Question

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 path described by them will be :

• A. $1:\sqrt{2}$
• B. $1:2$
• C. $1:3$
• D. $1:\sqrt{3}$

Answer 1

The correct option is A $1:\sqrt{2}$

$KE=q\Delta V$

$r=\frac{\sqrt{2mq\Delta V}}{qB}$

$r\propto \sqrt{\frac{m}{q}}$

$\frac{r_p}{r_{\propto }}=\frac{1}{\sqrt{2}}$