Q. A proton, a deuteron and an $\alpha -$ particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to $B.$ What is the ratio of the radii of their circular paths?
A) $1\ :\ \sqrt{2}\ :\sqrt{2}$
B) $1\ :\ \sqrt{2}\ :1$
C) $\sqrt{2}\ :1\ :1$
D) $\sqrt{2}\ :\ \sqrt{2}\ :1$
Explanation
(b) $\frac{mv^{2}}{R} = qvB$
For proton, $R_{p} = \frac{mv}{Bq} = \frac{\sqrt{}2M_{p}E}{q_{p}B}$
Similarly for deuteron and $\alpha$-particle
$R_{d} = \frac{\sqrt{2M_{d}E}}{q_{p}B}$ and $R_{\alpha} = \ \frac{\sqrt{2M_{\alpha}E}}{q_{\alpha}B}$
According to the question
$\therefore R_{p}\ :R_{d}\ :R_{\alpha}$
or $\frac{\sqrt{M_{p}}}{q_{p}}\ :\ \frac{\sqrt{M_{d}}}{q_{d}}\ :\ \frac{\sqrt{M_{\alpha}}}{q_{a}}\ $
$\therefore\frac{\sqrt{1}}{1}\ :\frac{\sqrt{2}}{1}\ :\frac{\sqrt{4}}{2}\ or\ \ 1\ :\ \sqrt{2}\ :1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.