A positively charged sigma particle (symbol Σ^+) produced in a particle physics experiment decays very quickly into a neutron and positively charged pion before either its energy or momentum can be measured. The neutron and pion are observed to move in the same direction as the Σ^+ was originally

Question

A positively charged sigma particle (symbol [latex]\Sigma^{+}[/latex]) produced in a particle physics experiment decays very quickly into a neutron and positively charged pion before either its energy or momentum can be measured. The neutron and pion are observed to move in the same direction as the [latex]\Sigma^{+}[/latex] was originally moving, with momenta of 4702 MeV/c and 169 MeV/c, respectively. What was the kinetic energy of the [latex]\Sigma^{+}[/latex] and its mass?

Strategy The decay reaction is

[latex]\Sigma^{+} \rightarrow n+\pi^{+}[/latex]

where n is a neutron. Obviously the [latex]\Sigma^{+}[/latex] has more mass than the sum of the masses of n and [latex]\pi^{+}[/latex], or the decay would not occur. We have to conserve both momentum and energy for this reaction. We use Equation (2.70) to find the total energy of the neutron and positively charged pion, but in order to determine the rest energy of [latex]\Sigma^{+}[/latex], we need to know the momentum. We can determine the [latex]\Sigma^{+}[/latex] momentum from the conservation of momentum.

[latex]E^{2}=p^{2} c^{2}+E_{0}^{2}[/latex] (2.70)

Accepted Answer

The rest energies of n and [latex]\pi^{+}[/latex] are 940 MeV and 140 MeV, respectively. The total energies of [latex]E_{n} ext { and } E_{\pi^{+}}[/latex] are, from [latex]E=\sqrt{p^{2} c^{2}+E_{0}^{2}}[/latex],

[latex]E_{n}=\sqrt{(4702 MeV )^{2}+(940 MeV )^{2}}=4795 MeV[/latex]

[latex]E_{\pi^{+}}=\sqrt{(169 MeV )^{2}+(140 MeV )^{2}}=219 MeV[/latex]

The sum of these energies gives the total energy of the reaction, 4795 MeV + 219 MeV = 5014 MeV, both before and after the decay of [latex]\Sigma^{+}[/latex]. Because all the momenta are along the same direction, we must have

[latex]\begin{aligned}p_{\Sigma^{+}}=p_{n}+p_{\pi^{+}} &=4702 MeV / c+169 MeV / c \&=4871 MeV / c\end{aligned}[/latex]

This must be the momentum of the [latex]\Sigma^{+}[/latex] before decaying, so
now we can find the rest energy of [latex]\Sigma^{+}[/latex] from Equation (2.70).

[latex]\begin{aligned}E_{0}^{2}\left(\Sigma^{+} ight)=E^{2}-p^{2} c^{2} &=(5014 MeV )^{2}-(4871 MeV )^{2} \&=(1189 MeV )^{2}\end{aligned}[/latex]

The rest energy of the [latex]\Sigma^{+}[/latex] is 1189 MeV, and its mass is 1189 [latex]MeV / c^{2}[/latex].

We find the kinetic energy of [latex]\Sigma^{+}[/latex] from Equation (2.65).

[latex]E=\gamma m c^{2}=\frac{m c^{2}}{\sqrt{1-u^{2} / c^{2}}}=\frac{E_{0}}{\sqrt{1-u^{2} / c^{2}}}=K+E_{0}[/latex] (2.65)

[latex]K=E-E_{0}=5014 MeV -1189 MeV =3825 MeV[/latex]

Question 2.18: A positively charged sigma particle (symbol Σ^+) produced in...

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