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Chapter 30: Problem 16

A proton of mass \(0.938 \mathrm{GeV} / c^{2}\) and an antiproton, at rest relative to an observer, annihilate each other as described by \(\mathrm{p}+\overline{\mathrm{p}} \rightarrow \pi^{-}+\pi^{+} .\) What are the kinetic energies of the two pions, each of which has mass $0.14 \mathrm{GeV} / \mathrm{c}^{2} ?$

Short Answer

Expert verified
Answer: The kinetic energies of both pions are 0.368 GeV.

Step by step solution

01

Note the given masses

We are given the following masses: The mass of the proton \(m_p = 0.938 \mathrm{GeV}/c^2\) The mass of the antiproton \(m_{\overline{p}} = 0.938 \mathrm{GeV}/c^2\) The mass of the pions \(m_\pi = 0.14 \mathrm{GeV}/c^2\)
02

Conservation of Energy

Since total energy is conserved before and after the collision, we write the expression: \(E_{before} = E_{after}\) The energy before the collision is given by the rest energies of the proton and antiproton, each of which is equal to its rest mass times the speed of light squared (\(mc^2\)): \(E_{before} = m_p c^2 + m_{\overline{p}} c^2\) The energy after the collision is given by the total energy of the two pions: \(E_{after} = E_{\pi^{-}} + E_{\pi^{+}} = \sqrt{(m_{\pi^{-}}c^2)^2 + (p_{\pi^-}c)^2} + \sqrt{(m_{\pi^{+}}c^2)^2 + (p_{\pi^+}c)^2}\)
03

Conservation of Momentum

Since total momentum is conserved before and after the interaction, we write the expression: \(\vec{P}_{before} = \vec{P}_{after}\) Because the proton and antiproton are initially at rest, their total momentum before the collision is zero. \(\vec{P}_{before} = 0\) The momentum after the collision is given by the sum of the momenta of the two pions. Since their masses are equal, their magnitudes of momenta are also equal and opposite: \(p_{\pi^{-}} = -p_{\pi^{+}}\)
04

Solve for the momentum magnitudes

From the conservation of energy equation, we can eliminate the momentum terms in \(E_{after}\) and solve for the magnitude of the pion momenta: \(p_{\pi^{-}}^2 c^2 = (\sqrt{E_{before}^2 - m_{\pi^{-}}^2 c^4} - m_{\pi^{+}} c^2)^2\) Solving for \(p_{\pi^{-}}\), we find: \(p_{\pi^{-}} = 0.814 \mathrm{GeV}/c\) Because the pion momenta are equal in magnitude, it follows that \(p_{\pi^{+}} = 0.814 \mathrm{GeV}/c\) as well.
05

Calculate the kinetic energies of the pions

Now, we can calculate the kinetic energy of each pion using the formula: \(T = \sqrt{E^2 - m^2c^4} - mc^2\) For the negative pion: \(T_{\pi^{-}} = \sqrt{(p_{\pi^{-}}c)^2 + (m_{\pi^{-}}c^2)^2} - m_{\pi^{-}}c^2\) \(T_{\pi^{-}} = 0.368 \mathrm{GeV}\) For the positive pion: \(T_{\pi^{+}} = \sqrt{(p_{\pi^{+}}c)^2 + (m_{\pi^{+}}c^2)^2} - m_{\pi^{+}}c^2\) \(T_{\pi^{+}} = 0.368 \mathrm{GeV}\) So, the kinetic energies of both pions are \(0.368 \mathrm{GeV}\).

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Most popular questions from this chapter

Which fundamental force is responsible for each of the decays shown here? [Hint: In each case, one of the decay products reveals the interaction force.] (a) \(\pi^{+} \rightarrow\) \(\mu^{+}+v_{\mu},\) (b) $\pi^{0} \rightarrow \gamma+\gamma(\mathrm{c}) \mathrm{n} \rightarrow \mathrm{p}^{+}+\mathrm{e}^{-}+\bar{v}_{\mathrm{e}}$
Two factors that can determine the distance over which a force can act are the mass of the exchange particle that carries the force and the Heisenberg uncertainty principle \([\mathrm{Eq} .(28-3)] .\) Assume that the uncertainty in the energy of an exchange particle is given by its rest energy and that the particle travels at nearly the speed of light. What is the range of the weak force carried by the \(\mathbf{Z}\) particle that has a mass of $92 \mathrm{GeV} / \mathrm{c}^{2} ?$ Compare this with the range of the weak force given in Table 30.3
What is the quark content of an antiproton? [Hint: Replace each of the three quarks that compose a proton with its corresponding antiquark.]
At the Stanford Linear Accelerator, electrons and positrons collide together at very high energies to create other elementary particles. Suppose an electron and a positron, each with rest energies of \(0.511 \mathrm{MeV},\) collide to create a proton (rest energy \(938 \mathrm{MeV}\) ), an electrically neutral kaon \((498 \mathrm{MeV}),\) and a negatively charged sigma baryon \((1197 \mathrm{MeV}) .\) The reaction can be written as: $$\mathrm{e}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{p}^{+}+\mathrm{K}^{0}+\Sigma^{-}$$ (a) What is the minimum kinetic energy the electron and positron must have to make this reaction go? Assume they each have the same energy. (b) The sigma can decay in the reaction \(\Sigma^{-} \rightarrow \mathrm{n}+\pi^{-}\) with rest energies of \(940 \mathrm{MeV}\) (neutron) and \(140 \mathrm{MeV}\) (pion). What is the kinetic energy of each decay particle if the sigma decays at rest?
In the Cornell Electron Storage Ring, electrons and positrons circulate in opposite directions with kinetic energies of 6.0 GeV each. When an electron collides with a positron and the two annihilate, one possible (though unlikely) outcome is the production of one or more proton-antiproton pairs. What is the maximum possible number of proton-antiproton pairs that could be formed?
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