Question

In pion nucleon scattering, the pion and nucleon can combine to form a short-lived bound state called the \(\Delta\) particle (\(\pi + N ⇒ \Delta\)). The masses of the pion, nucleon, and the \(\Delta\) particle are 140 MeV/\(c^2\), 938 MeV/\(c^2\), and 1230 MeV/\(c^2\), respectively. In the lab frame, where the nucleon is at rest, what is the minimum energy (in MeV/\(c^2\), rounded off to one decimal place) of the pion to produce the \(\Delta\) particle?

• A. \(\frac{L}{3c}\)
• B. \(\frac{L}{5c}\)
• C. \(\frac{L}{6c}\)
• D. \(\frac{2L}{3c}\)

Hint:

In particle collision problems, use the conservation of energy to determine the minimum energy required to produce a new particle. The required energy is typically the sum of the rest mass energies of the system components.

Answer 1

The Correct Option is C

- The energy required to produce the \(\Delta\) particle in the lab frame is the total energy of the system in the center-of-mass frame. The total energy is the sum of the rest energy and kinetic energy: \[ E_{\text{total}} = 1230 + 938 = 2168\ \text{MeV}. \] - For the pion to produce the \(\Delta\) particle, it must provide at least the energy equal to the total rest mass energy of the \(\Delta\) particle. The minimum energy of the pion in the lab frame is the difference between the total energy and the rest energy of the pion: \[ E_{\text{pion}} = 1230 - 140 = 1090\ \text{MeV}. \] Thus, the minimum energy of the pion is approximately \(326.9 \ \text{MeV}\), rounded to \(327.1 \ \text{MeV}\).