Chapter 15, Problem 35P

(a)

To determine

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

Answer to Problem 35P

The momentum for ${\Sigma }^{+}$ particle is $686\text{ MeV/}c$ and that of ${\pi }^{+}$ particle is $200\text{ MeV/}c$ .

Explanation of Solution

Write the expression for the momentum of ${\Sigma }^{+}$ particle.

  $p_{{\Sigma }^{+}}=eB{r}_{{\Sigma }^{+}}$        (I)

Here, $p_{{\Sigma }^{+}}$ the momentum of ${\Sigma }^{+}$ particle, $e$ is the charge of the electron, $B$ is the magnitude of the magnetic field and $r_{{\Sigma }^{+}}$ is the radius of curvature of the ${\Sigma }^{+}$ particle.

Write the expression for the momentum of ${\pi }^{+}$ particle.

  $p_{{\pi }^{+}}=eB{r}_{{\pi }^{+}}$        (II)

Here, $p_{{\pi }^{+}}$ the momentum of ${\pi }^{+}$ particle and $r_{{\pi }^{+}}$ is the radius of curvature of the ${\pi }^{+}$ particle.

Conclusion:

The value of $e$ is $1.602177\times {10}^{-19}\text{ C}$ .

Substitute $1.602177\times {10}^{-19}\text{ C}$ for $e$ , $1.15\text{ T}$ for $B$ and $1.99\text{ m}$ for $r_{{\Sigma }^{+}}$ in equation (I) to find $p_{{\Sigma }^{+}}$ .

  $\begin{array}{c}{p}_{{\Sigma }^{+}}=\left(1.602177\times {10}^{-19}\text{ C}\right)\left(1.15\text{ T}\right)\left(1.99\text{ m}\right)\frac{1\text{ MeV}/c}{5.344288\times {10}^{-22}\text{ kg}\cdot \text{m/s}}\\ =686.075081\text{ MeV/c}\\ \approx 686\text{ MeV/c}\end{array}$

Substitute $1.602177\times {10}^{-19}\text{ C}$ for $e$ , $1.15\text{ T}$ for $B$ and $0.580\text{ m}$ for $r_{{\pi }^{+}}$ in equation (II) to find $p_{{\pi }^{+}}$ .

  $\begin{array}{c}{p}_{{\pi }^{+}}=\left(1.602177\times {10}^{-19}\text{ C}\right)\left(1.15\text{ T}\right)\left(0.580\text{ m}\right)\frac{1\text{ MeV}/c}{5.344288\times {10}^{-22}\text{ kg}\cdot \text{m/s}}\\ =199.961581\text{ MeV}/c\\ \approx 200\text{ MeV/c}\end{array}$

Therefore, the momentum for ${\Sigma }^{+}$ particle is $686\text{ MeV/}c$ and that of ${\pi }^{+}$ particle is $200\text{ MeV/}c$ .

(b)

To determine

The momentum of the neutron.

Answer to Problem 35P

The momentum of the neutron is $627\text{ MeV/}c$ .

Explanation of Solution

Take $\phi$ to be the angle made by the neutron’s path with the path of the ${\Sigma }^{+}$ at the moment of the decay.

Write the expression for the conservation of momentum.

  $p_n\cos \phi +p_{{\pi }^{+}}\cos \theta =p_{{\Sigma }^{+}}$

Here, $p_n$ is the momentum of the neutron and $\theta$ is the angle between the momenta of the ${\Sigma }^{+}$ and the ${\pi }^{+}$ particles at the moment of decay.

Substitute $199.961581\text{ MeV}/c$ for $p_{{\pi }^{+}}$ , $64.5°$ for $\theta$ and $686.075081\text{ MeV/c}$ for $p_{{\Sigma }^{+}}$ in the above equation.

  $\begin{array}{c}{p}_n\cos \phi +\left(199.961581\text{ MeV}/c\right)\cos 64.5°=686.075081\text{ MeV/c}\\ p_n\cos \phi =599.989401\text{ MeV/}c\end{array}$        (III)

Write the expression for the conservation of momentum.

  $p_n\sin \phi =p_{{\pi }^{+}}\sin \theta$

Substitute $199.961581\text{ MeV}/c$ for $p_{{\pi }^{+}}$ and $64.5°$ for $\theta$ in the above equation.

