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$\alpha $ particle, proton and duetron enters in a uniform (transverse) magnetic field $'B'$ with same acceleration potential find ratio of radius of path followed by these particles.
$1\,:\,\sqrt 2 \,:\,\sqrt 2 $
$\sqrt 2 \,:\,1\,:\,\sqrt 2 $
$2\sqrt 2 \,:\,1\,:\,2$
$1 : 1 : 1$
Solution
$\mathrm{KE}=\frac{\mathrm{q}^{2} \mathrm{B}^{2} \mathrm{r}^{2}}{2 \mathrm{M}}$
$\mathrm{qV}=\frac{\mathrm{q}^{2} \mathrm{B}^{2} \mathrm{r}^{2}}{2 \mathrm{M}}$
$\frac{2 m V}{B^{2} q}=r^{2}$
$\mathrm{r}=\sqrt{\frac{2 \mathrm{mV}}{\mathrm{B}^{2} \mathrm{q}}}$
$r \propto \sqrt {\frac{m}{q}} $
$r_{\alpha}: r_{p} \cdot r_{d}=\sqrt{\frac{M_{\alpha}}{q_{\alpha}}}: \sqrt{\frac{M_{p}}{q_{p}}}: \sqrt{\frac{M_{d}}{q_{d}}}$
$=\sqrt{\frac{4 M_{p}}{2 q_{p}}}: \sqrt{\frac{M_{p}}{q_{p}}}: \sqrt{\frac{2 M_{p}}{q_{p}}}$
$=\sqrt{2}: 1: \sqrt{2}$