GATE 2023 PH – Question 60
In pion nucleon scattering, the pion and nucleon can combine to form a short lived bound state called the Δ particle (π+ N→Δ). The masses of the pion, nucleon and the Δ particle are 140 MeV/c2, 938 MeV/c2 and 1230 MeV/c2, respectively. In the lab frame, where the nucleon is at rest, what is the minimum energy (in MeV/c2, rounded off to one decimal place) of the pion to produce the Δ particle?
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Correct answer: 325.36 to 328.64
Explanation
Use the invariant mass of the pion-nucleon system. In the lab frame the nucleon is at rest (energy $m_N$) and the pion has energy $E_\pi$ and momentum $p_\pi$:
$$s=(E_\pi+m_N)^2-p_\pi^2=m_\pi^2+m_N^2+2E_\pi m_N .$$
To produce a $\Delta$ particle of mass $m_\Delta$ (a single particle), the invariant mass must equal $m_\Delta$:
$$m_\Delta^2=m_\pi^2+m_N^2+2E_\pi m_N .$$
**Solve for $E_\pi$:**
$$E_\pi=\frac{m_\Delta^2-m_\pi^2-m_N^2}{2m_N}=\frac{1230^2-140^2-938^2}{2\times938}=\frac{1\,512\,900-19\,600-879\,844}{1876}=\frac{613\,456}{1876}=\mathbf{327.0\ MeV}.$$