Question

Can the reaction \(\pi^{0}+n \rightarrow K+\Sigma^{+}\) occur?

Short answer

Answer: No, the reaction \(\pi^{0}+n \rightarrow K+\Sigma^{+}\) cannot occur as it violates charge conservation.

Step-by-step solution

Charge Conservation

First, we examine the charge conservation in the reaction. On the left-hand side of the reaction, we have a neutral pion (\(\pi^0\)) and a neutron (n), both with no charge. On the right-hand side, we have a kaon (K) and a sigma (\(\Sigma^+\)) with total charge given as follows: For the kaon (K): normally, K refers to a charged K meson (\(K^+\) or \(K^-\)) but a neutral kaon can be denoted as (\(K^0\)) For the sigma particle: \(\Sigma^+\) has a charge of +1. Let's consider if K represents the neutral kaon (\(K^0\)) since there is no charge for kaon K mentioned in the reaction: Total charge on left-side = 0 (for \(\pi^0\)) + 0 (for n) = 0 Total charge on right-side = 0 (for \(K^0\)) + 1 (for \(\Sigma^+\)) = 1 Charge conservation is violated in this case. Thus, the reaction \(\pi^{0}+n \rightarrow K+\Sigma^{+}\) cannot occur. However, if it was mentioned in the question that K represents charged kaon (\(K^+\) or \(K^-\)) this reaction would be valid.