A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an -particle to describe a circle of same radius in the same field?
- 2 MeV
- 1 MeV
- 0.5 MeV
- 4 MeV
Answer
(2) 1 MeV
Solution
r = mvqB = 2mKqB K = q2B2r22m q2m (same r, B)
For α-particle: q = 2e, m = 4mp
K Kp = (2e)2/4mpe2/mp = 1
K = 1 MeV
Answer: (2)