Chapter 46, Problem 26P

(a)

To determine

The magnitudes of the momenta of the ${\Sigma }^{+}$ and the ${\pi }^{+}$ particle in units of $\text{MeV}/c$.

Answer to Problem 26P

The magnitudes of the momenta of the ${\Sigma }^{+}$ particle is $686\text{MeV}/c$ and the ${\pi }^{+}$ particle is $200\text{MeV}/c$.

Explanation of Solution

Write the formula for force

    ${\displaystyle \sum F}=ma$                                                                             (I)

Write the formula for centripetal force

    ${\displaystyle \sum F}=qvB\sin \theta$                                                                  (II)

Here, $F$ is the centripetal force, $m$ is the mass, $a$ is the centripetal acceleration $a=\frac{v^2}{r}$, $q$ is the charge, $v$ is the velocity, $B$ is the magnetic field and $\theta$ is the angle.

Equating equation (I) and (II) and substitute $\theta =90°$

$\begin{array}{c}{\displaystyle \sum F}=ma\\ qvB\sin 90°=\frac{m{v}^2}{r}\\ mv=qBr\end{array}$

     $p=qBr$                                                                 (III)

To convert $\text{kg}\cdot \text{m}/\text{s}$ into $\text{MeV}/c$, multiply and divide by $c$

$\begin{array}{c}\left(\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)=\left(\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)\left(\frac{c}{c}\right)\\ =\left(\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)\left(2.998\times {10}^8\text{ m/s}\right)\left(\frac{1}{c}\right)\\ =\left(2.998\times {10}^8\frac{\text{kg}\cdot {\text{m}}^{\text{2}}}{{\text{s}}^{\text{2}}}\right)\left(\frac{1}{c}\right)\\ =2.998\times {10}^8\text{ J}\left(\frac{1}{c}\right)\left(\frac{1\text{ MeV}}{1.602\times {10}^{-13}\text{ J}}\right)\end{array}$

$\left(\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)=1.871\times {10}^{21}\text{MeV}/c$

Conclusion:

Substitute $1.602\times {10}^{-19}\text{ C}$ for $e$, $1.15\text{ T}$ for $B$ and $1.99\text{ m}$ for $r_{{\Sigma }^{+}}$ in equation (III) to find the value of $p_{{\Sigma }^{+}}$

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

Thus, the magnitudes of the momenta of the ${\Sigma }^{+}$ particle is $686\text{MeV}/c$.

Substitute $1.602\times {10}^{-19}\text{ C}$ for $e$, $1.15\text{ T}$ for $B$ and $0.580\text{ m}$ for $r_{{\pi }^{+}}$ in equation (III) to find the value of $p_{{\pi }^{+}}$

$\begin{array}{c}{p}_{{\pi }^{+}}=eB{r}_{{\pi }^{+}}\\ =\left(1.602\times {10}^{-19}\text{ C}\right)\left(1.15\text{ T}\right)\left(0.\text{580 m}\right)\left(\frac{1.871\times {10}^{21}\text{MeV}/c}{\text{kg}\cdot \text{m/s}}\right)\\ =200\text{MeV}/c\end{array}$

Thus, the magnitudes of the momenta of the ${\pi }^{+}$ particle is $200\text{MeV}/c$.

(b)

To determine

The magnitude of the momentum of the neutron.

Answer to Problem 26P

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

Explanation of Solution

Taking the direction of the momentum of the ${\Sigma }^{+}$ particle as an axis of reference and let $\varphi$ be the angle made of neutron’s path with respect to the ${\Sigma }^{+}$ path at the momentum of its decay. Given that the momentum of the pion makes an angle of $64.5°$ with respect to the axis of reference. The total momentum is equal to the momentum of the ${\Sigma }^{+}$ particle.

Using the conservation of momentum parallel to the original momentum,

$\begin{array}{c}{p}_{{\Sigma }^{+}}=p_n\cos \varphi +p_{{\pi }^{+}}\cos 64.5°\\ p_n\cos \varphi =p_{{\Sigma }^{+}}-p_{{\pi }^{+}}\cos 64.5°\end{array}$

$p_n\cos \varphi =686\text{MeV}/c-\left(200\text{MeV}/c\right)\cos 64.5°$                                      (IV)

Using the conservation of momentum perpendicular to the original momentum,

$\begin{array}{c}0=p_n\sin \varphi -\left(200\text{MeV}/c\right)\sin 64.5°\\ p_n\sin \varphi =\left(200\text{MeV}/c\right)\sin 64.5°\end{array}$                                              (V)

Conclusion:

Squaring and adding equation (IV) and (V)

$\begin{array}{c}{p}_n=\sqrt{{\left(p_n\cos \varphi \right)}^2+{\left(p_n\sin \varphi \right)}^2}\\ =626\text{MeV}/c\end{array}$

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

(c)

To determine

The total energy of the ${\pi }^{+}$ particle and the neutron from relativistic energy-momentum relation.