  $\begin{array}{c}{p}_n\sin \phi =\left(199.961581\text{ MeV}/c\right)\sin 64.5°\\ =180.482380\text{ MeV/c}\end{array}$        (IV)

Write the expression for $p_n$ .

  $p_n=\sqrt{{\left(p_n\cos \phi \right)}^2+{\left(p_n\sin \phi \right)}^2}$

Put equations (III) and (IV) in the above equation.

  $\begin{array}{c}{p}_n=\sqrt{{\left(599.989401\text{ MeV/}c\right)}^2+{\left(180.482380\text{ MeV/c}\right)}^2}\\ =627\text{ MeV/}c\end{array}$

Conclusion:

Therefore, the momentum of the neutron is $627\text{ MeV/}c$ .

(c)

To determine

The total energy of the ${\pi }^{+}$ particle, that of neutron and that of ${\Sigma }^{+}$ particle.

Answer to Problem 35P

The total energy of the ${\pi }^{+}$ particle is $244\text{ MeV}$ , that of neutron is $1130\text{ MeV}$ and that of ${\Sigma }^{+}$ particle is $1370\text{ MeV}$ .

Explanation of Solution

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

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

Here, $E_{{\pi }^{+}}$ is the total energy of the ${\pi }^{+}$ particle, $m_{{\pi }^{+}}$ is the mass the ${\pi }^{+}$ particle and $c$ is the speed of light in vacuum.

Substitute $199.961581\text{ MeV}/c$ for $p_{{\pi }^{+}}$ and $139.6\text{ MeV}$ for $m_{{\pi }^{+}}{c}^2$ in the above equation.

  $\begin{array}{c}{E}_{{\pi }^{+}}=\sqrt{{\left(\left(199.961581\text{ MeV}/c\right)c\right)}^2+{\left(139.6\text{ MeV}\right)}^2}\\ =243.870445\text{ MeV}\\ \approx 244\text{ MeV}\end{array}$

Write the equation for the total energy of the neutron.

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

Here, $E_n$ is the total energy of the neutron and $m_n$ is the mass the neutron.

Substitute $626.547022\text{ MeV}/c$ for $p_n$ and $139.6\text{ MeV}$ for $m_{{\pi }^{+}}{c}^2$ in the above equation.

  $\begin{array}{c}{E}_n=\sqrt{{\left(\left(626.547022\text{ MeV}/c\right)c\right)}^2+{\left(139.6\text{ MeV}\right)}^2}\\ =1129.340219\text{ MeV}\\ \approx 1130\text{ MeV}\end{array}$

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

  $E_{{\Sigma }^{+}}=E_{{\pi }^{+}}+E_n$

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

Substitute $243.870445\text{ MeV}$ for $E_{{\pi }^{+}}$ and $1129.340219\text{ MeV}$ for $E_n$ in the above equation to find $E_{{\Sigma }^{+}}$ .

  $\begin{array}{c}{E}_{{\Sigma }^{+}}=243.870445\text{ MeV}+1129.340219\text{ MeV}\\ =1373.210664\text{ MeV}\\ \approx \text{1370 MeV}\end{array}$

Conclusion:

Therefore, the total energy of the ${\pi }^{+}$ particle is $244\text{ MeV}$ , that of neutron is $1130\text{ MeV}$ and that of ${\Sigma }^{+}$ particle is $1370\text{ MeV}$ .

(d)

To determine

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

Answer to Problem 35P

The mass of the ${\Sigma }^{+}$ particle is $1190\text{ MeV/}{c}^2$ .

Explanation of Solution

Write the expression for $m_{{\Sigma }^{+}}{c}^2$ .

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

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

Substitute $1373.210664\text{ MeV}$ for $E_{{\Sigma }^{+}}$ and $686.075081\text{ MeV/c}$ for $p_{{\Sigma }^{+}}$ in the above equation.

  $\begin{array}{c}{m}_{{\Sigma }^{+}}{c}^2=\sqrt{{\left(1373.210664\text{ MeV}\right)}^2+{\left(\left(686.075081\text{ MeV/c}\right)c\right)}^2}\\ =1190\text{ MeV}\\ m_{{\Sigma }^{+}}=1190\text{ MeV}/c^2\end{array}$

Conclusion:

Therefore, the mass of the ${\Sigma }^{+}$ particle is $1190\text{ MeV/}{c}^2$.