Chapter 31, Problem 28P

(a)

To determine

The magnitude of momentum the particle ${\Sigma }^{+}$ and ${\pi }^{+}$.

Answer to Problem 28P

The magnitude of momentum the particle ${\Sigma }^{+}$ and ${\pi }^{+}$ is $\underset{\_}{686\text{MeV/c}}$ and $\underset{\_}{200\text{MeV/c}}$.

Explanation of Solution

Write the expression for the momentum.

$p=mv$        (I)

Here, $p$ is the momentum, $m$ is the mass, $v$ is the particle speed.

Write the expression for the momentum for a circular motion by using equation (I).

$\begin{array}{l}qvB=\frac{m{v}^2}{r}\\ mv=qBr\\ p=qBr\end{array}$        (II)

Here, $q$ is the charge, $B$ is the magnetic field and $r$ is the radius of the path.

Conclusion:

For the particle ${\Sigma }^{+}$

Substitute, $1.602\times {10}^{-19}\text{C}$ for $q$, $1.15\text{T}$ for $B$, $1.99\text{m}$ for $r$ in equation (II) to find $p$.

$\begin{array}{c}p=\left[\left(1.602\times {10}^{-19}\text{C}\right)\left(1.15\text{T}\right)\left(1.99\text{m}\right)\right]\left(\frac{1.871\times {10}^{21}\text{MeV/c}}{\text{kg-m/s}}\right)\\ =686\text{MeV/c}\end{array}$

For the particle ${\pi }^{+}$

Substitute, $1.602\times {10}^{-19}\text{C}$ for $q$, $1.15\text{T}$ for $B$, $0.580\text{m}$ for $r$ in equation (II) to find $p$.

$\begin{array}{c}p=\left[\left(1.602\times {10}^{-19}\text{C}\right)\left(1.15\text{T}\right)\left(0.580\text{m}\right)\right]\left(\frac{1.871\times {10}^{21}\text{MeV/c}}{\text{kg-m/s}}\right)\\ =200\text{MeV/c}\end{array}$

Thus, the magnitude of momentum the particle ${\Sigma }^{+}$ and ${\pi }^{+}$ is $\underset{\_}{686\text{MeV/c}}$ and $\underset{\_}{200\text{MeV/c}}$.

(b)

To determine

The magnitude of the momentum of neutron.

Answer to Problem 28P

The magnitude of the momentum of neutron is $\underset{\_}{626\text{MeV/c}}$.

Explanation of Solution

Write the expression for the magnitude of the momentum of the neutron.

$p_n=\sqrt{{\left(p_n\cos \varphi \right)}^2+{\left(p_{n}sin\varphi \right)}^2}$        (III)

Here, $p_n$ is the magnitude of the momentum of neutron, $p_n\cos \varphi$ is the x-component of momentum and $p_n\sin \varphi$ is y-component of momentum.

Write the expression for the x-component of neutron momentum from the conservation of momentum along parallel to original momentum

$\begin{array}{l}{p}_{{\Sigma }^{+}}=p_n\cos \varphi +p_{{\pi }^{+}}\cos \theta \\ p_n\cos \varphi =p_{{\Sigma }^{+}}-p_{{\pi }^{+}}\cos \theta \end{array}$        (IV)

Write the expression for the y-component of neutron momentum from the conservation of momentum along perpendicular to original momentum

$p_n\sin \varphi =p_{{\pi }^{+}}\cos \theta$        (V)

Write the expression from (I) by using (IV) and (V).

$p_n=\sqrt{{\left(p_{{\Sigma }^{+}}-p_{{\pi }^{+}}\cos \theta \right)}^2+{\left(p_{{\pi }^{+}}sin\theta \right)}^2}$        (VI)

Conclusion:

Substitute, $\left(686\text{MeV/c}\right)$ for $p_{{\Sigma }^{+}}$, $\left(200\text{MeV/c}\right)$ for $p_{{\pi }^{+}}$, $64.5°$ for $\theta$ in equation (VI) to find $p_n$.

$\begin{array}{c}{p}_n=\sqrt{{\left(\left(686\text{MeV/c}\right)-\left[\left(200\text{MeV/c}\right)\cos \left(64.5°\right)\right]\right)}^2+{\left(\left(200\text{MeV/c}\right)sin\left(64.5°\right)\right)}^2}\\ =626\text{MeV/c}\end{array}$

Thus, the magnitude of the momentum of neutron is $\underset{\_}{626\text{MeV/c}}$.

(c)

To determine

The total energy of the ${\pi }^{+}$ and neutron.

Answer to Problem 28P

The total energy of the ${\pi }^{+}$ and neutron is $\underset{\_}{244\text{MeV}}$ and $\underset{\_}{1.13G\text{eV}}$.

Explanation of Solution

Write the expression for the total energy of ${\pi }^{+}$.

