A proton, a deuteron and an

$\mathrm{F}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\frac{\mathrm{q}}{\mathrm{m}}(\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{B}})$

$\Rightarrow \mathrm{F} \propto \frac{\mathrm{q}}{\mathrm{m}}$

thus $\mathrm{F}_{1}: \mathrm{F}_{2}: \mathrm{F}_{3}=\frac{\mathrm{q}_{1}}{\mathrm{~m}_{1}}: \frac{\mathrm{q}_{2}}{\mathrm{~m}_{2}}: \frac{\mathrm{q}_{3}}{\mathrm{~m}_{3}}$

$=\frac{\mathrm{e}}{\mathrm{m}_{\mathrm{p}}}: \frac{\mathrm{e}}{2 \mathrm{~m}_{\mathrm{p}}}: \frac{2 \mathrm{e}}{4 \mathrm{~m}_{\mathrm{p}}}$

$=\frac{1}{1}: \frac{1}{2}: \frac{2}{4}$

$=2: 1: 1$

Now for speed calculation

$P=$ constant $\Rightarrow v \propto \frac{1}{m}$

thus $v_{1}: v_{2}: v_{3}=\frac{1}{m_{p}}: \frac{1}{2 m_{p}}: \frac{1}{4 m_{p}}$

$=\frac{1}{1}: \frac{1}{2}: \frac{1}{4}$

$=4: 2: 1$