Answer to Problem 26P

The total energy of the ${\pi }^{+}$ particle is $244\text{ MeV}$ and the neutron is $1.13\text{ GeV}$.

Explanation of Solution

Write the relativistic energy-momentum relation

    $E=\sqrt{{\left(pc\right)}^2+{\left(mc\right)}^2}$                                                         (VI)

Here, $p$ is the momentum and $c$ is the speed of light.

Conclusion:

Substitute $200\text{MeV}/c$ for $p_{{\pi }^{+}}$ and $139.6\text{MeV}/c$ for $m_{{\pi }^{+}}$ in equation (VI) to find the value of $E_{{\pi }^{+}}$

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

Thus, the total energy of the ${\pi }^{+}$ particle is $244\text{ MeV}$.

Substitute $626\text{MeV}/c$ for $p_n$ and $939.6\text{MeV}/c$ for $m_n$ in equation (VI) to find the value of $E_n$

$\begin{array}{c}{E}_n=\sqrt{{\left(p_{n}c\right)}^2+{\left(m_n{c}^2\right)}^2}\\ =\sqrt{{\left(626\text{ MeV}\right)}^2+{\left(939.6\text{ MeV}\right)}^2}\\ =1129\text{ MeV}\\ =1.13\text{ GeV}\end{array}$

Thus, the total energy of the neutron is $1.13\text{ GeV}$.

(d)

To determine

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

Answer to Problem 26P

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

Explanation of Solution

The total momentum is equal to the momentum of the ${\Sigma }^{+}$ particle. Thus, $E_{{\Sigma }^{+}}=E_{{\pi }^{+}}+E_n$.

Substitute $244\text{ MeV}$ for $E_{{\pi }^{+}}$ and $1129\text{ MeV}$ for $E_n$ in the above relation to find the value of $E_{{\Sigma }^{+}}$

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

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

(e)

To determine

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

Answer to Problem 26P

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

Explanation of Solution

Using equation (VI), the relativistic energy-momentum relation to calculate the mass of the ${\Sigma }^{+}$ particle

Substitute $1373\text{ MeV}$ for $E_{{\Sigma }^{+}}$ and $686\text{MeV}/c$ for $p_{{\Sigma }^{+}}$ in equation (VI) to find the value of $m_{{\Sigma }^{+}}$

$\begin{array}{c}{m}_{{\Sigma }^{+}}{c}^2=\sqrt{E_{{\Sigma }^{+}}^2-{\left(p_{{\Sigma }^{+}}c\right)}^2}\\ =\sqrt{{\left(1373\text{ MeV}\right)}^2-{\left(686\text{ MeV}\right)}^2}\\ =1189\text{ MeV}\end{array}$

$\begin{array}{c}{m}_{{\Sigma }^{+}}=1189\text{MeV}/c^2\\ =1.19\text{GeV}/c^2\end{array}$

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

(f)

To determine

Compare the mass of the ${\Sigma }^{+}$ particle with the value in Table 46.2.

Answer to Problem 26P

The mass of the ${\Sigma }^{+}$ particle is within $0.0504\%$ of the value in Table 46.2.

Explanation of Solution

The mass of the ${\Sigma }^{+}$ particle from Table 46.2 is $1.19\text{GeV}/c^2$ and the mass of the ${\Sigma }^{+}$ particle calculated in part (e) is $1189\text{ MeV}$.

The percentage of difference in their mass is $\frac{\Delta m}{m}$

$\begin{array}{c}\frac{\Delta m}{m}=\frac{1.19\times {10}^3\text{MeV}/c^2-1189.4\text{MeV}/c^2}{1189.4\text{MeV}/c^2}\times 100\%\\ =0.0504\%\end{array}$

Thus, the mass of the ${\Sigma }^{+}$ particle is within $0.0504\%$ of the value in Table 46.2.