A proton and an `alpha`-particle, accelerated through same potential difference, enter a region of uniform magnetic field with their velocities perpendicular to the field. Compare the radii of circular paths followed by them.

Explanation

Let mass of proton = m, charge of proton = e
Now, Mass of `alpha`-particle = 4m
Chanrge of `alpha`-particle = 2e
When a charge q is accelerated by V volts, it acquires a kinetic energy `E_(k) = qV`
`:.` Momentum is given by `mv = sqrt(2m E_(k)) = sqrt(2mqV)`
Radius `r = (mv)/(qB) or r = (sqrt(2mqV))/(qB) = sqrt((2mV)/(qB^(2))`
Thus `(r_(p))/(r_(alpha)) = sqrrt((2mV)/(eB^(2))) xx sqrt((2eB^(2))/(2(4m) V)) = (1)/(sqrt2)`