Chapter 15, Problem 33P

(a)

To determine

The energy released during the beta decay.

Answer to Problem 33P

The energy released during the beta decay is $0.78\text{ MeV}$.

Explanation of Solution

Write the given reaction of beta decay

    $n\to p+e^{-}+\overline{v}$        (I)

Here, $n$ is the neutron, $p$ is the proton, $e^{-}$ is the electron and $\overline{\nu }$ is the antineutrino.

Write the expression for energy released in this reaction

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

Here, $\Delta E$ is the energy released, $m_{\text{n}}$ the mass of the neutron, $m_{\text{p}}$ the mass of the proton, $m_e$ the mass of the electron and $c$ is the speed of light.

Conclusion:

Substitute $1.008665\text{u}$ for $m_{\text{n}}$, $1.007276\text{u}$ for $m_{\text{p}}$, $0.000549\text{u}$ for $m_{\text{e}}$ and $931.5\text{ Mev/u}$ for $c^2$ (II) to find $\Delta E$

    $\begin{array}{c}\Delta E=\left[\left(1.008665\text{u}\right)-\left(1.007276\text{u}\right)-\left(0.000549\text{u}\right)\right]\left(931.5\text{ Mev/u}\right)\\ =0.78\text{ MeV}\end{array}$

Therefore, the energy released during the beta decay is $0.78\text{ MeV}$.

(b)

To determine

The speed of the proton and the electron.

Answer to Problem 33P

The speed of the proton and electron is $0.00126c$ and $0.918c$ respectively.

Explanation of Solution

Write the expression for conservation of momentum

    $p_n=p_p+p_e$        (III)

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

The neutron is initially at rest and hence the momentum of the neutron in zero.

Rewrite equation (III)

    $\begin{array}{c}{p}_p=-p_e\\ \left|p_p\right|=\left|p_e\right|\\ {\left(p_{p}c\right)}^2={\left(p_{e}c\right)}^2=p^2{c}^2\end{array}$        (IV)

Write the expression for conservation of energy

    $m_n{c}^2=\sqrt{p_p{}^2{c}^2+{\left(m_p{c}^2\right)}^2}+\sqrt{p_e{}^2{c}^2+{\left(m_e{c}^2\right)}^2}$

Rewrite the above expression using (IV)

    $\begin{array}{c}{m}_n{c}^2-\sqrt{p^2{c}^2+{\left(m_p{c}^2\right)}^2}=\sqrt{p^2{c}^2+{\left(m_e{c}^2\right)}^2}\\ m_n{}^2{c}^4-2m_n{c}^2\sqrt{p^2{c}^2+{\left(m_p{c}^2\right)}^2}+p^2{c}^2+{\left(m_p{c}^2\right)}^2=p^2{c}^2+{\left(m_e{c}^2\right)}^2\\ {\left(\frac{m_n{}^2{c}^4+m_p{}^2{c}^4-m_e{}^2{c}^4}{2m_n{c}^2}\right)}^2-m_p{}^2{c}^4=p^2{c}^2\\ \left[{\left(\frac{m_n{}^2+m_p{}^2-m_e{}^2}{2m_n}\right)}^2-m_p{}^2\right]c^4=p^2{c}^2\end{array}$        (V)

Write the expression for the relativistic momentum of the proton

    $p_p=\frac{m_p{v}_p}{\sqrt{1-\frac{v_p{}^2}{c^2}}}$

Here, $v_p$ is the speed of the proton.

Rewrite the above expression

    $\begin{array}{c}{p}_p{}^2\left(1-\frac{v_p{}^2}{c^2}\right)=m_p{}^2{v}_p{}^2\\ p_p{}^2=v_p{}^2\left(m_p{}^2+\frac{p_p{}^2}{c^2}\right)\\ v_p=\sqrt{\frac{p_p{}^2{c}^2}{m_p{}^2{c}^2+p_p{}^2}}\end{array}$        (VI)

Similarly write the expression for speed of the electron

    $v_e=\sqrt{\frac{p_e{}^2{c}^2}{m_e{}^2{c}^2+p_e{}^2}}$        (VII)

Here, $v_e$ is the speed of the electron.

Conclusion:

Substitute $1.008665\text{u}$ for $m_{\text{n}}$, $1.007276\text{u}$ for $m_{\text{p}}$, $0.000549\text{u}$ for $m_{\text{e}}$ and $931.5\text{ Mev/u}$ for $c^2$ in equation (V) to find $p^2$

    $\begin{array}{c}{p}^2{c}^2=\left[{\left(\frac{{\left(1.008665\text{u}\right)}^2+{\left(1.007276\text{u}\right)}^2-{\left(0.000549\text{u}\right)}^2}{2\left(1.008665\text{u}\right)}\right)}^2-{\left(1.007276\text{u}\right)}^2\right]{\left(931.5\text{ Mev/u}\right)}^2\\ p^2=\frac{1.41{\left(\text{MeV}\right)}^2}{c^2}\end{array}$

Substitute $\frac{1.41{\left(\text{MeV}\right)}^2}{c^2}$ for $p_p{}^2$, $1.007276\text{u}$ for $m_{\text{p}}$ and $931.5\text{ Mev/u}$ for $c^2$ in equation (VI) to find $v_{\text{p}}$

    $\begin{array}{c}{v}_p=\sqrt{\frac{\frac{1.41{\left(\text{MeV}\right)}^2}{c^2}{c}^2}{{\left(1.007276\text{u}\right)}^2{c}^2+\frac{1.41{\left(\text{MeV}\right)}^2}{c^2}}}\\ =\sqrt{\frac{1.41{\left(\text{MeV}\right)}^2{c}^2}{{\left(1.007276\text{u}\right)}^2{c}^4+1.41{\left(\text{MeV}\right)}^2}}\\ =\sqrt{\frac{1.41{\left(\text{MeV}\right)}^2{c}^2}{{\left(1.007276\text{u}\right)}^2{\left(931.5\text{ Mev/u}\right)}^2+1.41{\left(\text{MeV}\right)}^2}}\\ =0.00126c\end{array}$

Substitute $\frac{1.41{\left(\text{MeV}\right)}^2}{c^2}$ for $p_e{}^2$, $0.000549\text{u}$ for $m_{\text{e}}$ and $931.5\text{ Mev/u}$ for $c^2$ in equation (VII) to find $v_{\text{e}}$

$\begin{array}{c}{v}_e=\sqrt{\frac{\frac{1.41{\left(\text{MeV}\right)}^2}{c^2}{c}^2}{{\left(0.000549\text{u}\right)}^2{c}^2+\frac{1.41{\left(\text{MeV}\right)}^2}{c^2}}}\\ =\sqrt{\frac{1.41{\left(\text{MeV}\right)}^2{c}^2}{{\left(0.000549\text{u}\right)}^2{c}^4+1.41{\left(\text{MeV}\right)}^2}}\\ =\sqrt{\frac{1.41{\left(\text{MeV}\right)}^2{c}^2}{{\left(0.000549\text{u}\right)}^2{\left(931.5\text{ Mev/u}\right)}^2+1.41{\left(\text{MeV}\right)}^2}}\\ =0.918c\end{array}$

Therefore, the speed of the proton and electron is $0.00126c$ and $0.918c$ respectively.

(c)

To determine

The particle moving at relativistic speed.

Answer to Problem 33P

Electron is moving at relativistic speed.

Explanation of Solution

If the speed of the particle is comparable to that of the speed of light, then it is said to relativistic.

Conclusion:

From part (b), the speed of the proton and electron is $0.00126c$ and $0.918c$ respectively.

Therefore, electron is moving at relativistic speed.