A proton of energy `8 eV` is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be

Options

A. `4eV`
B. `2eV`
C. `8eV`
D. `6eV`

Answer

3

Explanation

`Bqv=(mv^(2))/(r),Bq=(mv)/(r)=(p)/(r),E_(q)=(q^(2)r^(2)B^(2))/(2m)`
`E_(k)prop(q^(2))/(m),(E_(k_(a)))/(E_(K_(p)))prop(q_(alpha)^(2))/(q_(p)^(2))xx(m_(p))/(m_(d))`