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Chapter 30: Problem 11

The \(K^{0}\) meson can decay to two pions: \(K^{0} \rightarrow \pi^{+}+\pi^{-}\) The rest energies of the particles are: \(K^{0}=497.7 \mathrm{MeV}\) \(\pi^{+}=\pi^{-}=139.6 \mathrm{MeV} .\) If the \(K^{0}\) is at rest before it decays, what are the kinetic energies of the \(\pi^{+}\) and the \(\pi^{-}\) after the decay?

Short Answer

Expert verified
Answer: The kinetic energies of the π+ and π- pions after the decay of a K0 meson at rest are each 109.25 MeV.

Step by step solution

01

Calculate the total energy of the pion system after the decay

Since the \(K^{0}\) is at rest before it decays, its total energy is equal to its rest energy: \(E_{K^{0}}=mc^2=497.7\,\mathrm{MeV}\) After the decay, the total energy of the pions is conserved, so: \(E_{\text{total pions}} = E_{K^{0}} = 497.7\,\mathrm{MeV}\)
02

Calculate the rest energies of the pions

The rest energies of the \(\pi^{+}\) and \(\pi^{-}\) pions are given as: \(E_{\pi^{+}}=E_{\pi^{-}}=mc^2=139.6\,\mathrm{MeV}\)
03

Calculate the total kinetic energy of the pions

Before calculating the kinetic energy, we must determine the total energy of the pions after the decay. Since there are two pions, we can add their rest energies together: \(E_{\text{total rest pions}} = E_{\pi^{+}} + E_{\pi^{-}} = 2 \times 139.6\,\mathrm{MeV} = 279.2\,\mathrm{MeV}\) Now, we can find the total kinetic energy by subtracting the total rest energy from the total energy: \(E_{\text{total kinetic pions}} = E_{\text{total pions}} - E_{\text{total rest pions}} = 497.7\,\mathrm{MeV} - 279.2\,\mathrm{MeV} = 218.5\,\mathrm{MeV}\)
04

Determine the individual kinetic energies of the pions

Since the decay is symmetric, the two pions will have the same kinetic energy. Therefore, we can divide the total kinetic energy by 2 to find the kinetic energy of each pion: \(E_{\mathrm{kinetic}\,\pi^{+}}=E_{\mathrm{kinetic}\,\pi^{-}}=\frac{E_{\text{total kinetic pions}}}{2}=\frac{218.5\,\mathrm{MeV}}{2}=109.25\,\mathrm{MeV}\) So, the kinetic energies of the \(\pi^{+}\) and \(\pi^{-}\) after the decay are each \(109.25\,\mathrm{MeV}\).

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Most popular questions from this chapter

A proton in Fermilab's Tevatron is accelerated through a potential difference of 2.5 MV during each revolution around the ring of radius \(1.0 \mathrm{km} .\) In order to reach an energy of 1 TeV, how many revolutions must the proton make? How far has it traveled?
In an accelerator, two protons with equal kinetic energies collide head-on. The following reaction takes place: $\mathrm{p}+\mathrm{p} \rightarrow \mathrm{p}+\mathrm{p}+\mathrm{p}+\overline{\mathrm{p}} .$ What is the minimum possible kinetic energy of each of the incident proton beams?
A sigma baryon at rest decays into a lambda baryon and a photon: $\Sigma^{0} \rightarrow \Lambda^{0}+\gamma .$ The rest energies of the baryons are given by \(\Sigma^{0}=1192 \mathrm{MeV}\) and \(\Lambda^{0}=1116 \mathrm{MeV} .\) What is the photon wavelength? [Hint: Use relativistic formulas and be sure momentum is conserved as well as energy.]
When a proton and an antiproton annihilate, the annihilation products are usually pions. (a) Suppose three pions are produced. What combination(s) of \(\pi^{+}, \pi^{-},\) and \(\pi^{0}\) are possible? (b) Suppose five pions are produced. What combination(s) of \(\pi^{+}, \pi^{-},\) and \(\pi^{0}\) are possible? (c) What is the maximum number of pions that could be produced if the kinetic energies of the proton and antiproton are negligibly small? The mass of a charged pion is \(0.140 \mathrm{GeV} / c^{2}\) and the mass of a neutral pion is \(0.135 \mathrm{GeV} / c^{2}.\)
Show that the charge of the neutron and the charge of the proton can be derived from their constituent quark content.
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