A proton and an alpha particle are accelerated through the same potential difference $V$ and enter a uniform magnetic field $B$ perpendicular to their velocity. The ratio of the radii of their circular paths $r_p/r_{\alpha }$ is:

Solution

1. The radius of a circular path for a particle accelerated through potential $V$ is given by $r=\frac{\sqrt{2mV}}{qB}$.
2. For a proton, $m_p=m$ and $q_p=e$; for an alpha particle, $m_{\alpha }=4m$ and $q_{\alpha }=2e$.
3. The ratio of the radii is $$\begin{array}{rl}\frac{r_p}{r_{\alpha }}& =\frac{\sqrt{m_p}/q_p}{\sqrt{m_{\alpha }}/q_{\alpha }}\\ & =\frac{\sqrt{m}/e}{\sqrt{4m}/2e}\end{array}$$.
4. Simplifying the expression gives $$\begin{array}{rl}\frac{r_p}{r_{\alpha }}& =\frac{1}{1}\times \frac{2}{2}\\ & =\frac{1}{\sqrt{2}}\end{array}$$.
5. Hence the answer is (C).