A proton and an α-particle (with their masses in the ratio of 1:4 and charges in the ratio of 1:2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii rp:rα of the circular path described by them will be :
1:2
1:3
1:√3
1:√2
Solution:
The correct option is A 1:√2 KE=qΔV r=√(2mqΔV)/(qB) r ∝ √(m/q) rp/rα = √[(mp/qp)/(mα/qα)] = √[(1/1)/(4/2)] = √(1/2) = 1/√2 = 1:√2