Q64 / 90 · JEE Main 2021 (18 Mar Shift 2) · Physics

A proton and an $α$ -particle, having kinetic energies $K_{p}$ and $K_{α}$ , respectively, enter into a magnetic field at right angles. The ratio of the radii of the trajectory of proton to that of $α$ -particle is $2 : 1$ . The ratio of $K_{p} : K_{α}$ is:

Solution

$r = \frac{m v}{q B} = \frac{p}{q B} \frac{m_{α}}{m_{p}} = 4$

$\frac{r_{p}}{r_{α}} = \frac{p_{p}}{q_{p}} \frac{q_{α}}{p_{α}} = \frac{2}{1}$

$\frac{p_{p}}{p_{a}} = \frac{2 q_{p}}{q_{a}} = 2 (\frac{1}{2})$

$\frac{p_{p}}{p_{α}} = 1$

$\frac{K_{p}}{K_{α}} = \frac{p_{p}^{2}}{p_{α}^{p}} \frac{m_{α}}{m_{p}} = (1) (4)$

Hence, the correct answer is option: 'D'.