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An $\alpha$-particle and a proton moving with the same kinetic energy enter a region of uniform magnetic field at right angles to the field. The ratio of the radii of the paths of $\alpha$-particle to that of the proton is
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The correct answer is:
$1: 1$
Radius of circular path of charged particle in uniform magnetic field
$$
\begin{aligned}
\mathrm{r} &=\frac{\mathrm{P}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mK}}}{\mathrm{qB}} \\
\therefore \quad \mathrm{r}: \mathrm{r}_{2} &=\frac{\sqrt{\mathrm{m}_{1}}}{\mathrm{q}_{1}}: \frac{\sqrt{\mathrm{m}_{2}}}{\mathrm{q}_{2}} \\
&=\frac{\sqrt{4}}{2}: \frac{\sqrt{1}}{1}=1: 1
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{r} &=\frac{\mathrm{P}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mK}}}{\mathrm{qB}} \\
\therefore \quad \mathrm{r}: \mathrm{r}_{2} &=\frac{\sqrt{\mathrm{m}_{1}}}{\mathrm{q}_{1}}: \frac{\sqrt{\mathrm{m}_{2}}}{\mathrm{q}_{2}} \\
&=\frac{\sqrt{4}}{2}: \frac{\sqrt{1}}{1}=1: 1
\end{aligned}
$$
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