Q 12-04-169JEE MainJEE Main 2021 (20 Jul, Shift 1)Easy
A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_d$ and $r_\alpha$ be their respective radii of circular path. The value of $\dfrac{r_d}{r_\alpha}$ is equal to:
Answer: (B) $\sqrt2$
$r = \dfrac{\sqrt{2mK}}{qB}$. Deuteron: $m_d = 2m$, $q = e$. Alpha: $m_\alpha = 4m$, $q = 2e$.
$$\frac{r_d}{r_\alpha} = \frac{\sqrt{2m}/e}{\sqrt{4m}/2e} = \frac{2\sqrt2}{2} = \sqrt2$$
Solution by Sreeraj P, M.Sc Physics