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:
JEE MAIN 2024, Diffcult
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In uniform magnetic field,
$\mathrm{R}=\frac{\mathrm{m} v}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{~K} \cdot \mathrm{E})}}{\mathrm{qB}}$
Since same $K.E$
$\mathrm{R} \propto \frac{\sqrt{\mathrm{m}}}{\mathrm{q}}$
$\therefore \frac{\mathrm{R}_{\text {deutron }}}{\mathrm{R}_{\text {proton }}}=\sqrt{\frac{\mathrm{m}_{\mathrm{d}}}{\mathrm{m}_{\mathrm{p}}}} \times \frac{\mathrm{q}_{\mathrm{p}}}{\mathrm{q}_{\mathrm{d}}}$
$=\sqrt{2} \times 1$
$\therefore \gamma_{\mathrm{d}}: \gamma_{\mathrm{p}}=\sqrt{2}: 1$
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