Question:
In a hadronic interaction, \(\pi^0\)'s are produced with different momenta, and they immediately decay into two photons with an opening angle \(\theta\) between them. Assuming that all these decays occur in one plane, which one of the following figures depicts the behaviour of \(\theta\) as a function of the \(\pi^0\) momentum \(p\)?
In a hadronic interaction, \(\pi^0\)'s are produced with different momenta, and they immediately decay into two photons with an opening angle \(\theta\) between them. Assuming that all these decays occur in one plane, which one of the following figures depicts the behaviour of \(\theta\) as a function of the \(\pi^0\) momentum \(p\)?
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For high-momentum decays, the photons become nearly collinear due to relativistic effects, which means that the opening angle \(\theta\) decreases as the momentum increases.
Updated On: Aug 30, 2025
A
B
C
D
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The Correct Option is A
Solution and Explanation
- The decay of \(\pi^0\) mesons into two photons is governed by the conservation of energy and momentum, leading to a relationship between the opening angle \(\theta\) and the momentum \(p\).
- As the momentum increases, the opening angle decreases due to the relativistic motion of the photons. The correct representation of this relationship is given by option (A), where the opening angle decreases as the momentum increases.
- As the momentum increases, the opening angle decreases due to the relativistic motion of the photons. The correct representation of this relationship is given by option (A), where the opening angle decreases as the momentum increases.
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