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A proton and an $\alpha -$ particle (with their masses in the ratio of $1 : 4$ and charges in the ratio of $1:2$ are accelerated from rest through a potential difference $V$. If a uniform magnetic field $(B)$ is set up perpendicular to their velocities, the ratio of the radii $r_p : r_{\alpha }$ of the circular paths described by them will be
$1: \sqrt 2$
$1 : 2$
$1 : 3$
$1: \sqrt 3$
Solution
${{\text{m}}_p} = {\text{m}}$ ${{\text{q}}_{\text{p}}} = {\text{q}}$ ${{\text{k}}_{\text{p}}} = {\text{qV}} = {\text{k}}$
${{\text{m}}_\alpha } = {\text{4m}}$ ${q_\alpha } = 2q$ ${k_\alpha } = 2qV = 2k$
Radius of circular path,
$r=\frac{m v}{q B}=\frac{\sqrt{2 k m}}{q B}$
$ \Rightarrow $ $\frac{{{r_p}}}{{{r_\alpha }}} = \frac{1}{{\sqrt 2 }}$
