Discussion
The following nuclear reaction
\[ \pi^{+} + n \rightarrow K^{\circ} + P \]
is examined on the basis of conservation laws of charge, Baryon number, strangeness and third component of Isospin. It is observed that the reaction cannot be induced due to non-conservation of the :
Answer: D
The reaction fails on strangeness and the third component of isospin.
Check each quantity, initial against final, for \( \pi^{+} + n \rightarrow K^{0} + p \):
- Charge: \( +1 + 0 = 0 + 1 \), conserved.
- Baryon number: \( 0 + 1 = 0 + 1 \), conserved.
- Strangeness: \( 0 + 0 \ne +1 + 0 \), violated.
- Isospin \( I_{3} \): \( +1 - \frac{1}{2} = +\frac{1}{2} \) against \( -\frac{1}{2} + \frac{1}{2} = 0 \), violated.
Since charge and baryon number are both conserved, every option naming either of them is wrong. A strong reaction must conserve both strangeness and I3, so it cannot occur.
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