Question

A proton and ` alpha` - particle are accelerated with same potential difference and they enter in the region of constant magnetic field `B` perpendicular to the velocity of particles. Find the ratio of radius of curvature of ` alpha` - particle`.

Answer 1

Correct Answer - `(r_(p))/(r_(alpha)) = sqrt((m_(p)/(m_(alpha)).(q_(alpha))/(q_(p))) = (1)/(sqrt2)`
`r = (mv)/(qB) = (p)/(qB) = sqrt(2mE)/(qB) = (sqrt2mqV)/(qB)`
`(r_(p))/(r_(a)) = sqrt((m_(p))/(m_alpha) xx (q_alpha)/(q_p)) = sqrt((1)/(4) xx (2)/(1)) = (1)/(sqrt2)` .