A proton, a deuteron and an α particle are moving with same...

$F = q\left(\overset{→}{v} \times \overset{→}{B}\right) = \dfrac{q}{m} \left(\overset{→}{P} \times \overset{→}{B}\right)</p><p>$⇒ F ∝ $\dfrac{q}{m}</p><p>thus,<br />${F}_{1}: {F}_{2}: {F}_{3} = $\dfrac{q1}{{m}_{1}}$: $\dfrac{q2}{{m}_{2}}$: $\dfrac{q3}{{m}_{3}}= \dfrac{e}{mp}$: $\dfrac{e}{2{m}_{p}}$: $\dfrac{2e}{4{m}_{p}}= \dfrac{1}{1}$: $\dfrac{1}{2}$: $\dfrac{2}{4}=2: 1: 1$

Now for speed calculation, $P = constant ⇒ v ∝ \dfrac{1}{m}</p><p>thus${v}_{1}: {v}_{2}: {v}_{3} = $\dfrac{1}{{m}_{p}}$: $\dfrac{1}{2{m}_{p}}$: $\dfrac{1}{4{m}_{p}}= \dfrac{1}{1}$: $\dfrac{1}{2}$: $\dfrac{1}{4}=4: 2: 1$.