If a proton, deutron and $\alpha - $ particle on being accelerated by the same potential difference enters perpendicular to the magnetic field, then the ratio of their kinetic energies is
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(d) Kinetic energy in magnetic field remains constant and it is $K = q\,V\;\; \Rightarrow \;K \propto q\;$($V$ = constant)
$\therefore \;{K_p}\;:\;{K_d}\;:\;{K_\alpha }\; = {q_p}\;:\;{q_d}\;:\;{q_a} = 1:1:2$
$\therefore \;{K_p}\;:\;{K_d}\;:\;{K_\alpha }\; = {q_p}\;:\;{q_d}\;:\;{q_a} = 1:1:2$
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