A proton and a deutron ( \(\mathrm{q}=+\mathrm{e}, m=2.0 \mathrm{u})\) having same kinetic energies enter a region of uniform magnetic field \(\vec{B}\), moving perpendicular to \(\vec{B}\). The ratio of the radius \(r_d\) of deutron path to the radius \(r_p\) of the proton path is _______.

\(\mathrm{R}=\frac{\mathrm{m} v}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{~K} \cdot \mathrm{E})}}{\mathrm{qB}}\)

Since same \(K.E\)

\(\mathrm{R} \propto \frac{\sqrt{\mathrm{m}}}{\mathrm{q}}\)

\(\therefore \frac{\mathrm{R}_{\text {deutron }}}{\mathrm{R}_{\text {proton }}}=\sqrt{\frac{\mathrm{m}_{\mathrm{d}}}{\mathrm{m}_{\mathrm{p}}}} \times \frac{\mathrm{q}_{\mathrm{p}}}{\mathrm{q}_{\mathrm{d}}}\)

\(=\sqrt{2} \times 1\)

\(\therefore \gamma_{\mathrm{d}}: \gamma_{\mathrm{p}}=\sqrt{2}: 1\)