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 paths described by them will be :

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

Answer 1

Answer: (A)

$1 : \sqrt{2}$

Answer 2

The correct option is(A): $1:\sqrt2​$

$KE =q\Delta V$
$r =\frac{\sqrt{2mq\Delta V}}{qB}$
$r \propto \sqrt{\frac{m}{q}}$
$\frac{r_{p}}{r_{\infty}} \sqrt{\frac{1}{4}\times\frac{2}{1}}= \frac{1}{\sqrt{2}}$