Question

An unstable particle, initially at rest, decays into a proton (rest energy 938.3 MeV) and a negative pion (rest energy 139.5 MeV). A uniform magnetic field of 0.250 T exists perpendicular to the velocities of the created particles. The radius of curvature of each track is found to be 1.33 m. What is the mass of the original unstable particle?

Answer 1

the mass of the original unstable particle is 1115.08 MeV/c²

Step-by-step explanation:

The momentum of a particle is determined by:

p = e B R

where

• B is the magnetic field
• R is the radius of curvature
• e is the energy of the particle

Therefore,

p = e B R kg · m/s

We can transform the units to MeV/c and we do that by taking:

e = 0.511 MeV and

c = 3 × 10⁸ m/s

Therefore,

p = 300 B R MeV/c

p = 300(0.250 T)(1.33 m) MeV/c

p = 99.75 MeV/c

The energy of the unstable decayed particle is determined as:

E = √ [m²c⁴ + p²c²]

where

• m is the mass of the particle
• c is the speed of light
• p is the particle's momentum

Therefore,

E = E_p + E_(π⁻)

E = √[ (938.3)² + (99.75)² ] + √[ (139.5)² + (99.75)² ]

E = 1115.08 MeV

Since the particle was initially at rest, its energy is only rest-mass energy so its mass will be 1115.08 MeV/c²