Chapter 13, Problem 48P

(a)

To determine

The maximum kinetic energy of the electron when a neutron decays at rest .

Answer to Problem 48P

The maximum kinetic energy of the electron when a neutron decays at rest is $\underset{\_}{0.782\text{MeV}}$ .

Explanation of Solution

Write the expression for the decays reaction,

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

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

Write the expression for the energy conservation of the above equation,

    $E_n=E_p+E_{e^{-}}+E_{\overline{\nu }}$        (II)

Here, $E_n$ is the energy of neutron, $E_p$ is the energy of proton, $E_{e^{-}}$ is the energy of electron and $E_{\overline{\nu }}$ is the energy of antineutrino.

In this case, the energy of antineutrino,

    $E_{\overline{\nu }}\simeq 0$        (III)

Rewrite the above equation,

    $m_n{c}^2\simeq m_p{c}^2+K_p+E_e$        (IV)

    $\begin{array}{c}{K}_p+E_e=m_n{c}^2-m_p{c}^2\\ =c^2\left(m_n-m_p\right)\\ =1.293\text{MeV}\end{array}$        (V)

Write the kinetic energy of the proton,

    $K_p=\frac{p_p^2}{2m_p}$        (VI)

Here, $K_p$ is the energy of the proton, $p_p$ is the momentum of the proton and $m_p$ is the mass of the proton.

Conclusion:

Since, $K_p$ is negligible compared to $K_e$ .

So the kinetic energy values,

    $\begin{array}{c}{K}_p+K_e+m_e{c}^2=1.293\text{MeV}\\ K_e+m_e{c}^2=1.293\text{MeV}\\ K_e=1.293\text{MeV}-m_e{c}^2\end{array}$

Substitute $0.511\text{MeV}$ for $m_e{c}^2$ in the above equation,

    $\begin{array}{c}{K}_e=1.293\text{MeV}-0.511\text{MeV}\\ =0.783\text{MeV}\end{array}$

Therefore, the maximum kinetic energy of the electron when a neutron decays at rest is $\underset{\_}{0.782\text{MeV}}$ .

(b)

To determine

The kinetic energy of the proton .

Answer to Problem 48P

The kinetic energy of the proton is $\underset{\_}{7.52\times {10}^{-4}\text{MeV}}$ .

Explanation of Solution

Write the kinetic energy of the proton,

    $K_p=\frac{p_p^2}{2m_p}$        (VI)

Here, $K_p$ is the energy of the proton, $p_p$ is the momentum of the proton and $m_p$ is the mass of the proton.

In this case, the magnitude of the momentum of the electron is equal to the magnitude of the momentum of proton.

    $\left|p_e\right|=\left|p_p\right|$        (VII)

So,

    $p_e^2{c}^2=E_e^2-{\left(m_e{c}^2\right)}^2$        (VIII)

Conclusion:

Substitute $1.293\text{MeV}$ for $E_e$ and $0.511\text{MeV}$ for $m_e{c}^2$ in (VIII),

    $\begin{array}{c}{p}_e^2{c}^2={\left(1.293\text{MeV}\right)}^2-{\left(0.511\text{MeV}\right)}^2\\ =1.411{\left(\text{MeV}\right)}^2\end{array}$

Substitute $1.411{\left(\text{MeV}\right)}^2$ for $p_p^2$ and $938.3\text{MeV}$ for $m_p{c}^2$ in (VI),

    $\begin{array}{c}{K}_p=\frac{1.411{\left(\text{MeV}\right)}^2}{2\left(938.3\text{MeV}\right)}\\ =7.52\times {10}^{-4}\text{MeV}\end{array}$

Therefore, the kinetic energy of the proton is $\underset{\_}{7.52\times {10}^{-4}\text{MeV}}$.

(c)

To determine

The maximum kinetic energy and momentum of the anti-neutrino and the kinetic energy of the proton .

Answer to Problem 48P

The maximum kinetic energy and momentum of the anti-neutrino and the kinetic energy of the proton are $\underset{\_}{0.782\text{MeV},\frac{0.782\text{MeV}}{c}\&3.26\times {10}^{-4}\text{MeV}}$ .

Explanation of Solution

In this case, the kinetic energy of the electron is zero.

    $K_e=0$

So the expression for the energy of the given reaction,

    $m_n{c}^2=K_p+m_p{c}^2+m_e{c}^2+E_{\overline{\nu }}$        (IX)

Write the expression for the energy of the anti-neutrino,

    $E_{\overline{\nu }}=\left(m_n-m_p-m_e\right)c^2-K_p$        (X)

Write the expression for the momentum of the antineutrino,

    $\begin{array}{c}{E}_{\overline{\nu }}=p_{\overline{\nu }}c\\ p_{\overline{\nu }}=\frac{E_{\overline{\nu }}}{c}\end{array}$        (XI)

Here, $p_{\overline{\nu }}$ is momentum of the antineutrino.

Write the expression for the kinetic energy of proton,

    $\begin{array}{c}{K}_p=\frac{{\left(p_{\overline{\nu }}\right)}^2}{2m_p}\\ =\frac{{\left(p_{\overline{\nu }}c\right)}^2}{2m_p{c}^2}\end{array}$        (XII)

Conclusion:

Substitute $0.782\text{MeV}$ for $\left(m_n-m_p-m_e\right)c^2$ in (X),

    $E_{\overline{\nu }}=0.782\text{MeV}-K_p$

Since, $K_p<<0.782\text{MeV}$

So,

    $E_{\overline{\nu }\max }\cong 0.782\text{MeV}$

Substitute $0.782\text{MeV}$ for $E_{\overline{\nu }}$ in (XI),

    $p_{\overline{\nu }\max }=\frac{0.782\text{MeV}}{c}$

Substitute $\frac{0.782\text{MeV}}{c}$ for $p_{\overline{\nu }}$ and $938.3\text{MeV}$ for $m_p{c}^2$ in (XII)

    $\begin{array}{c}{K}_p=\frac{{\left(\frac{0.782\text{MeV}}{c}\right)}^2{c}^2}{2\left(938.3\text{MeV}\right)}\\ =3.26\times {10}^{-4}\text{MeV}\end{array}$

Therefore, the maximum kinetic energy and momentum of the anti-neutrino and the kinetic energy of the proton are $\underset{\_}{0.782\text{MeV},\frac{0.782\text{MeV}}{c}\&3.26\times {10}^{-4}\text{MeV}}$.