Question

A proton carrying `1 MeV` kinetic energy is moving in a circular path of radius `R` in unifrom magentic field. What should be the energy of an `alpha-` particle to describe a circle of the same radius in the same field?
A. `2 MeV`
B. `1 MeV
C. `0.5 MeV
D. `4 MeV

Answer 1

Correct Answer - B
For proton `r = (sqrt(2m(KE)))/(qB)`
So `q prop sqrt(m(KE))`
Hence `(e)/(2e) = sqrt(((m_(p)) (1 meV))/((4m_(p)) (KE)))`
`(1)/(4) = (1 MeV)/(4KE)`
`KE = 1 MeV`