Question

A proton of mass m and charge `+e` is moving in a circular orbit in a magnetic field with energy `1 MeV`. What should be the energy of alpha-particle (mass=`4m` and charge=`+2e`), so that it can revolve in the path of same radius?
A. 1MeV
B. 2MeV
C. 1.5MeV
D. 4MeV

Answer 1

Correct Answer - A
`r=(mv)/(q B)(E_k=(1)/(2)mv^2 therefore v=sqrt((2E_k)/(m)))`
`r=(msqrt((2E_k)/(m)))/(qB)=sqrt((2mE_k)/(q B)`
`therefore E_k=(q^2B^2r^2)/(2m)`
As B and r, are same
`E_p=(e^2B^2r^2)/(2m) and E_alpha=((2e)^2B^2r^2)/(2(4m))`
i.e., `(E_p)/(E_alpha)=1 therefore E_p=E_alpha=1Me V`.