Chapter 39: Problem 37

Which of the reactions (a) \(\Lambda^{0} \rightarrow \pi^{+}+\pi^{-}\) and (b) \(K^{0} \rightarrow \pi^{+}+\pi^{-}\) is not possible, and why not?

Short Answer

Expert verified

Reaction (a), \(\Lambda^{0} \rightarrow \pi^{+}+\pi^{-}\), is impossible because it violates the conservation of strangeness.

Step by step solution

Identify the Quantum Numbers of all Particles

The primary task at hand is to identify the quantum numbers of the particles involved. These include the charge (Q), baryon number (B), strangeness (S), and total spin (J). For the reactions given, the quantum numbers are as follows: for \(\Lambda^{0}\), Q=0, B=1, S=-1, J=1/2; for \(\pi^{+}\), Q=+1, B=0, S=0, J=0; for \(\pi^{-}\), Q=-1, B=0, S=0, J=0; for \(K^{0}\), Q=0, B=0, S=1, J=0.

Test the Conservation Laws

Next, test the conservation laws of the quantum numbers for each reaction. Check if the sum of the quantum numbers of the final state equals that of the initial state. For reaction (a), \(\Lambda^{0} \rightarrow \pi^{+}+\pi^{-}\), the strangeness is not conserved (initial S=-1, final S=0), indicating that this decay is not possible. For reaction (b), \(K^{0} \rightarrow \pi^{+}+\pi^{-}\), all quantum numbers are conserved.

Determine the Impossible Reaction

Based on the findings from step 2, one can conclude that reaction (a), \(\Lambda^{0} \rightarrow \pi^{+}+\pi^{-}\), is not possible due to the violation of strangeness conservation.

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Which of the reactions (a) \(\Lambda^{0} \rightarrow \pi^{+}+\pi^{-}\) and (b) \(K^{0} \rightarrow \pi^{+}+\pi^{-}\) is not possible, and why not?

Short Answer

Step by step solution

Identify the Quantum Numbers of all Particles

Test the Conservation Laws

Determine the Impossible Reaction

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