^2\\ {\left(\frac{v_e}{c}\right)}^2\left[1+{\left(\frac{m_e{c}^2}{p_{e}c}\right)}^2\right]=1\end{array}$

  $\frac{v_{\text{e}}}{c}=\frac{1}{\sqrt{1+{\left(m_e{c}^2/p_{e}c\right)}^2}}$                                               (IV)

Conclusion:

Substitute $939.6\text{ MeV}$ for $m_n{c}^2$, $0.511\text{ MeV}$ for $m_e{c}^2$ and $938.3\text{ MeV}$ for $m_p{c}^2$ in equation III.

    $\begin{array}{c}{p}^2{c}^2={\left[\frac{{\left(939.6\text{ MeV}\right)}^2-{\left(938.3\text{ MeV}\right)}^2+{\left(0.511\text{ MeV}\right)}^2}{2\left(939.6\text{ MeV}\right)}\right]}^2-{\left(0.511\text{ MeV}\right)}^2\\ pc=1.19\text{ MeV}\end{array}$

Substitute $1.19\text{ MeV}$ for $pc$, and $0.511\text{ MeV}$ for $m_e{c}^2$ in equation IV.

    $\begin{array}{c}\frac{v_{\text{e}}}{c}=\frac{1}{\sqrt{1+{\left(0.511\text{ MeV}/1.19\text{ MeV}\right)}^2}}\\ v_e=0.919c\end{array}$ 

For finding the speed of proton replace mass of electron with mass of proton in equation IV.

    $\begin{array}{c}{v}_{\text{p}}=\frac{c}{\sqrt{1+{\left(m_{\text{p}}{c}^2/p_{\text{e}}c\right)}^2}}\\ =\frac{2.998\times {10}^8\text{ m/s}}{\sqrt{1+{\left(938.3\text{ MeV}/1.19\text{ MeV}\right)}^2}}\\ =3.82\times {10}^5\text{ m/s}\\ =382\text{ km/s}\end{array}$

Therefore, the speed of electron and the proton after the reaction is $0.919c$ and $382\text{ km/s}$ respectively.

(c)

To determine

Does any of these particle moving at a relativistic speed.

Answer to Problem 69CP

The electron is moving at relativistic speed whereas proton is not moving at relativistic speed.

Explanation of Solution

From the previous part the electron is moving at more than one tenth of speed of light so electron is moving with relativistic speed.

Whereas the speed of proton is less than one tenth of the speed of light so the proton is not moving with relativistic speed.