Proton and deuteron are placed in a uniform electric field and allowed to move for same time. The ratio of their kinetic energies will be

acceleration of the charge particle:  $\mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}$

and, kinetic energy: ${\mathrm{E}}_{\mathrm{K}}=\frac{1}{2}{\left(\mathrm{m}\left)\right(\mathrm{at}\right)}^2=\frac{1}{2}\mathrm{m}{\left(\frac{\mathrm{qE}}{\mathrm{m}}\mathrm{t}\right)}^2= \frac{{\mathrm{q}}^2{\mathrm{E}}^2{\mathrm{t}}^2}{2\mathrm{m}}$

$\Rightarrow {\mathrm{E}}_{\mathrm{K}} \propto \frac{1}{\mathrm{m}} \left[\because \mathrm{both} \mathrm{have} \mathrm{same} \mathrm{charges}, \mathrm{and} \mathrm{E} \mathrm{and} \mathrm{t} \mathrm{are} \mathrm{same}\right]$

$\Rightarrow \frac{{\mathrm{E}}_{\mathrm{K},\mathrm{p}}}{{\mathrm{E}}_{\mathrm{K},\mathrm{d}}}= \frac{2}{1}$