$E_{{\pi }^{+}}=\sqrt{{\left(p_{{\pi }^{+}}c\right)}^2+{\left(m_{{\pi }^{+}}{c}^2\right)}^2}$        (VII)

Here, $E_{{\pi }^{+}}$ is the total energy of , ${\pi }^{+}$ $c$ is the light of speed, $m_{{\pi }^{+}}$ is the mass of ${\pi }^{+}$, $p_{{\pi }^{+}}$ is the momentum of ${\pi }^{+}$.

Write the expression for the total energy of neutron.

$E_n=\sqrt{{\left(p_{n}c\right)}^2+{\left(m_n{c}^2\right)}^2}$        (VIII)

Here, $E_n$ is the total energy of neutron, $c$ is the light of speed, $m_n$ is the mass of neutron, $p_n$ is the momentum of neutron.

Conclusion:

Substitute, $200\text{MeV}$ for $p_{{\pi }^{+}}c$, $139.6\text{MeV}$ for $m_{{\pi }^{+}}{c}^2$ in equation (VII) to find $E_{{\pi }^{+}}$.

$\begin{array}{c}{E}_{{\pi }^{+}}=\sqrt{{\left(200\text{MeV}\right)}^2+{\left(139.6\text{MeV}\right)}^2}\\ =244\text{MeV}\end{array}$

Substitute, $626\text{MeV}$ for $p_{n}c$, $939.6\text{MeV}$ for $m_n{c}^2$ in equation (VIII) to find $E_n$.

$\begin{array}{c}{E}_n=\sqrt{{\left(626\text{MeV}\right)}^2+{\left(939.6\text{MeV}\right)}^2}\\ =1129\text{MeV}\\ \text{=1}\text{.13}\text{GeV}\end{array}$

Thus, the total energy of the ${\pi }^{+}$ and neutron is $\underset{\_}{244\text{MeV}}$ and $\underset{\_}{1.13G\text{eV}}$.

(d)

To determine

The total energy of the ${\Sigma }^{+}$.

Answer to Problem 28P

The total energy of the ${\Sigma }^{+}$ is $\underset{\_}{1.37\text{GeV}}$.

Explanation of Solution

Write the expression for the total energy of ${\Sigma }^{+}$.

$E_{{\Sigma }^{+}}=E_{{\pi }^{+}}+E_n$        (IX)

Here, $E_{{\Sigma }^{+}}$ is the total energy of ${\Sigma }^{+}$.

Conclusion:

Substitute, $244\text{MeV}$ for $E_{{\pi }^{+}}$, $1129\text{MeV}$ for $E_n$ in equation (IX) to find $E_{{\Sigma }^{+}}$.

$\begin{array}{c}{E}_{{\Sigma }^{+}}=\left(244\text{MeV}\right)+\left(1129\text{MeV}\right)\\ =1373\text{MeV}\\ =1.37\text{GeV}\end{array}$

Thus, The total energy of the ${\Sigma }^{+}$ is $\underset{\_}{1.37\text{GeV}}$.

(e)

To determine

The mass of ${\Sigma }^{+}$.

Answer to Problem 28P

The mass of ${\Sigma }^{+}$ is $1.19{\text{GeV/c}}^{\text{2}}$.

Explanation of Solution

Write the expression for the mass of ${\Sigma }^{+}$ by using rest mass energy.

$m_{{\Sigma }^{+}}=\frac{1}{c^2}\left[E_{{\Sigma }^{+}}^2-{\left(p_{{\Sigma }^{+}}c\right)}^2\right]$        (X)

Here, $m_{{\Sigma }^{+}}$ is the mass.

Conclusion:

Substitute, $1373\text{MeV}$ for $E_{{\Sigma }^{+}}$, $686\text{MeV}$ for $p_{{\Sigma }^{+}}c$ in equation (X) to find $m_{{\Sigma }^{+}}$.

$\begin{array}{c}{m}_{{\Sigma }^{+}}=\frac{1}{c^2}\left[{\left(1373\text{MeV}\right)}^2-{\left(686\text{MeV}\right)}^2\right]\\ =1189{\text{MeV/c}}^2\\ =1.19{\text{GeV/c}}^{\text{2}}\end{array}$

Thus, the mass of ${\Sigma }^{+}$ is $1.19{\text{GeV/c}}^{\text{2}}$.

(f)

To determine

Compare the calculated and given value.

Answer to Problem 28P

The mass of ${\Sigma }^{+}$ is $1.19{\text{GeV/c}}^{\text{2}}$.

Explanation of Solution

Write the expression for the comparison between the given and the calculated value of mass.

$\left[\frac{\left(1.19\times {10}^3{\text{MeV/c}}^{\text{2}}\right)-\left(1189.4{\text{MeV/c}}^{\text{2}}\right)}{\left(1189.4{\text{MeV/c}}^{\text{2}}\right)}\right]\times 100\%=0.0504\%$

The calculated value differs from 0.05% from given value.