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An $\alpha$ particle and a proton travel with same velocity in a magnetic field perpendicular to the direction of their velocities, find the ratio of the radii of their circular path
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$2: 1$
As \(\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}\)
Hence, \(\frac{r_\alpha}{r_p}=\frac{m_\alpha}{m_p} \times \frac{q_p}{q_a}\)
We know, \(\mathrm{m}_\alpha=4 \mathrm{~m}_{\mathrm{p}}\) and \(\mathrm{q}_\alpha=2 \mathrm{q}_{\mathrm{p}}\)
\(\frac{r_\alpha}{r_p}=\left(\frac{4}{1}\right) \times\left(\frac{1}{2}\right)=\frac{2}{1}=2: 1\)
Hence, \(\frac{r_\alpha}{r_p}=\frac{m_\alpha}{m_p} \times \frac{q_p}{q_a}\)
We know, \(\mathrm{m}_\alpha=4 \mathrm{~m}_{\mathrm{p}}\) and \(\mathrm{q}_\alpha=2 \mathrm{q}_{\mathrm{p}}\)
\(\frac{r_\alpha}{r_p}=\left(\frac{4}{1}\right) \times\left(\frac{1}{2}\right)=\frac{2}{1}=2: 1\)
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