Q 12-04-168JEE MainJEE Main 2021 (18 Mar, Shift 2)Medium
A proton and an $\alpha$-particle, having kinetic energies $K_p$ and $K_\alpha$, respectively, enter into a magnetic field at right angles. The ratio of the radii of the trajectory of proton to that of $\alpha$-particle is 2 : 1. The ratio of $K_p : K_\alpha$ is:
Answer: (D) 4 : 1
$r = \dfrac{\sqrt{2mK}}{qB}$. With $m_\alpha = 4m_p$ and $q_\alpha = 2e$:
$$\frac{r_p}{r_\alpha} = \frac{\sqrt{m_pK_p}/e}{\sqrt{4m_pK_\alpha}/2e} = \sqrt{\frac{K_p}{K_\alpha}} = 2 \Rightarrow \frac{K_p}{K_\alpha} = \frac41$$
Solution by Sreeraj P, M.Sc Physics