91影视

The given data can be listed below,

• The mass of the sigma-minus is,${\underset{}{鈭�}}^{-}=1196\mathrm{MeV}/{\mathrm{c}}^2$ .
• The mass of the neutron is,$\mathrm{n}=939\mathrm{MeV}/{\mathrm{c}}^2$ .
• The mass of the negative pion is,${\mathrm{蟺}}^{-}=140\mathrm{MeV}/{\mathrm{c}}^2$

The momentum is the measurement of the volume of mass 鈥� movement, reflecting the ability to keep its tendency of movement state.

The total energy of the decay particle is given by,

$E=\sqrt{{\left(pc\right)}^2+{\left(m{c}^2\right)}^2}$

Here, pis the momentum of the particle, mis the mass of the particle and cis the speed of light.

For given three particles, energy of the particles is given by,

$E_{{\underset{}{鈭�}}^{-}}=E_n+E_{\mathrm{蟺}鈥�}$

Here,$E_n$is the energy of the neutron and$E_{\mathrm{蟺}-}$is the energy of negative pion.

Substitute the values according to energy equation.

$\sqrt{{\left(p_{{\underset{}{鈭�}}^{-}}c\right)}^2+{\left(m_{{\underset{}{鈭�}}^{-}}{c}^2\right)}^2}=\sqrt{{\left(p_{n}c\right)}^2+{\left(m_n{c}^2\right)}^2}+\sqrt{{\left(p_{\mathrm{蟺}-}c\right)}^2+{\left(m_{\mathrm{蟺}-}{c}^2\right)}^2}$

The initial momentum of the sigma negative is zero because the particle is at rest. So, the final momentum of the system will also equal to zero.

So, the momentum of rest two particle is given by,

$$\begin{gathered} {\stackrel{鈫�}{p}}_n+{\stackrel{鈫�}{p}}_x=0 \\ \left|{\stackrel{鈫�}{p}}_n\right|=\left|{\stackrel{鈫�}{p}}_{\mathrm{蟺}-}\right| \\ =p \end{gathered}$$

Rewrite the energy equation.

$m_{{\underset{}{鈭�}}^{-}}{c}^2=\sqrt{{\left(pc\right)}^2+{\left(m_n{c}^2\right)}^2}+\sqrt{{\left(pc\right)}^2+{\left(m_{\mathrm{蟺}-}{c}^2\right)}^2}$

The energy of the particles like pion and neutron is only dependent on mass and momentum.so from above, two equations can be written. One for momentum and other for energy of pion and neutron which can be given below as,

$$\begin{gathered} m_{{\underset{}{鈭�}}^{-}}{c}^2=\sqrt{{\left(p_{n/{\mathrm{蟺}}^{-}}c\right)}^2+{\left(m_n{c}^2\right)}^2}+\sqrt{{\left(p_{n/{\mathrm{蟺}}^{-}}c\right)}^2+{\left(m_{{\mathrm{蟺}}^{-}}{c}^2\right)}^2} \\ E=\sqrt{{\left(p_{n/{\mathrm{蟺}}^{-}}c\right)}^2+{\left(m_{n/{\mathrm{蟺}}^{-}}{c}^2\right)}^2} \end{gathered}$$

Thus, the equations for energy and momentum f pion and neutron are

$$\begin{gathered} m_{{\underset{}{鈭�}}^{-}}{c}^2=\sqrt{{\left(p_{n/{\mathrm{蟺}}^{-}}c\right)}^2+{\left(m_n{c}^2\right)}^2}+\sqrt{{\left(p_{n/{\mathrm{蟺}}^{-}}c\right)}^2+{\left(m_{{\mathrm{蟺}}^{-}}{c}^2\right)}^2} \\ E=\sqrt{{\left(p_{n/{\mathrm{蟺}}^{-}}c\right)}^2+{\left(m_{n/{\mathrm{蟺}}^{-}}{c}^2\right)}^2} \end{gathered}$$