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A proton, a deuteron and an $α$ – particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field, If $r_{p}$ , $r_{d}$ and $r_{a}$ denote respectively the radii of the trajectories of these particles, then

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r_{a}

r_{p}

r_{d}

r_{a}

r_{p}

r_{d}

r_{a}

r_{p}

r_{d}

r_{a}

r_{p}

r_{d}

answer is A.

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Detailed Solution

r_{p}

r_{d}

and

r_{a}

denote respectively the radii of the trajectories of these particles, then

r_{a}

r_{p}

r_{d}

The centripetal force is provided by magnetic force

\frac{m v^{2}}{r} = q \times v \times B

Where,

m = mass

v = velocity of charged particle

q = charge

r = radius

B = strength of magnetic field

\frac{mv}{r} = qB

r = \frac{mv}{qB} = \frac{p}{qB} (where p - mv)

But

KE = \frac{p^{2}}{2 m}

p = \sqrt{2 mKE}

where

KE = kinetic energy

B = strength of magnetic field

Here, KE and B are same for the three particles r directly proportional to

\frac{\sqrt{m}}{q}

r_{p} : r_{d} : r_{a} = \frac{\sqrt{1}}{2} : \frac{\sqrt{2}}{1} : \frac{\sqrt{4}}{2}

= 1 : \sqrt{2} : 1

r_{p}, r_{d}, r_{α}

is the radius of proton, deuteron and α−particle respectively

r_{α} = r_{p} < r_{d}

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