Chapter 31, Problem 11P

(a)

To determine

The energy released in the reaction.

Answer to Problem 11P

The energy released in the reaction is $\underset{\_}{0.782\text{MeV}}$.

Explanation of Solution

Write the expression for the energy released in the reaction.

$\left(\Delta E\right)=\left(m_n-m_p-m_e\right)c^2$        (I)

Here, $\left(\Delta E\right)$ is the energy, $m_n,m_p,m_e$ are the masses of the neutron, proton and electron, $c$ is the speed of light.

Conclusion:

Mass of electron is negligible.

Substitute, $1.008665\text{u}$ for $m_n$, $1.007825\text{u}$ for $m_p$, $931.5\text{MeV/u}$ for $c^2$ in equation (I) to find $\left(\Delta E\right)$.

$\begin{array}{c}\left(\Delta E\right)=\left(1.008665\text{u}-1.007825\text{u}\right)\left(931.5\text{MeV/u}\right)\\ =0.783\text{MeV}\end{array}$

Thus, the energy released in the reaction is $\underset{\_}{0.782\text{MeV}}$.

(b)

To determine

The speed of electron and proton after reaction.

Answer to Problem 11P

The speed of electron and proton after reaction is $v_e=0.919c$ and $v_p=382\text{km/s}$.

Explanation of Solution

Write the expression for the conservation momentum.

$p_e=p_p$        (II)

Here, $p_e,p_p$ are the momentum of electron and proton.

Write the expression for the conservation energy by using (II).

$\begin{array}{l}\sqrt{{\left(m_p{c}^2\right)}^2+p_p^2{c}^2}+\sqrt{{\left(m_e{c}^2\right)}^2+p_e^2{c}^2}=m_n{c}^2\\ p_e^2{c}^2={\left\{\frac{\left[{\left(m_n{c}^2\right)}^2-{\left(m_p{c}^2\right)}^2+{\left(m_e{c}^2\right)}^2\right]}{2m_n{c}^2}\right\}}^2-{\left(m_e{c}^2\right)}^2\end{array}$        (III)

Write the expression for the gamma factor.

$\gamma =\frac{1}{\sqrt{1-{\left(v_e/c\right)}^2}}$        (IV)

Here, $\gamma$ is the gamma factor, $v_e$ is the speed of the electron.

Write the expression for the velocity of electron by using conservation of momentum and equation (IV).

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

Here, $m_e$ is the mass of electron.

Write the expression for the velocity of proton by using conservation of momentum and equation (II) and (IV).

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

Here, $m_p$ is the mass of proton and $v_p$ is the velocity of proton.

Conclusion:

Substitute, $939.6\text{MeV}$ for $m_n{c}^2$, $938.3\text{MeV}$ for $m_p{c}^2$, $0.511\text{MeV}$ for $m_e{c}^2$ in equation (II) to find $pc$.

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

Substitute, $1.19\text{MeV}$ for $p_{e}c$, $0.511\text{MeV}$ for $m_e{c}^2$ in equation (V) to find $v_e$.

$\begin{array}{c}{v}_e=\left[\frac{1}{\sqrt{1+{\left(0.511\text{MeV}/1.19\text{MeV}\right)}^2}}\right]c\\ =0.919c\end{array}$

Substitute, $1.19\text{MeV}$ for $p_{e}c$, $938.3\text{MeV}$ for $m_p{c}^2$, $2.998\times {10}^8\text{m/s}$ for $c$ in equation (VI) to find $v_p$.

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

Thus, the speed of electron and proton after reaction is $v_e=0.919c$ and $v_p=382\text{km/s}$.

(c)

To determine

The particles which is in relativistic speed.

Answer to Problem 11P

The speed of the electron is relativistic.

Explanation of Solution

The speed of electron is $v_e=0.919c$. The speed proton after reaction $v_p=382\text{km/s}$.

The speed of light is $2.998\times {10}^8\text{m/s}$. Hence the speed of proton is less than one-tenth of light speed. On the other side, the speed of electron is almost one-tenth of light speed. Thus, it is clear that electron is having relativistic speed.