A proton, a deuteron and an $α$ -particle is accelerated through the same potential difference from rest, enters a region of uniform magnetic field, moving at right angles to it. What is the ratio of their kinetic energies?

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Solution

We know that work done by the magnetic force on the charged particle is zero.

Therefore, by work energy theorem, there will be no change in kinetic energy of the charged particle due to magnetic field.

The charged particle is accelerated through the same potential difference from rest.

So, $K E = q V$

For same potential difference,

$K E \propto q$

For -
Proton, $q_{p} = 1 e$
Deuteron, $q_{d} = 1 e$
$α -$ particle, $q_{α} = 2 e$

Hence,

$K E_{p} : K E_{d} : K E_{α} = q_{p} : q_{d} : q_{α} = 1 : 1 : 2$

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