A narrow beam of protons and deuterons, each having the same...

Since, the radius of circular path of a charged particle in magnetic field is $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\rho}{\mathrm{qB}}$ Now, the radius of circular path of charged particle of given momentum $\rho$ and magnetic field $\mathrm{B}$ is given by $\mathrm{r} \propto \frac{1}{\mathrm{q}}$ But charge on both charged particles, protons and deuterons, is same. Therefore, $\frac{\mathrm{r}_{\mathrm{p}}}{\mathrm{r}_{\mathrm{D}}}=\frac{\mathrm{q}_{\mathrm{D}}}{\mathrm{q}_{\rho}}=\frac{1}{\mathrm{l}}$