91影视

The given data can be listed below,

• The momentum of incoming pion is 3GeV/c
• Angle of scattering is 40潞
• Rest mass energy of pion is 140MeV
• Rest mass energy of proton is 938 MeV
• Measured final momentum is 1510 MeV/c

The energy momentum relation is relativistic case is given as

$\mathbf{E}\mathbf{=}\sqrt{{\mathbf{\left(}\mathbf{pc}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}{\mathbf{m}}_{\mathbf{0}}{\mathbf{c}}^{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}}}$

Here, E denotes the total energy, c is the speed of light, p is the momentum of the particle and m0 is the rest mass of the particle.

The initial momentum of particle is given by,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{蟿蟿},\mathrm{i}}=\frac{3\mathrm{GeV}}{\mathrm{c}}\left(\frac{1000\mathrm{MeV}}{1\mathrm{GeV}}\right) \\ =3000\mathrm{MeV}/\mathrm{c} \end{gathered}$$

By conservation of momentum in x direction.

$P_{p,i}+P_{\mathrm{蟺},\mathrm{i}}=P_{\mathrm{蟺},\mathrm{f}}\cos {40}^{掳}+P_x\cos 胃$

Here, $P_{p,i}$is the initial momentum of the proton,$P_{\mathrm{蟺},\mathrm{f}}$is the final momentum of the $蟺$particle and $P_x$is the momentum of the unknown particle X.

$P_{p,i}=0$, as the proton is initially at rest.

So,

$P_{\mathrm{蟺颈}}=P_{\mathrm{蟺},\mathrm{f}}\cos {40}^{掳}+P_x\cos 胃$ ....(i)

Similarly, conserving momentum in y direction-

$$\begin{gathered} 0=P_{\mathrm{蟺},\mathrm{f}}\sin {40}^{掳}-P_x\sin 胃 \\ {P}_x=\frac{P_{\mathrm{蟺},\mathrm{f}}\sin {40}^{掳}}{\sin 胃} \end{gathered}$$ ...(ii)

Substituting the $P_x$expression of equation (ii) in equation (i) and solving-

$P_{\mathrm{蟺颈}}=P_{\mathrm{蟺蹿}}\cos {40}^{掳}+\frac{P_{\mathrm{蟺},\mathrm{f}}\sin {40}^{掳}}{\tan 胃}$

Solving for and substituting all the values

$$\begin{gathered} \tan 胃=\frac{(1510\mathrm{Mev}/\mathrm{c})\sin {40}^{掳}}{3000\mathrm{MeV}/\mathrm{c}-1510\mathrm{MeV}/\mathrm{c}\cos {40}^{掳}} \\ \tan 胃=\frac{970.6\mathrm{MeV}/\mathrm{c}}{1843.3\mathrm{Mev}/\mathrm{c}} \\ 胃={\tan }^{-1}(0.53) \\ =27.8 \end{gathered}$$

This is the scattering angle of X+

For momentum of X⁺ particle using equation (ii)

$$\begin{gathered} P_x=\frac{(1510\mathrm{MeV}/\mathrm{c})\sin {40}^{掳}}{\sin (27.8^{掳})} \\ =2082.8\mathrm{MeV}/\mathrm{c} \end{gathered}$$

Now, the energy expression for the given reaction is-

${\mathrm{E}}_{\mathrm{蟿蟿},\mathrm{i}}+{\mathrm{E}}_{\mathrm{p},\mathrm{re}\mathrm{st}}={\mathrm{E}}_{\mathrm{蟿蟿},\mathrm{f}}+{\mathrm{E}}_{\mathrm{x}}$

Solving for E_X 鈥�

${\mathrm{E}}_{\mathrm{x}}={\mathrm{E}}_{\mathrm{蟿蟿},\mathrm{i}}+{\mathrm{E}}_{\mathrm{p}.\mathrm{rest}}-{\mathrm{E}}_{\mathrm{蟿蟿}.\mathrm{f}}$

Using relativistic energy equation

$$\begin{gathered} {\mathrm{E}}_{\mathrm{蟿蟿},\mathrm{i}}=\sqrt{{{\mathrm{E}}_{\mathrm{蟿蟿},\mathrm{rest}}}^2+{\left({\mathrm{P}}_{\mathrm{蟿蟿},\mathrm{i}}\mathrm{c}\right)}^2} \\ =\sqrt{{(140\mathrm{MeV})}^2+\frac{(3000\mathrm{MeVc})2}{\mathrm{c}}} \\ =3003.3\mathrm{MeV} \\ {\mathrm{E}}_{\mathrm{蟿蟿},\mathrm{f}}=\sqrt{{{\mathrm{E}}_{\mathrm{蟿蟿},\mathrm{rest}}}^2+{\left({\mathrm{P}}_{\mathrm{蟿蟿},\mathrm{f}}\mathrm{c}\right)}^2} \\ =\sqrt{{(140\mathrm{MeV})}^2+\frac{{(1510\mathrm{MeVc})}^2}{\mathrm{c}}} \\ =1516.5\mathrm{MeV} \end{gathered}$$

Therefore,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{x}}=3003.3\mathrm{MeV}+38\mathrm{MeV}-1516.5\mathrm{MeV} \\ =2424.8\mathrm{MeV} \end{gathered}$$

Finally, the rest mass of the unknown particle X⁺ is-

$$\begin{gathered} {\mathrm{m}}_{\mathrm{x}}=\sqrt{\frac{{{\mathrm{E}}_{\mathrm{x}}}^2-{{\mathrm{P}}_{\mathrm{x}}}^2{\mathrm{c}}^2}{{\mathrm{c}}^4}} \\ =\sqrt{\frac{{(2424.8\mathrm{MeV})}^2-{(2082.8\mathrm{MeV}/\mathrm{c})}^2{\mathrm{c}}^2}{{\mathrm{c}}^4}} \\ =1241.6\mathrm{MeV}/{\mathrm{c}}^2 \end{gathered}$$

Thus, the rest mass energy of the unknown particle is$1241.6\mathrm{MeV}/{\mathrm{c}}^2$